POJ 3204 Ikki's Story I - Road Reconstruction(最小割+残余网络)

题目大意：给出一张有向图，现在要求你增大一条边的容量，使得最大流增大

解题思路：增大的边肯定是割边了，所以，先找到割边
接着枚举割边，增容，在残余网络上跑最大流，如果有流的话，就表示增加了改边可以使最大流增大

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
const int MAXNODE = 1010;
const int MAXEDGE = 100010;
typedef int Type;
const Type INF = 0x3f3f3f3f;

struct Edge{
    int u, v, next;
    Type cap, flow;
    Edge() {}
    Edge(int u, int v, Type cap, Type flow, int next) : u(u), v(v), cap(cap), flow(flow), next(next){}
};

struct Dinic{
    int n, m, s, t;
    Edge edges[MAXEDGE];
    int head[MAXNODE];
    bool vis[MAXNODE];
    Type d[MAXNODE];
    vector<int> cut;

    void init(int n) {
        this->n = n;
        memset(head, -1, sizeof(head));
        m = 0;
    }

    void AddEdge(int u, int v, Type cap) {
        edges[m] = Edge(u, v, cap, 0, head[u]);
        head[u] = m++;
        edges[m] = Edge(v, u, 0, 0, head[v]);
        head[v] = m++;
    } 

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

        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            for (int i = head[u]; ~i; i = edges[i].next) {
                Edge &e = edges[i];
                if (!vis[e.v] && e.cap > e.flow) {
                    vis[e.v] = true;
                    d[e.v] = d[u] + 1;
                    Q.push(e.v);
                }
            }
        }
        return vis[t];
    }

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

        Type flow = 0, f;
        for (int i = head[u]; ~i; i = edges[i].next) {
            Edge &e = edges[i];
            if (d[u] + 1 == d[e.v] && (f = DFS(e.v, min(a, e.cap - e.flow))) > 0) {
                e.flow += f;
                edges[i ^ 1].flow -= f;
                flow += f;
                a -= f;
                if (a == 0) break;
            }
        }
        return flow;
    }

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

    void Residue() {
        for (int i = 0; i < m; i++) 
            edges[i].cap -= edges[i].flow;
    }

    void Clear() {
        for (int i = 0; i < m; i++)
            edges[i].flow = 0;
    }

    void Mincut() {
        cut.clear();
        int ans = 0;
        for (int i = 0; i < m; i += 2) {
            if (vis[edges[i].u] && !vis[edges[i].v] && edges[i].cap > 0) 
                cut.push_back(i);
        }
    }

}dinic;

int n, m;

void solve() {
    int source = 0, sink = n - 1;
    dinic.init(sink + 1);

    int u, v, c;
    for (int i = 1; i <= m; i++) {
        scanf("%d%d%d", &u, &v, &c);
        dinic.AddEdge(u, v, c);
    }
    dinic.Maxflow(source, sink);
    dinic.Mincut();
    dinic.Residue();
    int ans = 0;
    for (int i = 0; i < dinic.cut.size(); i++) {
        dinic.Clear();
        Edge &e = dinic.edges[dinic.cut[i]];
        e.cap += 10;
        int tmp = dinic.Maxflow(source, sink);
        if (tmp) ans++;
        e.cap -= 10;
    }
    printf("%d\n", ans);
}

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