ZOJ 2770 Burn the Linked Camp 差分约束+SPFA

第一道正儿八经的差分约束题

有排成一列的n个点，首先告诉你每个点的值最多是多少(最少显然要大于0），然后告诉你m段i，j,k，表示第i个点到第j个点的值的和至少有k，问你总和至少为多少。

要注意的是，告诉你的所有关系式都不要忘记建边，一开始漏了大于0的条件调半天o(╯□╰)o

不等式的形式是a-b<=c这样的= =

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <algorithm>
 5 #include <climits>
 6 #include <string>
 7 #include <iostream>
 8 #include <map>
 9 #include <cstdlib>
10 #include <list>
11 #include <set>
12 #include <queue>
13 #include <stack>
14 
15 using namespace std;
16 
17 typedef long long LL;
18 const int maxn = 1000 + 5;
19 const int maxm = 50000 + 5;
20 const LL INF = LONG_LONG_MAX / 4;
21 int v[maxm],w[maxm],first[maxn],nxt[maxm];
22 int n,m,C,cnt[maxn],ecnt;
23 LL d[maxn];
24 bool inq[maxn];
25 
26 void solve() {
27     bool bad = false;
28     queue<int> q;
29     q.push(n);
30     for(int i = 0;i <= n;i++) {
31         inq[i] = false;
32         d[i] = INF;
33         cnt[i] = 0;
34     }
35     d[n] = 0;
36     inq[n] = true;
37     cnt[n] = 1;
38     while(!q.empty() && !bad) {
39         int x = q.front(); q.pop();
40         inq[x] = false;
41         for(int i = first[x];i != -1;i = nxt[i]) {
42             if(d[v[i]] > d[x] + w[i]) {
43                 d[v[i]] = d[x] + w[i];
44                 if(!inq[v[i]]) {
45                     q.push(v[i]);
46                     cnt[v[i]]++;
47                     inq[v[i]] = true;
48                     if(cnt[v[i]] > n + 1) {
49                         bad = true; break;
50                     }
51                 }
52             }
53         }
54     }
55     if(bad) puts("Bad Estimations");
56     else printf("%lld
",-d[0]);
57 }
58 
59 void adde(int _u,int _v,int _w) {
60     v[ecnt] = _v; w[ecnt] = _w;
61     nxt[ecnt] = first[_u];
62     first[_u] = ecnt;
63     ecnt++;
64 }
65 
66 int main() {
67     while(~scanf("%d%d",&n,&m)) {
68         ecnt = 0;
69         memset(first,-1,sizeof(first));
70         memset(nxt,-1,sizeof(nxt));
71         for(int i = 1;i <= n;i++) {
72            scanf("%d",&C);
73            adde(i - 1,i,C);
74            adde(i,i - 1,0);
75         }
76         for(int i = 1;i <= m;i++) {
77             int a,b,c; scanf("%d%d%d",&a,&b,&c);
78             adde(b,a - 1,-c);
79         }
80         solve();
81     }
82     return 0;
83 }

ZOJ 2770 Burn the Linked Camp 差分约束+SPFA

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