POJ-1459 Power Network

题目大意：

这道题目意思不好理解（可能只是我理解太差...）

一个电网，里面有一些结点，代表电站，消费者，调度站。电站不消耗电能，消费者不产生电能，调度站产生的电能和消耗的电能均为0

现在给出一个电网，有n个结点，np个电站(u)z[表示u号节点是能产生最多z电能的电站]，nc个消费者(u)z[表示u是最多消耗z电能的消费者]，m条输电线(u,v)z[表示从u到z最多能输送z电能]，现在问你最这个电网最多能消费多少电能。

解题思路：

这是标准的网络流模型= =

首先把电站跟源点连接，容量为z

再把想消费者与汇点连接，容量为z

再把m条输电线连接，容量为z

然后就是直接裸着写网络流就可以了= =

代码：

#include <queue>#include <cstdio>#include <cstring>using namespace std;typedef struct node{    int v, cap, nxt;    node(int a = 0, int b = 0, int c = 0){        v = a; cap = b; nxt = c;    }}Edge;const int maxm = 150;const int maxn = 1e5 + 5;const int INF = 0x3f3f3f3f;int s, t, tot;Edge edge[maxn];int head[maxn], dis[maxm];inline int Min(int a, int b){    return (a < b ? a : b);}void add(int u, int v, int cap){    edge[tot] = Edge(v, cap, head[u]);    head[u] = tot++;    edge[tot] = Edge(u, 0, head[v]);    head[v] = tot++;}int bfs(){    queue<int> q;    while(!q.empty()) q.pop();    memset(dis, 0, sizeof(dis));    q.push(s); dis[s] = 1;    while(!q.empty()){        int x = q.front(); q.pop();        for(int i = head[x]; ~i; i = edge[i].nxt){            Edge &e = edge[i];            if(e.cap && dis[e.v] == 0){                dis[e.v] = dis[x] + 1;                q.push(e.v);            }        }    }    return dis[t];}int dfs(int x, int f){    if(x == t) return f;    int sum = 0;    for(int i = head[x]; ~i; i = edge[i].nxt){        Edge &e = edge[i];        if(e.cap && dis[e.v] == dis[x] + 1){            int ret = dfs(e.v, Min(f, e.cap));            sum += ret; f -= ret;            e.cap -= ret; edge[i^1].cap += ret;        }    }    return sum;}int dinic(){    int ret = 0;    while(bfs()) ret += dfs(s, INF);    return ret;}int main(){    int u, v, z, n, np, nc, m;    while(~scanf("%d%d%d%d", &n, &np, &nc, &m)){        tot = 0; s = n + 1; t = s + 1;        memset(head, -1, sizeof(head));        for(int i = 0; i < m; ++i){            scanf(" (%d,%d)%d", &u, &v, &z);            add(u, v, z);        }        for(int i = 0; i < np; ++i){            scanf(" (%d)%d", &u, &z);            add(s, u, z);        }        for(int i = 0; i < nc; ++i){            scanf(" (%d)%d", &u, &z);            add(u, t, z);        }        printf("%d\n", dinic());    }    return 0;}