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.
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).
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
将近一个小时的成果==早上5点多睡不着突然想到的做法居然真的有bug 唉
题意：为顾客服务食物和饮料，每个人喜欢某些食物和某些饮料，上了她不喜欢的东西他就走了，不同食物饮料有特定的供应量，问最多多少人满意==最初的想法当然是左边加一个源点，右边加一个汇点，根据已知条件连边，然而需要注意的是：刚开始不理解为啥拆点，想一个极端的例子，只有一个人，所有食物和饮料他都喜欢，但是最大流一定只是1！！所以要通过拆点限制人
#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mm=1000000;
const int mn=22222;
const int oo=1000000000;
int node,src,dest,edge;
int reach[mm],flow[mm],nxt[mm];
int head[mn],work[mn],dis[mn],q[mn];
inline int min(int a,int b)
{
return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
node=_node,src=_src,dest=_dest;
for(int i=0;i<node;++i)head[i]=-1;
edge=0;
}
inline void addedge(int u,int v,int c1,int c2)
{
reach[edge]=v,flow[edge]=c1,nxt[edge]=head[u],head[u]=edge++;
reach[edge]=u,flow[edge]=c2,nxt[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0;i<node;++i)dis[i]=-1;
dis[q[r++]=src]=0;
for(l=0;l<r;++l)
for(i=head[u=q[l]];i>=0;i=nxt[i])
if(flow[i]&&dis[v=reach[i]]<0)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)return 1;
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)return exp;
for(int &i=work[u],v,tmp;i>=0;i=nxt[i])
if(flow[i]&&dis[v=reach[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
{
flow[i]-=tmp;
flow[i^1]+=tmp;
return tmp;
}dis[u]--;
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while(Dinic_bfs())
{
for(i=0;i<node;++i)work[i]=head[i];
while(delta=Dinic_dfs(src,oo))ret+=delta;
}
return ret;
}
int main()
{
// freopen("cin.txt","r",stdin);
int n,d,f;
while(~scanf("%d%d%d",&n,&f,&d))
{
prepare(n*2+f+d+2,0,f+n*2+d+1);
for(int i=1;i<=f;i++)
{
int a;
scanf("%d",&a);
addedge(src,i,a,0);
}
for(int i=f+n*2+1;i<=f+n*2+d;i++)
{
int a;
scanf("%d",&a);
addedge(i,dest,a,0);
}
for(int i=f+1;i<=f+n;i++)
addedge(i,i+n,1,0);
char str[300];
for(int i=f+1;i<=f+n;i++)
{
scanf("%s",str);
int len=strlen(str);
for(int j=0;j<len;j++)
if(str[j]=='Y')
addedge(j+1,i,1,0);
}
for(int i=f+1+n;i<=f+n*2;i++)
{
scanf("%s",str);
int len=strlen(str);
for(int j=0;j<len;j++)
if(str[j]=='Y')
addedge(i,j+1+f+n*2,1,0);
}
printf("%d\n",Dinic_flow());
}
return 0;
}
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