POJ-1459 Power Network(最大流模板)
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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;
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