hdu4292 网络流 —最大流

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4364    Accepted Submission(s): 1475

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 31 1 11 1 1YYNNYYYNYYNYYNYYYNYYNNNY

 

Sample Output

3
题意：有N头牛，F种食物，D种饮料；
有2*N组输入，第一组1-N表示第i头牛对F种食物的喜欢与否，Y表示喜欢没N表示不喜欢，第二组1-N表示对饮料的喜欢与否；
问最多有几头牛可以吃上自己喜欢的食物，并喝上自己喜欢的饮料；
最大流算法，  S——F（food）——N（n头牛）——N（n头牛）——D（饮料）——T；S表示超级源点，T表示超级汇点；
#include <iostream>#include <queue>#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <stdlib.h>const int inf=0x3f3f3f3f;const int MAXN=1e5+10;const int MAXM=400010;using namespace std;struct Node{    int to,next,cap;}edge[MAXM];int tol;int head[MAXN];int gap[MAXN],dis[MAXN],pre[MAXN],cur[MAXN];void init(){    tol=0;    memset(head,-1,sizeof(head));}void addedge(int u,int v,int w,int rw=0){    edge[tol].to=v;edge[tol].cap=w;edge[tol].next=head[u];head[u]=tol++;    edge[tol].to=u;edge[tol].cap=rw;edge[tol].next=head[v];head[v]=tol++;}int sap(int start,int end0,int nodenum){    memset(dis,0,sizeof(dis));    memset(gap,0,sizeof(gap));    memcpy(cur,head,sizeof(head));    int u=pre[start]=start,maxflow=0,aug=-1;    gap[0]=nodenum;    while(dis[start]<nodenum)    {        loop:        for(int  &i=cur[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(edge[i].cap&&dis[u]==dis[v]+1)            {                if(aug==-1||aug>edge[i].cap)                    aug=edge[i].cap;                pre[v]=u;                u=v;                if(v==end0)                {                    maxflow+=aug;                    for(u=pre[u];v!=start;v=u,u=pre[u])                    {                        edge[cur[u]].cap-=aug;                        edge[cur[u]^1].cap+=aug;                    }                    aug=-1;                }                goto loop;            }        }        int mindis=nodenum;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(edge[i].cap&&mindis>dis[v])            {                cur[u]=i;                mindis=dis[v];            }        }        if((--gap[dis[u]])==0)break;        gap[dis[u]=mindis+1]++;        u=pre[u];    }    return maxflow;}int S,T;char str[555];int N,F,D;int food[MAXN],dirk[MAXN];int main(){    while(scanf("%d%d%d",&N,&F,&D)!=-1)    {        S=0;        T=F+D+N*2+1;        init();        for(int i=1; i<=F; i++)        {             scanf("%d",&food[i]);            // cin>>food[i];             addedge(S,i,food[i]);        }        for(int i=1; i<=D; i++)        {            scanf("%d",&dirk[i]);             // cin>>dirk[i];               addedge(i+F+N*2,T,dirk[i]);        }        for(int i=1; i<=N; i++)            addedge(i+F,i+N+F,1);        for(int i=1; i<=N; i++)        {            scanf("%s",str);            //cin>>str;            int l=strlen(str);            for(int j=0; j<l; j++)            {                if(str[j]=='Y')                    addedge(j+1,F+i,1);            }        }        for(int i=1; i<=N; i++)        {            scanf("%s",str);            //cin>>str;            int l=strlen(str);            for(int j=0; j<l; j++)            {                if(str[j]=='Y')                    addedge(F+N+i,F+N*2+j+1,1);            }        }        printf("%d\n",sap(S,T,T+1));        //cout<<max_flow(0,hui)<<endl;    }    return 0;}