POJ 2125 Destroying The Graph(二分图最小点权覆盖)

题目大意：给出一个有向图。对于每个点，定义两种操作：

操作A：删除点u的所有出边，花费为Cu

操作B：删除点u的所有入边，花费为du

要使所有的边都被删除，最小的花费是多少，并输出所有操作

解题思路：一看就知道是二分图的最小点权覆盖了，二分图的最小点权覆盖，胡伯涛:算法合集之《最小割模型在信息学竞赛中å的应用》 里面有讲，里面的内容挺重要的，推荐自己理解

这里讲一下如果找出所有的操作

首先，先dfs找出S集的点

接着判断一下哪些边是割边，割边就是我们所要的操作了(因为割边将所有的s-u-v-t的边全部破坏掉了，且花费的代价是最小的)

如何判断是否是割边呢？前面我们dfs的时候已经标记出了所有属于S集合的点了，而那些没标记的点肯定是T集的点了，所以判断一条边是不是割边的依据就是该边的起点是属于S集的，终点是属于T集的，且容量-流量==0

[code]#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge{
    int from, to, cap, flow;
    Edge() {}
    Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow) {}
};

struct Dinic{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G
;
    bool vis
;
    int d
, cur
;

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++) {
            G[i].clear();
        }
        edges.clear();
    }

    void AddEdge(int from, int to, int cap) {
        edges.push_back(Edge(from, to, cap, 0));
        edges.push_back(Edge(to, from, 0, 0));
        int m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    } 

    bool BFS() {
        memset(vis, 0, sizeof(vis));
        queue<int> Q;
        Q.push(s);
        vis[s] = 1;
        d[s] = 0;

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (!vis[e.to] && e.cap > e.flow) {
                    vis[e.to] = true;
                    d[e.to] = d[u] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a) {
        if (x == t || a == 0)
            return a;

        int flow = 0, f;
        for (int i = cur[x]; i < G[x].size(); i++) {
            Edge &e = edges[G[x][i]];
            if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[G[x][i] ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0)
                    break;
            }
        }
        return flow;
    }

    int Maxflow(int s, int t) {
        this->s = s; this->t = t;
        int flow = 0;
        while (BFS()) {
            memset(cur, 0, sizeof(cur));
            flow += DFS(s, INF);
        }
        return flow;
    }
};

Dinic dinic;

#define maxn 110
int n, m, source, sink, cnt;
int in[maxn], out[maxn], ans
;
bool vis[maxn * 2];

void init() {
    source = 0, sink = 2 * n + 1;
    dinic.init(sink);
    int c;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &c);
        dinic.AddEdge(i + n, sink, c);
    }

    for (int i = 1; i <= n; i++) {
        scanf("%d", &c);
        dinic.AddEdge(source, i, c);
    }

    int u, v;
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &u, &v);
        dinic.AddEdge(u, n + v, INF);
    }
}

void dfs(int u) {
    vis[u] = true;
    for (int i = 0; i < dinic.G[u].size(); i++) {
        Edge &e = dinic.edges[dinic.G[u][i]];
        if (!vis[e.to] && e.cap - e.flow > 0) dfs(e.to);
    }
}

void solve() {
    int maxflow = dinic.Maxflow(source, sink);
    memset(vis, 0, sizeof(vis));
    cnt = 0;
    dfs(source);
    for (int i = 0; i < 4 * n; i += 2) {
        Edge &e = dinic.edges[i];
        int u = e.from, v = e.to;
        if (vis[u] && !vis[v] && e.flow - e.cap == 0) 
            ans[cnt++] = i;
    }

    printf("%d\n%d\n", maxflow, cnt);
    for (int i = 0; i < cnt; i++) {
        int u = dinic.edges[ans[i]].from, v = dinic.edges[ans[i]].to;
        if (u == source) printf("%d -\n", v);
        else printf("%d +\n", u - n);
    }
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        init();
        solve();
    }
    return 0;
}