P4126 [AHOI2009]最小割

题目地址：P4126 [AHOI2009]最小割

最小割的可行边与必须边

首先求最大流，那么最小割的可行边与必须边都必须是满流。

• 可行边：在残量网络中不存在 \(x\) 到 \(y\) 的路径（强连通分量）；
• 必须边：在残量网络中 \(S\) 能到 \(x\) && \(y\) 能到 \(T\) 。

#include <bits/stdc++.h>
using namespace std;
const int N = 4e3 + 6, M = 6e4 + 6, inf = 1e9;
int n, m, s, t, d[N], f[N];
int Head[N], Edge[M<<1], Leng[M<<1], Next[M<<1], tot = 1;
struct E {
    int x, y, z;
} e[M<<1];
int dfn[N], low[N], num, st[N], top, ins[N], c[N], cnt;

inline void add(int x, int y, int z) {
    Edge[++tot] = y;
    Leng[tot] = z;
    Next[tot] = Head[x];
    Head[x] = tot;
}

inline bool bfs() {
    memset(d, 0, sizeof(d));
    queue<int> q;
    d[s] = 1;
    q.push(s);
    while (q.size()) {
        int x = q.front();
        q.pop();
        for (int i = Head[x]; i; i = Next[i]) {
            int y = Edge[i], z = Leng[i];
            if (d[y] || !z) continue;
            d[y] = d[x] + 1;
            q.push(y);
            if (y == t) return 1;
        }
    }
    return 0;
}

int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow;
    for (int i = Head[x]; i && rest; i = Next[i]) {
        int y = Edge[i], z = Leng[i];
        if (d[y] != d[x] + 1 || !z) continue;
        int k = dinic(y, min(z, rest));
        if (!k) d[y] = 0;
        else {
            Leng[i] -= k;
            Leng[i^1] += k;
            rest -= k;
        }
    }
    return flow - rest;
}

void dfs(int x, int k) {
    f[x] = k;
    for (int i = Head[x]; i; i = Next[i]) {
        int y = Edge[i], z = Leng[i^(k-1)];
        if (f[y] || !z) continue;
        dfs(y, k);
    }
}

void tarjan(int x) {
    dfn[x] = low[x] = ++num;
    st[++top] = x;
    ins[x] = 1;
    for (int i = Head[x]; i; i = Next[i]) {
        int y = Edge[i], z = Leng[i];
        if (!z) continue;
        if (!dfn[y]) {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        } else if (ins[y])
            low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
        ++cnt;
        int y;
        do {
            y = st[top--];
            ins[y] = 0;
            c[y] = cnt;
        } while (x != y);
    }
}

int main() {
    cin >> n >> m >> s >> t;
    for (int i = 1; i <= m; i++) {
        scanf("%d %d %d", &e[i].x, &e[i].y, &e[i].z);
        add(e[i].x, e[i].y, e[i].z);
        add(e[i].y, e[i].x, 0);
    }
    while (bfs())
        while (dinic(s, inf));
    dfs(s, 1);
    dfs(t, 2);
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i);
    for (int i = 1; i <= m; i++) {
        int k = i << 1;
        if (Leng[k]) puts("0 0");
        else printf("%d %d\n", c[e[i].x] != c[e[i].y], f[e[i].x] == 1 && f[e[i].y] == 2);
    }
    return 0;
}