标签：target 思路 show 题意 == src 消费 min sizeof
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。
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。
#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; }
标签：target 思路 show 题意 == src 消费 min sizeof
原文地址：https://www.cnblogs.com/kangkang-/p/11329976.html
