zoj 3792 Romantic Value

题目链接

求最小割的值， 以及割边最少的情况的边数。

先求一遍最小割， 然后把所有割边的权值变为1， 其他边变成inf， 在求一遍最小割， 此时求出的就是最少边数。

Inf打成inf  WA了好几发............

 #include<bits/stdc++.h>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, a, n) for(int i = a; i<n; i++)
 #define ull unsigned long long
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = 1e4;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 2e5+;
 int num, q[maxn*], head[maxn*], dis[maxn*], s, t;
 struct node
 {
     int to, nextt, c;
     node(){}
     node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
 }e[maxn*];
 int bfs() {
     mem(dis);
     int st = , ed = ;
     dis[s] = ;
     q[ed++] = s;
     while(st<ed) {
         int u = q[st++];
         for(int i = head[u]; ~i; i = e[i].nextt) {
             int v = e[i].to;
             if(!dis[v]&&e[i].c) {
                 dis[v] = dis[u]+;
                 if(v == t)
                     return ;
                 q[ed++] = v;
             }
         }
     }
     return ;
 }
 int dfs(int u, int limit) {
     int cost = ;
     if(u == t)
         return limit;
     for(int i = head[u]; ~i; i = e[i].nextt) {
         int v = e[i].to;
         if(e[i].c&&dis[v] == dis[u]+) {
             int tmp = dfs(v, min(e[i].c, limit-cost));
             if(tmp>) {
                 e[i].c -= tmp;
                 e[i^].c += tmp;
                 cost += tmp;
                 if(cost == limit)
                     break;
             } else {
                 dis[v] = -;
             }
         }
     }
     return cost;
 }
 int dinic() {
     int ans = ;
     while(bfs()) {
         ans += dfs(s, inf);
     }
     return ans;
 }
 void add(int u, int v, int c) {
     e[num] = node(v, head[u], c); head[u] = num++;
     e[num] = node(u, head[v], c); head[v] = num++;
 }
 void init() {
     mem1(head);
     num = ;
 }
 int main()
 {
     int T, m, n, x, y, q, p, z;
     cin>>T;
     while(T--) {
         init();
         int sum = ;
         scanf("%d%d%d%d", &n, &m, &p, &q);
         s = p, t = q;
         while(m--) {
             scanf("%d%d%d", &x, &y, &z);
             add(x, y, z);
             sum += z;
         }
         int ans = dinic();
         if(ans == ) {
             cout<<"Inf"<<endl;
             continue;
         }
         sum -= ans;
         for(int i = ; i<num; i+=) {
             if(e[i].c == ) {
                 e[i].c = ;
                 e[i^].c = inf;
             } else if(e[i^].c==) {
                 e[i].c = inf;
                 e[i^].c = ;
             } else {
                 e[i].c = e[i^].c = inf;
             }
         }
         ans = dinic();
         printf("%.2f\n", 1.0*sum/ans);
     }
 }