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BFS && DFS

分类: IT文章 • 2023-11-29 20:38:13

HDOJ 1312 Red and Black

http://acm.hdu.edu.cn/showproblem.php?pid=1312

很裸的dfs，在dfs里面写上ans++，能到几个点就调了几次dfs，最后ans就是答案

 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 char map[22][22];
 5 int n,m,si,sj,ans;
 6 int dir[4][2] = {{1,0}, {-1,0}, {0,-1}, {0,1}};
 7 int dfs(int x, int y)
 8 {
 9     if (x<=0||x>m||y<=0||y>n)
10     {
11         return 0;
12     }
13     ans++;
14     for (int i = 0; i < 4; ++i)
15     {
16         if (map[x+dir[i][0]][y+dir[i][1]] == '.')
17         {
18             map[x+dir[i][0]][y+dir[i][1]] = 'X';//把访问过的置为“墙”，以免重复计算
19             dfs(x+dir[i][0],y+dir[i][1]);
20         }
21     }
22 
23 }
24 int main()
25 {
26     while(scanf("%d%d", &n, &m) != EOF)
27     {
28         if (n == 0 && m == 0)
29         {
30             break;
31         }
32         for (int i = 1; i <= m; ++i)
33         {
34             for (int j = 1; j <= n; ++j)
35             {
36                 cin >> map[i][j];
37                 if (map[i][j] == '@')
38                 {
39                     si = i;
40                     sj = j;
41                 }
42             }
43         }
44         ans = 0;
45         dfs(si,sj);
46         cout << ans << endl;
47     }
48     return 0;
49 }
View Code

HDOJ 1241 Oil Deposits

http://acm.hdu.edu.cn/showproblem.php?pid=1241

简单dfs 求连通块 对地图中的每个@点，ans++调dfs 把和它连通的都标记了

调dfs的次数，就是连通油田的块数

 1 #include<stdio.h>
 2 int m,n;
 3 char map[105][105];
 4 int dir[8][2]={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
 5 void dfs(int x,int y)
 6 {
 7     map[x][y]='*';
 8     for(int i=0;i<8;i++)
 9     {
10         int fx=x+dir[i][0];
11         int fy=y+dir[i][1];
12         if(fx>=0&&fy>=0&&fx<m&&fy<n)
13         {
14             if(map[fx][fy]=='@')
15             {
16                 dfs(fx,fy);
17             }
18         }
19     }
20 }
21 int main()
22 {
23     while(~scanf("%d%d",&m,&n))
24     {
25         int ans=0;
26         if(m==0)  break;
27         getchar();
28         for(int i=0;i<m;i++)
29         {
30             for(int j=0;j<n;j++)
31             {
32                 scanf("%c",&map[i][j]);
33             }
34             getchar();
35         }
36         for(int i=0;i<m;i++)
37         {
38             for(int j=0;j<n;j++)
39             {
40                 if(map[i][j]=='@')
41                 {
42                     ans++;
43                     dfs(i,j);
44                 }
45             }
46         } 
47         printf("%d
",ans);
48     }
49     return 0;
50 }
View Code

COJ 1224 ACM小组的古怪象棋

http://122.207.68.93/OnlineJudge/problem.php?id=1224

马吃将最少要几步...而且这个将还是不会动的...

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 using namespace std;
 5 int n,m,x,y,si,sj,ans;
 6 int map[25][25];//这里map实际上起到的是vis的作用
 7 int dir[][2]={1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
 8 struct node
 9 {
10     int x,y,d;
11 };
12 int bfs()
13 {
14     queue<node> q;
15     while(!q.empty()) q.pop();
16     node now,next;
17     now.x=si;
18     now.y=sj;
19     now.d=0;
20     q.push(now);
21     map[now.x][now.y]=1;
22     while(!q.empty())
23     {
24         now = q.front();
25         if(now.x==x&&now.y==y) return now.d;
26         q.pop();
27         for (int i = 0; i < 8; ++i)
28         {
29             next.x=now.x+dir[i][0];
30             next.y=now.y+dir[i][1];
31             if(next.x>=0&&next.x<n&&next.y>=0&&next.y<m&&map[next.x][next.y]==0)
32             {
33                 next.d=now.d+1;
34                 q.push(next);
35                 map[next.x][next.y]=1;
36             }
37         }
38     }
39     return -1;
40 }
41 int main()
42 {
43     while(scanf("%d%d%d%d%d%d",&n,&m,&x,&y,&si,&sj)!=EOF)
44     {
45         memset(map,0,sizeof(map));
46         ans = bfs();
47         if(ans==-1) printf("-1
");
48         else printf("%d
", ans);
49     }
50     return 0;
51 }
View Code

拿双向BFS也写了一个 AC了  不过我总感觉怪怪的...

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 using namespace std;
 5 int n,m,si,sj,ei,ej;
 6 int dir[][2]={1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
 7 int vis[25][25],mi[25][25];//vis 0表示尚未访问过 1表示被正向bfs扫到过 2表示被反向bfs扫到
 8 struct node                //mi保存到达这个点的最小步数 以便两个bfs相遇时读出步数
 9 {                          //具体就是一层一层地扫了 遇到了(1遇见2 或者 2遇见1)就立即返回
10     int x,y,d;
11 };
12 bool check(int x,int y)
13 {
14     if(x>0&&x<=n&&y>0&&y<=m) return true;
15     return false;
16 }
17 int bfs()
18 {
19     queue<node> q1,q2;
20     while(!q1.empty()) q1.pop();
21     while(!q2.empty()) q2.pop();
22     node now,next;
23     now.x=si;
24     now.y=sj;
25     now.d=0;
26     q1.push(now);
27     mi[si][sj]=0;
28     now.x=ei;
29     now.y=ej;
30     now.d=0;
31     q2.push(now);
32     mi[ei][ej]=0;
33     while(!q1.empty()||!q2.empty())
34     {
35         if(!q1.empty())
36         {
37             now=q1.front();
38             q1.pop();
39             if(vis[now.x][now.y]==1) continue;
40             if(vis[now.x][now.y]==2) return now.d+mi[now.x][now.y];
41             vis[now.x][now.y]=1;
42             for(int i=0;i<8;i++)
43             {
44                 next.x=now.x+dir[i][0];
45                 next.y=now.y+dir[i][1];
46                 if(check(next.x,next.y)&&vis[next.x][next.y]!=1)
47                 {
48                     if(vis[next.x][next.y]==2) return now.d+mi[next.x][next.y]+1;
49                     else
50                     {
51                         next.d=now.d+1;
52                         mi[next.x][next.y]=next.d;
53                         q1.push(next);
54                     }
55                 }
56             }
57         }
58         //
59         if(!q2.empty())
60         {
61             now=q2.front();
62             q2.pop();
63             if(vis[now.x][now.y]==2) continue;
64             if(vis[now.x][now.y]==1) return now.d+mi[now.x][now.y];
65             vis[now.x][now.y]=2;
66             for(int i=0;i<8;i++)
67             {
68                 next.x=now.x+dir[i][0];
69                 next.y=now.y+dir[i][1];
70                 if(check(next.x,next.y)&&vis[next.x][next.y]!=2)
71                 {
72                     if(vis[next.x][next.y]==1) return now.d+mi[next.x][next.y]+1;
73                     else
74                     {
75                         next.d=now.d+1;
76                         mi[next.x][next.y]=next.d;
77                         q2.push(next);
78                     }
79                 }
80             }
81         }
82     }
83     return -1;
84 }
85 int main()
86 {
87     while(scanf("%d%d%d%d%d%d",&n,&m,&si,&sj,&ei,&ej)!=EOF)
88     {
89         memset(vis,0,sizeof(vis));
90         memset(mi,-1,sizeof(mi));
91         printf("%d
", bfs());
92     }
93     return 0;
94 }
View Code

COJ 1259 跳跳

http://122.207.68.93/OnlineJudge/problem.php?id=1259

为数字2到9的都建一个队列，读地图时候遇到了就加到各自的队列里

在bfs主体中，如果探索到了2到9的数字，就把相应队列里的结点顺便一并添加到bfs的主队列中...

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 #define REP(i,a,b) for(int i = a; i < b; i++)
 5 using namespace std;
 6 char map[105][105];
 7 int vis[105][105];
 8 int dir[][2]={1,0,-1,0,0,1,0,-1};
 9 int n,si,sj,ei,ej,ans;
10 queue<int> q[8];
11 struct node
12 {
13     int x,y,d;
14 }now,next;
15 int bfs()
16 {
17     queue<node> qq;
18     while(!qq.empty()) qq.pop();
19     now.x=si;
20     now.y=sj;
21     now.d=0;
22     qq.push(now);
23     while(!qq.empty())
24     {
25         now=qq.front();
26         qq.pop();
27         if(now.x==ei&&now.y==ej) return now.d;
28         if(vis[now.x][now.y]) continue;
29         vis[now.x][now.y]=1;
30         REP(i,0,4) {
31             next.x=now.x+dir[i][0];
32             next.y=now.y+dir[i][1];
33             if(next.x>=0&&next.x<n&&next.y>=0&&next.y<n&&map[next.x][next.y]!='1'&&vis[next.x][next.y]==0)
34             {
35                 if(map[next.x][next.y]=='0') 
36                 {
37                     next.d=now.d+1;
38                     qq.push(next);
39                 }else
40                 {
41                     int a=map[next.x][next.y]-'0';
42                     if(a>=2&&a<=9)
43                     {
44                         next.d=now.d+1;
45                         qq.push(next);
46                         while(!q[a-2].empty())
47                         {
48                             next.x=q[a-2].front();
49                             q[a-2].pop();
50                             next.y=q[a-2].front();
51                             q[a-2].pop();
52                             next.d=now.d+1;
53                             qq.push(next);
54                         }
55                     }
56                 }
57             }
58         }
59     }
60     return -1;
61 }
62 int main()
63 {
64     while(scanf("%d",&n)!=EOF)
65     {
66         REP(i,0,8) {
67             while(!q[i].empty()) q[i].pop();
68         }
69         getchar();
70         memset(vis,0,sizeof(vis));
71         REP(i,0,n) {
72             REP(j,0,n) {
73                 scanf("%c",&map[i][j]);
74                 if(map[i][j]=='S') 
75                 {
76                     si=i;
77                     sj=j;
78                 }else if(map[i][j]=='E')
79                 {
80                     ei=i;
81                     ej=j;
82                 }else if((map[i][j]-'0')>=2&&(map[i][j]-'0')<=9)
83                 {
84                     q[(map[i][j]-'0')-2].push(i);
85                     q[(map[i][j]-'0')-2].push(j);
86                 }
87             }
88             getchar();
89         }
90         map[si][sj]='1';
91         map[ei][ej]='0';
92         ans=bfs();
93         if(ans==-1) printf("Oh No!
");
94         else printf("%d
", ans);
95     }
96     return 0;
97 }
View Code

HDOJ 1242 Rescue

http://acm.hdu.edu.cn/showproblem.php?pid=1242

bfs+优先队列

实际上救援可以有多个，而公主只有一个，所以这题应该从公主开始bfs，遇到救援就停止，但是杭电这道题貌似只有一个救援...

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <vector>
 4 #include <queue>
 5 using namespace std;
 6 char map[201][201];
 7 bool flag[201][201];
 8 int dx[]={1,0,-1,0};
 9 int dy[]={0,1,0,-1};
10 int n,m;
11 struct node
12 {
13     int x;
14     int y;
15     int time;
16     friend bool operator<(node a,node b) //优先队列
17     {
18         return a.time>b.time;   //时间小的先出队
19     }
20 };
21 int BFS(int x,int y)
22 {
23     priority_queue<node> q;
24     node now,next;
25     int i;
26     now.x=x;
27     now.y=y;
28     now.time=0;
29     q.push(now);
30     flag[now.y][now.x]=true;
31     
32     while(!q.empty())
33     {
34         now=q.top();
35         for(i=0;i<4;i++)
36         {
37             next.x=now.x+dx[i];
38             next.y=now.y+dy[i];
39             if(next.x>=1&&next.x<=m&&next.y>=1&&next.y<=n&&map[next.y][next.x]!='#'&&flag[next.y][next.x]==false)
40             {
41                 flag[next.y][next.x]=true;
42                 next.time=now.time+1;
43                 if(map[next.y][next.x]=='x')
44                     next.time++;
45 
46                 q.push(next);   //next更新后在入队
47                 if(map[next.y][next.x]=='a')
48                     return next.time;
49             }
50         }
51         q.pop();
52     }
53     return -1;
54 }
55 int main()
56 {
57     int i,j,xe,ye;
58     while(scanf("%d%d",&n,&m)!=EOF)
59     {
60         vector<node> r;
61         r.clear();
62         for(i=1;i<=n;i++)
63         {
64             getchar();
65             for(j=1;j<=m;j++)
66             {
67                 scanf("%c",&map[i][j]);
68             }
69         }
70         for(i=1;i<=n;i++)
71         {
72             for(j=1;j<=m;j++)
73             {
74                 if(map[i][j]=='r')
75                 {
76                     node temp;
77                     temp.y=i;
78                     temp.x=j;
79                     temp.time=0;
80                     r.push_back(temp);  
81                 }
82             }
83         }
84         
85         int min=99999;
86         for(i=0;i<r.size();i++)
87         {
88             memset(flag,false,sizeof(flag));
89             int tem=BFS(r[i].x,r[i].y);
90             if(tem<min)
91                 min=tem;
92         }
93         if(min<0||r.size()==0)   //要判断是否有r，之前没判断，WA了几次
94             printf("Poor ANGEL has to stay in the prison all his life.
");
95         else
96             printf("%d
",min);
97     }
98     return 0;
99 }
View Code

HDOJ 1026 Ignatius and the Princess I

http://acm.hdu.edu.cn/showproblem.php?pid=1026

bfs+记录路径 多开一个path[][]用于记录路径 并在结构体里增加一对pre指向前一个节点的x，y坐标

然后从终点开始把各个节点入栈，出栈的时候就是从起点到终点了...

  1 #include <iostream>
  2 #include <queue>
  3 #include <stack>
  4 using namespace std;
  5 typedef struct Node{
  6     int x, y, cost;
  7     int prex, prey;
  8 }Node;
  9 int N, M;
 10 char maze[105][105];   // 记录初始输入
 11 Node path[105][105];   // 记录路径
 12 int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
 13 // 判断(x, y)是否可行
 14 bool isOK(int x, int y)
 15 {
 16     if(x>=0 && x<N && y>=0 && y<M && maze[x][y]!='X')
 17         return 1;
 18     else
 19         return 0;
 20 }
 21 void Init()
 22 {
 23     int i, j;
 24     for(i = 0; i < N; ++i)
 25         for(j = 0; j < M; ++j)
 26             path[i][j].cost = -1;
 27 }
 28 void backPath(int x, int y)
 29 {
 30     stack<Node> S;
 31     Node a, b;
 32     int cc = 1, tmp;
 33  
 34     cout << "It takes " << path[N - 1][M - 1].cost 
 35         << " seconds to reach the target position, let me show you the way." << endl;
 36     a = path[N - 1][M - 1];
 37     while(1)
 38     {
 39         if(a.x == 0 && a.y == 0)
 40             break;
 41         S.push(a);
 42         a = path[a.prex][a.prey];
 43     }
 44  
 45     a = path[0][0];
 46  
 47     while(!S.empty())
 48     {
 49         b = S.top();
 50         S.pop();
 51         if(maze[b.x][b.y] == '.')
 52             cout << cc++ << "s:(" << a.x << "," << a.y << ")->(" << b.x << "," << b.y << ")" << endl;
 53         else
 54         {
 55             cout << cc++ << "s:(" << a.x << "," << a.y << ")->(" << b.x << "," << b.y << ")" << endl;
 56             tmp = maze[b.x][b.y] - '0';
 57             while(tmp--)
 58                 cout << cc++ << "s:FIGHT AT (" << b.x << "," << b.y << ")" <<endl;
 59         }
 60         a = b;
 61     }
 62     cout<<"FINISH"<<endl;
 63 }
 64 int BFS(int x, int y)
 65 {
 66     queue<Node> Q;
 67     Node a, b;
 68     a.x = a.y = a.cost = a.prex = a.prey = 0;
 69     if(maze[0][0] != '.')
 70         a.cost = maze[0][0] - '0';
 71     path[0][0] = a;
 72     Q.push(a);
 73     while(!Q.empty())
 74     {
 75         a = Q.front();
 76         Q.pop();
 77         for(int i=0; i<4; ++i)
 78         {
 79             b.x = a.x + dir[i][0];
 80             b.y = a.y + dir[i][1];
 81             if(!isOK(b.x, b.y))
 82                 continue;
 83             if(maze[b.x][b.y] == '.')
 84                 b.cost = a.cost + 1;
 85             else
 86                 b.cost = a.cost + maze[b.x][b.y]-'0' + 1;
 87             if(b.cost < path[b.x][b.y].cost || path[b.x][b.y].cost == -1)
 88             {
 89                 b.prex = a.x; 
 90                 b.prey = a.y;
 91                 path[b.x][b.y] = b;
 92                 Q.push(b);
 93             }
 94         }
 95     }
 96     if(path[N - 1][M - 1].cost == -1)
 97     {
 98         cout << "God please help our poor hero." << endl;
 99         cout << "FINISH" << endl;
100         return 0;
101     }
102     backPath(N-1, M-1);
103 }
104 int main()
105 {
106     while(cin >> N >> M)
107     {
108         memset(maze, 0, sizeof(maze));
109         for(int i=0; i<N; ++i)
110             for(int j=0; j<M; ++j)
111                 cin >> maze[i][j];
112         Init();
113         BFS(0, 0);
114     }
115     return 0;
116 }
View Code

HDOJ 1195 Open the Lock

http://acm.hdu.edu.cn/showproblem.php?pid=1195

这题相当蛋疼，光是那四位数字在那转换就让我写的菊紧...

这题可以用双向bfs，不过这题数据比较弱，写了个bfs过了，也就没心情优化了...

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 #define REP(i,a,b) for(int i = a; i < b; i++)
 5 using namespace std;
 6 int T,b,start,end,tmp[4],cur[4],vis[10000];
 7 int cal(int n)
 8 {
 9     tmp[0]=n/1000;
10     tmp[1]=(n%1000)/100;
11     tmp[2]=(n%100)/10;
12     tmp[3]=n%10;
13     return 0;
14 }
15 int init()
16 {
17     cur[0]=tmp[0];
18     cur[1]=tmp[1];
19     cur[2]=tmp[2];
20     cur[3]=tmp[3];
21     return 0;
22 }
23 int ans()
24 {
25     return cur[0]*1000+cur[1]*100+cur[2]*10+cur[3];
26 }
27 struct node
28 {
29     int v,d;
30 }now,next;
31 int bfs()
32 {
33     queue<node> q;
34     while(!q.empty()) q.pop();
35     now.v=start;
36     now.d=0;
37     q.push(now);
38     while(!q.empty())
39     {
40         now=q.front();
41         if(now.v==end) return now.d;
42         q.pop();
43         if(vis[now.v]) continue;
44         vis[now.v]=1;
45         cal(now.v);
46         REP(i,0,4) {
47             init();
48             cur[i]++;
49             if(cur[i]==10) cur[i]=1;
50             next.v=ans();
51             if(vis[next.v]) continue;
52             next.d=now.d+1;
53             q.push(next);
54         }
55         REP(i,0,4) {
56             init();
57             cur[i]--;
58             if(cur[i]==0) cur[i]=9;
59             next.v=ans();
60             if(vis[next.v]) continue;
61             next.d=now.d+1;
62             q.push(next);
63         }
64         next.v=tmp[1]*1000+tmp[0]*100+tmp[2]*10+tmp[3];
65         next.d=now.d+1;
66         if(!vis[next.v]) q.push(next);
67         next.v=tmp[0]*1000+tmp[2]*100+tmp[1]*10+tmp[3];
68         next.d=now.d+1;
69         if(!vis[next.v]) q.push(next);
70         next.v=tmp[0]*1000+tmp[1]*100+tmp[3]*10+tmp[2];
71         next.d=now.d+1;
72         if(!vis[next.v]) q.push(next);
73     }
74     return 0;
75 }
76 int main()
77 {
78     scanf("%d",&T);
79     while(T--)
80     {
81         memset(vis,0,sizeof(vis));
82         scanf("%d",&start);
83         scanf("%d",&end);
84         printf("%d
", bfs());
85     }
86     return 0;
87 }
View Code

COJ 1336 Interesting Calculator

http://122.207.68.93/OnlineJudge/problem.php?id=1336

这题是湖南第九届省赛的题，bfs+优先队列 写不好的话空间可能会爆

我是多用了一个mi数组存达到这个数需要的最小步骤，如果扩展中遇到这个点，但是此时的d已经比保存的大了，就不扩展此结点...

看是觉得*0和*1 +0这些纯属没用的状态 WA好几次才发现 *0还是有用的...当第二数比第一个小时，要先*0  具体看代码吧

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 #define MAXN 100005
 5 #define REP(i,a,b) for(int i = a; i < b; i++)
 6 using namespace std;
 7 int start,end,T=0,min_c,min_d;
 8 int a[10],b[10],c[10];
 9 int mi[MAXN],vis[MAXN];
10 struct Node
11 {
12     int val,cost,dep;
13     friend bool operator<(Node a,Node b)
14     {
15         if(a.cost==b.cost) return a.dep>b.dep;
16         else return a.cost>b.cost;
17     }
18 }now,next;
19 int bfs()
20 {
21     priority_queue<Node> q;
22     while(!q.empty()) q.pop();
23     now.cost=0;
24     now.dep=0;
25     now.val=start;
26     mi[now.val]=0;
27     q.push(now);
28     while(!q.empty())
29     {
30         now=q.top();
31         q.pop();
32         if(vis[now.val]) continue;
33         vis[now.val]=1;
34         if(now.val==end)
35         {
36             min_c=now.cost;
37             min_d=now.dep;
38             return 1;
39         }
40         REP(i,0,10) {
41             next.val=now.val*10+i;
42             if(next.val<=end)
43             {
44                 next.cost=now.cost+a[i];
45                 next.dep=now.dep+1;
46                 if(!vis[next.val]&&mi[next.val]>next.cost)
47                 {
48                     q.push(next);
49                     mi[next.val]=next.cost;
50                 }
51             }
52         }
53         REP(i,0,10) {
54             next.val=now.val+i;
55             if(next.val<=end)
56             {
57                 next.cost=now.cost+b[i];
58                 next.dep=now.dep+1;
59                 if(!vis[next.val]&&mi[next.val]>next.cost)
60                 {
61                     q.push(next);
62                     mi[next.val]=next.cost;
63                 }
64             }
65         }
66         REP(i,0,10) {//开始这里i是从2开始的...WA好几发
67             next.val=now.val*i;
68             if(next.val<=end)
69             {
70                 next.cost=now.cost+c[i];
71                 next.dep=now.dep+1;
72                 if(!vis[next.val]&&mi[next.val]>next.cost)
73                 {
74                     q.push(next);
75                     mi[next.val]=next.cost;
76                 }
77             }
78         }
79     }
80     return 0;
81 }
82 int main()
83 {
84     while(scanf("%d%d",&start,&end)!=EOF)
85     {
86         T++;
87         memset(mi,0x3f,sizeof(mi));
88         memset(vis,0,sizeof(vis));
89         REP(i,0,10) scanf("%d",&a[i]);
90         REP(i,0,10) scanf("%d",&b[i]);
91         REP(i,0,10) scanf("%d",&c[i]);
92         bfs();
93         printf("Case %d: %d %d
",T,min_c,min_d);
94     }
95 }
View Code

POJ 1077 Eight

http://poj.org/problem?id=1077

经典八数码...据说不做此题，人生是不完整的...

判重就是一大难点：这题要用到全排列的变进制hash存储

然后算法的话 bfs可能险超时 双向bfs和A*是比较好的选择... 至于代码吗，还没写 哈哈...

持续更新中...

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