HDU-3987 Harry Potter and the Forbidden Forest（最大流）

题目：http://acm.hdu.edu.cn/showproblem.php?pid=3987
题意：求边数最少的割集
思路：与上文一样
代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;const ll INF = 1e18;const int N = 500005;int n,m,s,t,tot;int head[N],dep[N],cur[N];struct edge{    int to,nxt;    ll cap;    edge(int t = 0,int n = 0,ll c = 0):to(t),nxt(n),cap(c){}}E[N];void init(){    memset(head,-1,sizeof(head));    tot = 0;}void add(int u,int v,ll cap){    E[tot] = edge(v,head[u],cap);    head[u] = tot++;}bool bfs(int s,int t){    memset(dep,-1,sizeof(dep));    queue<int> q;    dep[s] = 0;    q.push(s);    while(!q.empty())    {        int u = q.front();q.pop();        for(int i = head[u];~i;i = E[i].nxt)        {            int v = E[i].to;            if(E[i].cap > 0 && dep[v] == -1)            {                dep[v] = dep[u] + 1;                q.push(v);            }        }    }    return dep[t] != -1;}ll dfs(int u,ll flow){    if(u == t)        return flow;    ll w ,used = 0;    for(int i = head[u];~i;i = E[i].nxt)    {        int v = E[i].to;        if(dep[v] == dep[u]+1)        {            w = flow - used;            w = dfs(v,min(w,E[i].cap));            E[i].cap -= w;            E[i^1].cap += w;            if(v) cur[u] = i;            used += w;            if(used == flow) return flow;        }    }    if(!used) dep[u] = -1;    return used;}ll dinic(int s,int t){    ll res = 0;    while(bfs(s,t))    {         for(int i = 1;i <= t;i++)            cur[i] = head[i];         res += dfs(s,INF);    }    return res;}int main(){    int T;    scanf("%d",&T);    for(int ca = 1;ca <= T;ca++)    {        init();        scanf("%d%d",&n,&m);        s = 1,t = n;        int u,v,d;        ll w;        while(m--)        {            scanf("%d%d%lld%d",&u,&v,&w,&d);            ++u;++v;            add(u,v,w*1000000+1);            if(d)                add(v,u,w*1000000+1);            else                add(v,u,0);        }        printf("Case %d: %lld\n",ca,dinic(s,t)%1000000);    }    return 0;}