HDU 4292 Food

有两个资源供应，而需求只有一个，把需求拆点，限制容量，放中间，其中一个点跟资源1相连，另外一个点与资源2相连，跑最大流。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 1<<30
#define maxn 10100
#define maxm 1000000
using namespace std;

int v[maxm],next[maxm],w[maxm];
int first[maxn],d[maxn],work[maxn],q[maxn];
int e,S,T;

void init(){
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,int c){
    //printf("add:%d to %d,cap = %d
",a,b,c);
    v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
    v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++;
}

int bfs(){
    int rear = 0;
    memset(d,-1,sizeof(d));
    d[S] = 0;q[rear++] = S;
    for(int i = 0;i < rear;i++){
        for(int j = first[q[i]];j != -1;j = next[j])
            if(w[j] && d[v[j]] == -1){
                d[v[j]] = d[q[i]] + 1;
                q[rear++] = v[j];
                if(v[j] == T)   return 1;
            }
    }
    return 0;
}

int dfs(int cur,int a){
    if(cur == T)    return a;
    for(int &i = work[cur];i != -1;i = next[i]){
        if(w[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i],min(a,w[i]))){
                w[i] -= t;w[i^1] += t;
                return t;
            }
    }
    return 0;
}

int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(work,first,sizeof(first));
        while(int t = dfs(S,INF))   ans += t;
    }
    return ans;
}

int main()
{
    int N,F,D;
    while(scanf("%d%d%d",&N,&F,&D) == 3){
        S = 0,T = F+D+N+N+1;
        init();
        int tmp;
        char str[210];
        for(int i = 1;i <= F;i++){
            scanf("%d",&tmp);
            add_edge(S,i,tmp);
        }
        for(int i = 1;i <= D;i++){
            scanf("%d",&tmp);
            add_edge(F+i,T,tmp);
        }
        for(int i = 1;i <= N;i++){
            scanf("%s",str);
            for(int j = 0;j < F;j++){
                if(str[j] == 'Y')
                    add_edge(j+1,F+D+i,1);
            }
        }
        for(int i = 1;i <= N;i++){
            scanf("%s",str);
            for(int j = 0;j < D;j++){
                if(str[j] == 'Y')
                    add_edge(F+D+N+i,F+j+1,1);
            }
        }
        for(int i = 1;i <= N;i++)
            add_edge(F+D+i,F+D+N+i,1);
        int ans = dinic();
        printf("%d
",ans);
    }
    return 0;
}