[POJ 1459] Power Network (网络流）

来源：互联网发布：jquery.reveal.js下载编辑：程序博客网时间：2024/05/24 04:53

Power Network

题目链接：http://poj.org/problem?id=1459

题目大意：

有一个电网，其中有N个结点，分为np个发电厂，每个发电厂发电。nc个用户，每个用户要用电。以及N-np-nc个中间站，只负责传送电量。有M条边连接着这些结点，每条边上有最大的电量限制。问用户最多在同一时间可以用多少电？

解题思路：

构造一个超级源点，加上每个源点到发电站的边，边的权值为发电站的发电量。构造一个超级汇点，每个用户到汇点有边，边的权值为用户的用电量。这样就成了一道普通的最大流问题。

/*ID: wuqi9395@126.comPROG:LANG: C++*/#include<map>#include<set>#include<queue>#include<stack>#include<cmath>#include<cstdio>#include<vector>#include<string>#include<fstream>#include<cstring>#include<ctype.h>#include<iostream>#include<algorithm>#define maxn 140#define maxm 100000#define INF (1<<30)#define PI acos(-1.0)#define mem(a, b) memset(a, b, sizeof(a))#define For(i, n) for (int i = 0; i < n; i++)typedef long long ll;using namespace std;struct node {    int v, f, nxt;}e[maxm * 2];int n, m, st, ed, g[maxn], cnt, np, nc;void add(int u, int v, int c) {    e[++cnt].v = v;    e[cnt].f = c;    e[cnt].nxt = g[u];    g[u] = cnt;    e[++cnt].v = u;    e[cnt].f = 0;    e[cnt].nxt = g[v];    g[v] = cnt;}void init() {    mem(g, 0);    cnt = 1;    int u, v, c;    char ch;    st = n, ed = n + 1;    for (int i = 0; i < m; i++) {        while(getchar() != '(') ;        scanf("%d,%d)%d", &u, &v, &c);        add(u, v, c);    }    for (int i = 0; i < np; i++) {        while(getchar() != '(') ;        scanf("%d)%d", &v, &c);        add(st, v, c);    }    for (int i = 0; i < nc; i++) {        while(getchar() != '(') ;        scanf("%d)%d", &u, &c);        add(u, ed, c);    }}int vis[maxn], dis[maxn], q[maxn];void bfs() {    mem(dis, 0);    int font = 0, rear = 1;    q[font] = st;    vis[st] = 1;    while(font < rear) {        int u = q[font++];        for (int i = g[u]; i; i = e[i].nxt) if (e[i].f && !vis[e[i].v]) {            q[rear++] = e[i].v;            vis[e[i].v] = 1;            dis[e[i].v] = dis[u] + 1;        }    }}int dfs(int u, int delta) {    if (u == ed) return delta;    else {        int ret = 0;        for (int i = g[u]; i && delta; i = e[i].nxt) if (e[i].f && dis[e[i].v] == dis[u] + 1) {            int dd = dfs(e[i].v, min(delta, e[i].f));            e[i].f -= dd;            e[i ^ 1].f += dd;            delta -= dd;            ret += dd;        }        return ret;    }}int maxflow() {    int ret = 0;    while(1) {        mem(vis, 0);        bfs();        if (!vis[ed]) return ret;        ret += dfs(st, INF);    }}int main (){    while(scanf("%d%d%d%d", &n, &np, &nc, &m) != EOF) {        init();        printf("%d\n", maxflow());    }    return 0;}

0 0