hdu3987 Harry Potter and the Forbidden Forest 最小割边数

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=3987

题意：

给一个网络，边有有向边和无向边两种，挑选一些边破坏，使得从0无法到达n-1，每条边都有一个破环的成本，求在成本最低的情况下需要破坏的最少边数

思路：

明显最小成本就是最小割，求边数的话，扩大边权然后取余即可

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1000 + 10;const ll INF = 0x3f3f3f3f3f3f3f3f;struct edge{    int to, next;    ll cap;}g[N*N*2];int level[N], cur[N], pre[N], gap[N];int cnt, head[N];int nv;void init(){    cnt = 0;    memset(head, -1, sizeof head);}void add_edge(int v, int u, ll cap){    g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;    g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;}ll sap(int s, int t){    memset(level, 0, sizeof level);    memset(gap, 0, sizeof gap);    memcpy(cur, head, sizeof head);    gap[0] = nv;    int v = pre[s] = s;    ll flow = 0, aug = INF;    while(level[s] < nv)    {        bool flag = false;        for(int &i = cur[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && level[v] == level[u] + 1)            {                flag = true;                pre[u] = v;                v = u;                aug = min(aug, g[i].cap);                if(v == t)                {                    flow += aug;                    while(v != s)                    {                        v = pre[v];                        g[cur[v]].cap -= aug;                        g[cur[v]^1].cap += aug;                    }                    aug = INF;                }                break;            }        }        if(flag) continue;        int minlevel = nv;        for(int i = head[v]; i != -1; i = g[i].next)        {            int u = g[i].to;            if(g[i].cap > 0 && level[u] < minlevel)                minlevel = level[u], cur[v] = i;        }        if(--gap[level[v]] == 0) break;        level[v] = minlevel + 1;        gap[level[v]]++;        v = pre[v];    }    return flow;}int main(){    int t, n, m, cas = 0;    scanf("%d", &t);    while(t--)    {        init();        scanf("%d%d", &n, &m);        int a, b, c, d;        for(int i = 1; i <= m; i++)        {            scanf("%d%d%d%d", &a, &b, &c, &d);            a++, b++;            add_edge(a, b, 1LL*c*(m+1) + 1);            if(d) add_edge(b, a, 1LL*c*(m+1) + 1);//边权扩大(m+1)倍再+1        }        int ss = 1, tt = n;        nv = n;        printf("Case %d: %lld\n", ++cas, sap(ss, tt) % (m+1));    }    return 0;}