POJ-1459 Power Network(最大流模板)

题目链接：POJ-1459 Power Network

题意

有$np$个发电站，$nc$个消费者，$m$条有向边，给出每个发电站的产能上限，每个消费者的需求上限，每条边的容量上限，问最大流量。

思路

很裸的最大流问题，源点向发电站连边，边权是产能上限，消费者向汇点连边，边权是需求上限，其余的连边按给出的$m$条边加上去即可。

代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using std::queue;
const int INF = 0x3f3f3f3f, N = 110, M = N * N * 2;
int head[N], d[N];
int s, t, tot, maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init(int n) {
    tot = 1, maxflow = 0;
    s = n, t = n + 1;
    memset(head, 0, sizeof(head));
}

int main() {
    int n, np, nc, m;
    while (~scanf("%d %d %d %d", &n, &np, &nc, &m)) {
        init(n);
        for (int i = 0, u, v, z; i < m; i++) {
            scanf(" %*c %d %*c %d %*c %d", &u, &v, &z);
            add(u, v, z);
        }
        for (int i = 0, u, z; i < np; i++) {
            scanf(" %*c %d %*c %d", &u, &z);
            add(s, u, z);
        }
        for (int i = 0, u, z; i < nc; i++) {
            scanf(" %*c %d %*c %d", &u, &z);
            add(u, t, z);
        }
        while (bfs()) maxflow += dinic(s, INF);
        printf("%d\n", maxflow);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/kangkang-/p/11329976.html