poj1459 Power Network (最大流)

题意：

一个电力网络，一共n个节点，m条边，np个发电站，nc个用户。

每条边有一个容量，每个发电站有一个最大负载，每一个用户也有一个最大接受量。

问最多能供给多少电力。

思路：

多源多汇的最大流，设一个超级源点，与每一个源点连一条容量为最大负载的边，

设一个超级汇点，每一个用户与超级汇点连一条容量为最大接受量的边，然后求最大流。

代码（1396K，766MS）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>

#define INF 0x3f3f3f3f
#define T (n + 1)

using namespace std;

struct Edge{
int from, to, cap, flow;
Edge() {}
Edge(int a, int b, int c, int d) : from(a), to(b), cap(c), flow(d) {}
};

int n, np, nc, m;
vector<Edge> edges;
vector<int> G[105];
void addEdge(int from, int to, int cap) {
edges.push_back(Edge(from, to, cap, 0));
edges.push_back(Edge(to, from, 0, 0));
int siz = edges.size();
G[from].push_back(siz - 2);
G[to].push_back(siz - 1);
}

int cur[105];
int layer[105];

bool build() {
memset(layer, -1, sizeof(layer));
queue<int> q;
q.push(0);
layer[0] = 0;
while (!q.empty()) {
int current = q.front();
q.pop();
for (int i = 0; i < G[current].size(); i++) {
Edge e = edges[G[current][i]];
if (layer[e.to] == -1 && e.cap > e.flow) {
layer[e.to] = layer[current] + 1;
q.push(e.to);
}
}
}
return layer[T] != -1;
}

int find(int x, int curflow) {
if (x == T || !curflow)
return curflow;
int f, flow = 0;
for (int &i = cur[x]; i < G[x].size(); i++) {
Edge &e = edges[G[x][i]];
if (layer[e.to] == layer[x] + 1
&& (f = find(e.to, min(curflow, e.cap - e.flow)))) {
e.flow += f;
edges[G[x][i] ^ 1].flow -= f;
flow += f;
curflow -= f;
if (!curflow) break;
}
}
return flow;
}

int dinic() {
int maxflow = 0;
while (build()) {
memset(cur, 0, sizeof(cur));
maxflow += find(0, INF);
}
return maxflow;
}

int main() {
while (~scanf("%d %d %d %d", &n, &np, &nc, &m)) {
char t;
int a, b, c;
for (int i = 0; i <= n; i++)
G[i].clear();
edges.clear();
for (int i = 0; i < m; i++) {
cin >> t >> a >> t >> b >> t >> c;
addEdge(a + 1, b + 1, c);
}
for (int i = 0; i < np; i++) {
cin >> t >> a >> t >> b;
addEdge(0, a + 1, b);
}
for (int i = 0; i < nc; i++) {
cin >> t >> a >> t >> b;
addEdge(a + 1, T, b);
}
int ans = dinic();
printf("%d\n", ans);
}
return 0;
}