HDU 1242 Rescue Rescue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12546 Accepted Submission(s): 4581
Problem Description
Angel was caught by the MOLIGPY! He was put in * by Moligpy. The * is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the *.
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.
You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)
Input
First line contains two integers stand for N and M.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.
Process to the end of the file.
Output
For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the * all his life."
Sample Input
7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........
Sample Output
13
Author
CHEN, Xue
Source
Recommend
Eddy
思路：优先队列就可以解决
优先级别的定义如下
struct Node
{
int x,y;
int time;
friend bool operator < (const Node &a,const Node &b)
{
return a.time > b.time;
}
};
{
int x,y;
int time;
friend bool operator < (const Node &a,const Node &b)
{
return a.time > b.time;
}
};
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <cstring>
using namespace std;
char map[210][210];
int hash[210][210];
int bfs[4][2] = {1,0,-1,0,0,1,0,-1};
int n,m;
int ax,ay,rx,ry;
struct Node
{
int x,y;
int time;
friend bool operator < (const Node &a,const Node &b)
{
return a.time > b.time;
}
};
int BFS()
{
int max = 0;
priority_queue <Node> q;
Node top;
top.x = ax;top.y = ay;top.time = 0;
q.push(top);
while(!q.empty())
{
Node temp;
temp = q.top();
//printf("%d %d have %d ",temp.x,temp.y,temp.time);
q.pop();
if(map[temp.x][temp.y] == 'r')
{
max = temp.time;
break;
}
for(int k = 0;k < 4;k ++)
{
int x = temp.x + bfs[k][0],y = temp.y + bfs[k][1];
int time = temp.time;
if(map[x][y] != '#' && hash[x][y] != 1 &&
x >= 1 && x <= n && y >= 1 && y <= m)
{
hash[x][y] = 1;
Node xin;
xin.x = x;xin.y = y;
if(map[x][y] == 'x')
xin.time = time + 2;
else
xin.time = time + 1;
q.push(xin);
}
}
}
return max;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(hash,0,sizeof(hash));
for(int i = 1;i <= n;i ++)
for(int j = 1;j <= m;j ++)
{
scanf(" %c",&map[i][j]);
if(map[i][j] == 'a')
{
ax = i;ay = j;hash[i][j] = 1;
}
}
int sss = BFS();
if(sss == 0)
printf("Poor ANGEL has to stay in the * all his life. ");
else
printf("%d ",sss);
}
return 0;
}
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <cstring>
using namespace std;
char map[210][210];
int hash[210][210];
int bfs[4][2] = {1,0,-1,0,0,1,0,-1};
int n,m;
int ax,ay,rx,ry;
struct Node
{
int x,y;
int time;
friend bool operator < (const Node &a,const Node &b)
{
return a.time > b.time;
}
};
int BFS()
{
int max = 0;
priority_queue <Node> q;
Node top;
top.x = ax;top.y = ay;top.time = 0;
q.push(top);
while(!q.empty())
{
Node temp;
temp = q.top();
//printf("%d %d have %d ",temp.x,temp.y,temp.time);
q.pop();
if(map[temp.x][temp.y] == 'r')
{
max = temp.time;
break;
}
for(int k = 0;k < 4;k ++)
{
int x = temp.x + bfs[k][0],y = temp.y + bfs[k][1];
int time = temp.time;
if(map[x][y] != '#' && hash[x][y] != 1 &&
x >= 1 && x <= n && y >= 1 && y <= m)
{
hash[x][y] = 1;
Node xin;
xin.x = x;xin.y = y;
if(map[x][y] == 'x')
xin.time = time + 2;
else
xin.time = time + 1;
q.push(xin);
}
}
}
return max;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
memset(hash,0,sizeof(hash));
for(int i = 1;i <= n;i ++)
for(int j = 1;j <= m;j ++)
{
scanf(" %c",&map[i][j]);
if(map[i][j] == 'a')
{
ax = i;ay = j;hash[i][j] = 1;
}
}
int sss = BFS();
if(sss == 0)
printf("Poor ANGEL has to stay in the * all his life. ");
else
printf("%d ",sss);
}
return 0;
}