HDU 3996 Gold Mine

题意：挖金矿。一个矿有n层，每层有m块金子。要挖每块金子都有一个代价cost，每块金子都有一个价值cost，另外每块金子还有w个关系，表示如果要挖它，必须先挖掉第i层的第j块金子，求最大获利。

最大闭合子图。

#include <cstdio>
#include <cstring>
#include <algorithm>
#define maxn 3010
#define maxm 400000
using namespace std;
typedef long long LL;
const LL INF = 1LL<<60;
int v[maxm],next[maxm];
LL w[maxm];
int first[maxn],q[maxn],work[maxn],d[maxn];
int e,S,T;

void init(){
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,LL c){
    v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
    v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++;
}

int bfs(){
    int rear = 0;
    memset(d,-1,sizeof(d));
    d[S] = 0;q[rear++] = S;
    for(int i = 0;i < rear;i++){
        for(int j = first[q[i]];j != -1;j = next[j])
            if(w[j] && d[v[j]] == -1){
                d[v[j]] = d[q[i]] + 1;
                q[rear++] = v[j];
                if(v[j] == T)   return 1;
            }
    }
    return 0;
}

LL dfs(int cur,LL a){
    if(cur == T)    return a;
    for(int &i = work[cur];i != -1;i = next[i]){
        if(w[i] && d[v[i]] == d[cur] + 1)
            if(LL t = dfs(v[i],min(a,w[i]))){
                w[i] -= t,w[i^1] += t;
                return t;
            }
    }
    return 0;
}

LL dinic(){
    LL ans = 0;
    while(bfs()){
        memcpy(work,first,sizeof(first));
        while(LL t = dfs(S,INF))    ans += t;
    }
    return ans;
}

int main()
{
    int n,m,t;
    scanf("%d",&t);
    for(int kase = 1;kase <= t;kase++){
        init();
        S = 0,T = 3000;
        LL sum = 0;
        scanf("%d",&n);
        for(int i = 1;i <= n;i++){
            scanf("%d",&m);
            for(int j = 1;j <= m;j++){
                int pos = i*26+j;
                LL cost,value;
                int ww;
                scanf("%I64d%I64d%d",&cost,&value,&ww);
                value -= cost;
                if(value > 0)   add_edge(S,pos,value),sum += value;
                if(value < 0)   add_edge(pos,T,-value);
                for(int k = 0;k < ww;k++){
                    int a,b;
                    scanf("%d%d",&a,&b);
                    add_edge(pos,a*26+b,INF);
                }
            }
        }
        printf("Case #%d: %I64d
",kase,sum-dinic());
    }
    return 0;
}