BZOJ 1000 A+B Problem

Description
　　输入两个数字，输出它们之和
　　

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【题目分析】
虽然这是一道水题，但是暗藏玄机。网络流才是正解，8ms迅速出解。其解法之快，令人感叹！！！！

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【代码】

#include <cstdio>#include <iostream>#include <cstring>#define INF 1000000using namespace std;int u[501],v[501],h[501],ne[501],map[501],f[501];int n,m,en=0,S,T;void add(int a,int b,int r){    u[en]=a;v[en]=b;ne[en]=h[a];f[en]=r;h[a]=en++;    u[en]=b;v[en]=a;ne[en]=h[b];f[en]=0;h[b]=en++;}bool tell(){    memset(map,-1,sizeof map);    int que[501];    int head=0,tail=0;    que[tail++]=S;    map[S]=0;    while (head<tail){        int u=que[head++];        for (int i=h[u];i!=-1;i=ne[i]){            if (map[v[i]]==-1&&f[i]){                map[v[i]]=map[u]+1;                que[tail++]=v[i];            }        }    }    if (map[T]!=-1) return true;    else return false;}int zeng (int k,int now){    if (k==T) return now;    int r=0;    for (int i=h[k];i!=-1&&now>r;i=ne[i]){        if (map[k]+1==map[v[i]]&&f[i]!=0){            int t=zeng(v[i],min(now-r,f[i]));            f[i]-=t;f[i^1]+=t;r+=t;        }    }    if (!r) map[k]=-1;    return r;}int dinic(){    int r=0,t;    while (tell()) while (t=zeng(S,INF)) r+=t;    return r; }int main(){    memset(h,-1,sizeof h);    S=0;T=2;    int a,b;    scanf("%d%d",&a,&b);    add(S,1,a),add(S,1,b),add(1,T,a),add(1,T,b);    cout<<dinic()<<endl;}