【最大流问题 dinic算法】Internet Bandwidth, ACM/ICPC World Finals 2000, UVa820

来源：互联网发布：树莓派网络设置编辑：程序博客网时间：2024/06/06 12:45

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>#include<queue>using namespace std;int n,s,t,m;const int MAXM = 160000;  const int MAXN = 400;  const int INF = 0x3f3f3f3f; struct edge{int to,cap,next;}e[MAXM];int head[MAXN],level[MAXN];int eid;void adde(int u,int v,int c){e[eid].to=v;e[eid].cap=c;e[eid].next=head[u];head[u] = eid++;}int bfs(){memset(level,-1,sizeof(level));queue<int> q;level[s] = 0;q.push(s);while(!q.empty()){int u = q.front();q.pop();for(int i=head[u];~i;i=e[i].next){int v = e[i].to;if(e[i].cap>0&&level[v]==-1){level[v] = level[u]+1;q.push(v);}}}return level[t]!=-1;}int dfs(int u,int f){if(u == t)return f;int tmp;for(int i=head[u];~i;i=e[i].next){int v = e[i].to;if(e[i].cap>0&&level[v]==level[u]+1){tmp = dfs(v,min(f,e[i].cap));if(tmp>0){e[i].cap -= tmp;e[i^1].cap += tmp;return tmp;}}}level[u] = -1;return 0;}int Dinic(){int ans = 0,tmp;while(bfs())while((tmp = dfs(s,INF))>0)ans +=tmp;return ans;}int main(void){int a,b,c;int kase = 0;while(scanf("%d",&n),n){memset(head,-1,sizeof(head));scanf("%d%d%d",&s,&t,&m);for(int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);adde(a,b,c);adde(b,a,c);}printf( "Network %d\nThe bandwidth is %d.\n\n",++kase, Dinic() );}return 0;}

朴素的EK增广路要TLE。

SAP、Dinic可过。

1 0