UVA - 10330 Power Transmission 最大流问题

来源：互联网发布：幼儿学前教育软件编辑：程序博客网时间：2024/06/05 04:28

题目大意：发电厂要求把电传输到另一个城市，有N个转换器和发电厂和城市构成了一幅图，每个转换器都有个容量，每条连线之间都有权值，边是有向边

解题思路：将点拆分成变，边的权值为点的容量，2*i表示该点的入点，2*i^1表示该点的出点，然后再构成图即可

#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;#define maxn 220#define maxm 81000#define INF 0x3f3f3f3fint u[maxm],v[maxm],next[maxm],flow[maxm];int head[maxn],d[maxn],e,N,q[maxm];void add(int a, int b, int len) {u[e] = a;v[e] = b;flow[e] = len;next[e] = head[a];head[a] = e;e++;}void init() {memset(head,-1,sizeof(head));e = 0;int M,D,B,w,st,ed;for(int i = 1; i <= N; i++) {scanf("%d",&w);add(2*i,2*i^1,w);add(2*i^1,2*i,0);}scanf("%d",&M);for(int i = 0; i < M; i++) {scanf("%d%d%d",&st,&ed,&w);add(st*2^1,2*ed,w);add(2*ed,st*2^1,0);}scanf("%d%d",&B,&D);for(int i = 0; i < B; i++) {scanf("%d",&w);add(0,2*w,INF);add(2*w,0,0);}for(int i = 0; i < D; i++) {scanf("%d",&w);add(2*w^1,1,INF);add(1,2*w^1,0);}}int bfs() {memset(d,-1,sizeof(d));d[0] = 0;int rear = 0;q[rear++] = 0;for(int i = 0; i < rear; i++)for(int j = head[q[i]]; j != -1; j = next[j]) {if(d[v[j]] == -1 && flow[j])  {d[v[j]] = d[q[i]] + 1;if(v[j] == 1)return 1;q[rear++] = v[j];}}return 0;}int dfs(int cur,int a) {if(cur == 1)return a;int i;for(i = head[cur]; i != -1; i = next[i]) {if(d[v[i]] == d[cur] + 1 && flow[i]) if(int t = dfs(v[i],a < flow[i] ?a:flow[i])) {flow[i] -= t;flow[i^1] += t;return t;}}if(i == -1)d[cur] = -1;return 0;}int EK() {int ans = 0,t;while(bfs()) {while(t = dfs(0,INF))ans += t;}return ans;}int main() {while(scanf("%d",&N) != EOF) {init();int ans = EK();printf("%d\n",ans);}return 0;}

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