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-8;
const int mod = 1e9+7;
const int inf = 1e4;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 2e5+5;
int num, q[maxn*5], head[maxn*2], dis[maxn*2], s, t;
struct node
{
    int to, nextt, c;
    node(){}
    node(int to, int nextt, int c):to(to), nextt(nextt), c(c){}
}e[maxn*2];
int bfs() {
    mem(dis);
    int st = 0, ed = 0;
    dis[s] = 1;
    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]+1;
                if(v == t)
                    return 1;
                q[ed++] = v;
            }
        }
    }
    return 0;
}
int dfs(int u, int limit) {
    int cost = 0;
    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]+1) {
            int tmp = dfs(v, min(e[i].c, limit-cost));
            if(tmp>0) {
                e[i].c -= tmp;
                e[i^1].c += tmp;
                cost += tmp;
                if(cost == limit)
                    break;
            } else {
                dis[v] = -1;
            }
        }
    }
    return cost;
}
int dinic() {
    int ans = 0;
    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 = 0;
}
int main()
{
    int T, m, n, x, y, q, p, z;
    cin>>T;
    while(T--) {
        init();
        int sum = 0;
        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 == 0) {
            cout<<"Inf"<<endl;
            continue;
        }
        sum -= ans;
        for(int i = 0; i<num; i+=2) {
            if(e[i].c == 0) {
                e[i].c = 1;
                e[i^1].c = inf;
            } else if(e[i^1].c==0) {
                e[i].c = inf;
                e[i^1].c = 1;
            } else {
                e[i].c = e[i^1].c = inf;
            }
        }
        ans = dinic();
        printf("%.2f
", 1.0*sum/ans);
    }
}