POJ 3469 Dinic （二元关系）

题意：

思路：

//By SiriusRen#include <queue>#include <cstdio>#include <cstring>using namespace std;#define mem(x,y) memset(x,y,sizeof(x))const int N=20005,M=880005;int first[N],vis[N],next[M],v[M],w[M],tot,xx,yy,zz,n,m,T,ans;void Add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}void add(int x,int y,int z){Add(x,y,z),Add(y,x,0);}bool tell(){    queue<int>q;q.push(0);    memset(vis,-1,sizeof(vis)),vis[0]=0;    while(!q.empty()){        int t=q.front();q.pop();        for(int i=first[t];~i;i=next[i])if(!~vis[v[i]]&&w[i])            vis[v[i]]=vis[t]+1,q.push(v[i]);    }return ~vis[T];}int zeng(int x,int y){    if(x==T)return y;    int r=0;    for(int i=first[x];~i&&y>r;i=next[i])        if(vis[v[i]]==vis[x]+1&&w[i]){            int t=zeng(v[i],min(y-r,w[i]));            w[i]-=t,w[i^1]+=t,r+=t;        }    if(!r)vis[x]=-1;    return r;}int main(){    mem(first,-1),scanf("%d%d",&n,&m),T=n+1;    for(int i=1;i<=n;i++)scanf("%d%d",&xx,&yy),add(0,i,xx),add(i,T,yy);    for(int i=1;i<=m;i++)scanf("%d%d%d",&xx,&yy,&zz),add(xx,yy,zz),add(yy,xx,zz);    while(tell())while(xx=zeng(0,0x3f3f3f3f))ans+=xx;    printf("%d\n",ans);}