hdu 1532——Drainage Ditches

Dinic演算法 數組版

#include<cstdio>#include<iostream>#include<cstring>#include<queue>using namespace std;#define INF 100001int G[205][205],dis[205];int cur[205];int n,m;int Min(int a,int b){    if(a<b)        return a;    return b;}int bfs(){    memset(dis,-1,sizeof(dis));    queue<int> q;    q.push(1);    dis[1]=0;    while(!q.empty())    {        int u=q.front();        q.pop();        for(int v=1;v<=n;v++)            if(dis[v]==-1&&G[u][v]>0)            {                dis[v]=dis[u]+1;                q.push(v);            }    }    if(dis[n]==-1)        return 0;    return 1;}int dfs(int u,int flow){    int d;    int f=0;    if(u==n||flow==0)        return flow;    for(int &v=cur[u];v<=n;v++)        if(dis[v]==dis[u]+1&&G[u][v]>0&&(d=dfs(v,Min(flow,G[u][v]))))        {            G[u][v]-=d;            G[v][u]+=d;            flow-=d;            f+=d;        }        return f;}int Dinic(){    int d;    int ans=0;    while(bfs())    {    for(int i=1;i<=n;i++)    cur[i]=1;        while((d=dfs(1,INF)))            ans+=d;    }    return ans;}int main(){    int u,v,c;    while(~scanf("%d%d",&m,&n))    {        memset(G,0,sizeof(G));            for(int i=1;i<=m;i++)        {            scanf("%d%d%d",&u,&v,&c);            G[u][v]+=c;        }        printf("%d\n",Dinic());    }    return 0;    }