HDU 4292 Food 最大流

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3658    Accepted Submission(s): 1246


Problem Description
　　You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many people as possible.
　　The issue comes up as people in your college are more and more difficult to serve with meal: They eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
　　You have prepared F (1 <= F <= 200) kinds of food and D (1 <= D <= 200) kinds of drink. Each kind of food or drink has certain amount, that is, how many people could this food or drink serve. Besides, You know there’re N (1 <= N <= 200) people and you too can tell people’s personal preference for food and drink.
　　Back to your goal: to serve as many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
 
Input
　　There are several test cases.
　　For each test case, the first line contains three numbers: N,F,D, denoting the number of people, food, and drink.
　　The second line contains F integers, the ith number of which denotes amount of representative food.
　　The third line contains D integers, the ith number of which denotes amount of representative drink.
　　Following is N line, each consisting of a string of length F. �e jth character in the ith one of these lines denotes whether people i would accept food j. “Y” for yes and “N” for no.
　　Following is N line, each consisting of a string of length D. �e jth character in the ith one of these lines denotes whether people i would accept drink j. “Y” for yes and “N” for no.
　　Please process until EOF (End Of File).
 
Output
　　For each test case, please print a single line with one integer, the maximum number of people to be satisfied.
 
Sample Input
4 3 3 1 1 1 1 1 1 YYN NYY YNY YNY YNY YYN YYN NNY
 
Sample Output
3
 
Source
2012 ACM/ICPC Asia Regional Chengdu Online
题解：
   建图：S-F-人-人-D-T
代码
//з&#1111;еп:1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef __int64 ll;
using namespace std;
const int inf = 10000000;
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

//*******************************
namespace NetFlow
{
    const int MAXN=100000,MAXM=1000000,inf=1e9;
    struct Edge
    {
        int v,c,f,nx;
        Edge() {}
        Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
    } E[MAXM];
    int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
    void init(int _n)
    {
        N=_n,sz=0; memset(G,-1,sizeof(G[0])*N);
    }
    void link(int u,int v,int c)
    {
        E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;
        E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;
    }
    int ISAP(int S,int T)
    {//S -> T
        int maxflow=0,aug=inf,flag=false,u,v;
        for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0;
        for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
        {
            for (int &it=cur[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+1)
                {
                    if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
                    pre[v]=u,u=v; flag=true;
                    if (u==T)
                    {
                        for (maxflow+=aug;u!=S;)
                        {
                            E[cur[u=pre[u]]].f+=aug;
                            E[cur[u]^1].f-=aug;
                        }
                        aug=inf;
                    }
                    break;
                }
            }
            if (flag) continue;
            int mx=N;
            for (int it=G[u];~it;it=E[it].nx)
            {
                if (E[it].c>E[it].f&&dis[E[it].v]<mx)
                {
                    mx=dis[E[it].v]; cur[u]=it;
                }
            }
            if ((--gap[dis[u]])==0) break;
            ++gap[dis[u]=mx+1]; u=pre[u];
        }
        return maxflow;
    }
    bool bfs(int S,int T)
    {
        static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N);
        dis[S]=0; Q[0]=S;
        for (int h=0,t=1,u,v,it;h<t;++h)
        {
            for (u=Q[h],it=G[u];~it;it=E[it].nx)
            {
                if (dis[v=E[it].v]==-1&&E[it].c>E[it].f)
                {
                    dis[v]=dis[u]+1; Q[t++]=v;
                }
            }
        }
        return dis[T]!=-1;
    }
    int dfs(int u,int T,int low)
    {
        if (u==T) return low;
        int ret=0,tmp,v;
        for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
        {
            if (dis[v=E[it].v]==dis[u]+1&&E[it].c>E[it].f)
            {
                if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
                {
                    ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp;
                }
            }
        }
        if (!ret) dis[u]=-1; return ret;
    }
    int dinic(int S,int T)
    {
        int maxflow=0,tmp;
        while (bfs(S,T))
        {
            memcpy(cur,G,sizeof(G[0])*N);
            while (tmp=dfs(S,T,inf)) maxflow+=tmp;
        }
        return maxflow;
    }
}
using namespace NetFlow;
int main()
{
 int n,f,d;
 int fs,ds;
    while(scanf("%d%d%d",&n,&f,&d)!=EOF){
            init(900);
        for(int i=1;i<=f;i++)
        {
            fs=read();
            link(801,i,fs);
        }
        for(int i=1;i<=d;i++)
        {
            ds=read();
            link(i+600,802,ds);
        }
        char ch[201];
        for(int i=1;i<=n;i++)
        {
            scanf("%s",ch);
            for(int j=0;j<f;j++)
            {
                if(ch[j]=='Y'){
                    link(j+1,200+i,1);
                }
            }
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",ch);
            for(int j=0;j<d;j++)
            {
                if(ch[j]=='Y'){
                   link(400+i,j+1+600,1);
                }
            }
        }//
        for(int i=1;i<=n;i++)link(200+i,400+i,1);
        printf("%d
",ISAP(801,802));//cout<<1111<<endl;
    }
    return 0;
}
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原文地址：https://www.cnblogs.com/zxhl/p/4738852.html