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 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY

 

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;
}