[HDOJ5855]Less Time, More profit（最大权闭合子图，二分，最大流）

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5855

题意：有n个工厂m个商店，建造工厂需要花费pay和时间t，商店需要指定的几个工厂供货，如果同时供货的话可以获利pro。要求时间最小的情况下获利最多，求最小时间和最小时间下最大获利。

 岐爷博客学习了：http://www.cnblogs.com/wuyiqi/archive/2012/03/12/2391960.html

这个图是这么建的。先读入数据后在区间(0~tmax)二分枚举时间time，代表当前建造时间不超过time的工厂可以同时建造。

之后就是求最大权闭合图，最大权闭合图的的权 = 原图中权值为正的点的和（所有用户的收益之和） - 最小割（最大流）

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge {
    int u, v, w, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 2020;

int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[20202020];
int S, T, N;

void init() {
    memset(dhead, -1, sizeof(dhead));
    for(int i = 0; i < maxn; i++) dedge[i].next = -1;
    S = 0; cnt = 0;
}

void adde(int u, int v, int w, int c1=0) {
    dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 
    dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
    dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 
    dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
}

bool bfs(int s, int t, int n) {
    queue<int> q;
    for(int i = 0; i < n; i++) dd[i] = inf;
    dd[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = dhead[u]; ~i; i = dedge[i].next) {
            if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
                dd[dedge[i].v] = dd[u] + 1;
                if(dedge[i].v == t) return 1;
                q.push(dedge[i].v);
            }
        }
    }
    return 0;
}

int dinic(int s, int t, int n) {
    int st[maxn], top;
    int u;
    int flow = 0;
    while(bfs(s, t, n)) {
        for(int i = 0; i < n; i++) cur[i] = dhead[i];
        u = s; top = 0;
        while(cur[s] != -1) {
            if(u == t) {
                int tp = inf;
                for(int i = top - 1; i >= 0; i--) {
                    tp = min(tp, dedge[st[i]].w);
                }
                flow += tp;
                for(int i = top - 1; i >= 0; i--) {
                    dedge[st[i]].w -= tp;
                    dedge[st[i] ^ 1].w += tp;
                    if(dedge[st[i]].w == 0) top = i;
                }
                u = dedge[st[top]].u;
            }
            else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
                st[top++] = cur[u];
                u = dedge[cur[u]].v;
            }
            else {
                while(u != s && cur[u] == -1) {
                    u = dedge[st[--top]].u;
                }
                cur[u] = dedge[cur[u]].next;
            }
        }
    }
    return flow;
}

inline bool scan_d(int &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
        num*=10,num+=in-'0';
    }
    if(IsN) num=-num;
    return true;
}

int n, m, L;
int pay[maxn], t[maxn], pro[maxn];
int k[maxn];
int G[maxn][maxn];
int maxx;

int main() {
    // freopen("in", "r", stdin);
    int tt, _ = 1;
    scan_d(tt);
    while(tt--) {
        maxx = -1;
        memset(G, 0, sizeof(G));
        scan_d(n); scan_d(m); scan_d(L);
        for(int i = 1; i <= n; i++) {
            scan_d(pay[i]); scan_d(t[i]);
            maxx = max(maxx, t[i]);
        }
        for(int i = 1; i <= m; i++) {
            scan_d(pro[i]); scan_d(k[i]);
            for(int j = 0; j < k[i]; j++) {
                scan_d(G[i][j]);
            }
        }
        int lo = 0, hi = maxx + 1;
        int ret1 = -1, ret2 = 0;
        while(lo <= hi) {
            init();
            S = 0, T = n + m + 1, N = T + 1;
            int mid = (lo + hi) >> 1;
            for(int i = 1; i <= n; i++) {
                if(mid >= t[i]) adde(S, i, pay[i]);
                else adde(S, i, inf);
            }
            int pos = 0;
            for(int i = 1; i <= m; i++) {
                for(int j = 0; j < k[i]; j++) {
                    adde(G[i][j], i+n, inf);
                }
                adde(i+n, T, pro[i]);
                pos += pro[i];
            }
            int flow = dinic(S, T, N);
            if(pos - flow >= L) {
                hi = mid - 1;
                ret1 = mid, ret2 = pos - flow;
            }
            else lo = mid + 1;
        }
        if(ret1 == -1) printf("Case #%d: impossible
", _++);
        else printf("Case #%d: %d %d
", _++, ret1, ret2);
    }
    return 0;
}