Loj #6000.「 网络流 24 题 」搭配飞行员

放图片是不是很骚气

解题思路

建立超级源点和超级汇点。将主驾驶雨源点相连，副驾驶与汇点相连，再把输入的有向边加进去，同时建反边。

跑$Dinic$，网络流模板不难，难的是建模QAQ

附上代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int maxnode = 105, maxedge = 30003, INF = 2147483647;
int n, m, s, t, head[maxnode], cnt = 1, Depth[maxnode], Ans;
struct Edge {
    int nxt, v, w;
}ed[maxedge];
inline void addedge(int u, int v, int w) {
    ed[++cnt].nxt = head[u];
    ed[cnt].v = v, ed[cnt].w = w;
    head[u] = cnt;
}
inline bool BFS() {
    queue <int> Q;
    memset(Depth, 0, sizeof(Depth));
    Depth[s] = 1, Q.push(s);
    int u;
    while (!Q.empty()) {
        u = Q.front();
        Q.pop();
        for(int i=head[u]; i; i=ed[i].nxt) {
            if(Depth[ed[i].v] == 0 && ed[i].w > 0) {
                Depth[ed[i].v] = Depth[u] + 1;
                Q.push(ed[i].v);
                if(ed[i].v == t) return true;
            }
        }
    }
    return false;
}
inline int Dinic(int u, int cap) {
    if(u == t) return cap;
    int delta;
    for(int i=head[u]; i; i=ed[i].nxt) {
        if(ed[i].w > 0 && Depth[ed[i].v] == Depth[u] + 1) {
            delta = Dinic(ed[i].v, min(cap, ed[i].w));
            if(delta > 0) {
                ed[i].w -= delta;
                ed[i^1].w += delta;
                return delta;
            }
        }
    }
    return 0;
}
int main() {
    scanf("%d%d", &n, &m);
    s = 0, t = n+1;
    for(int i=1; i<=m; i++) addedge(s, i, 1), addedge(i, s, 0);
    for(int i=m+1; i<=n; i++) addedge(i, t, 1), addedge(t, i, 0);
    static int x, y;
    while (scanf("%d%d", &x, &y) == 2) addedge(x, y, 1), addedge(y, x, 0);
    while (BFS()) Ans += Dinic(s, INF);
    printf("%d", Ans);
}

原文地址：https://www.cnblogs.com/bljfy/p/9543665.html