F - Food

https://vjudge.net/contest/402729#problem/F

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

Sponsor

代码:

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include<time.h>
using namespace std;
typedef long long ll;
const ll maxn=200000+10;
ll n,m,k,s,t,h[maxn],cur[maxn],cnt=1,vis[maxn];
ll F,D;
char x[maxn];
struct node{
    ll to,nt,w;
}e[maxn];
void add(ll u,ll v,ll w)
{
    e[++cnt]={v,h[u],w};
    h[u]=cnt;
}
queue<ll>q;
bool bfs()
{
    memset(vis,0,sizeof(vis));
    vis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        ll u=q.front();
        q.pop();
        cur[u]=h[u];
        for(ll i=h[u];i;i=e[i].nt)
        {
            ll v=e[i].to,w=e[i].w;
            if(w&&!vis[v])
            {
                vis[v]=vis[u]+1;
                q.push(v);
            }
        }
    }
    return vis[t];

}
ll dfs(ll u,ll flow)
{
    if(u==t)
        return flow;
    ll res=flow;
    for(ll i=cur[u];i;i=e[i].nt)
    {
        ll v=e[i].to,w=e[i].w;
        if(w&&vis[u]+1==vis[v])
        {
            ll now=dfs(v,min(res,w));
            if(!now)
                vis[v]=1;
            else
            {
                e[i].w-=now;
                e[i^1].w+=now;
                res-=now;
            }
        }
        if(!res)
        return flow;
    }
    return flow-res;
}
int main()
{
    while(~scanf("%lld %lld %lld",&n,&F,&D))
    {
        memset(cur,0,sizeof(cur));
        memset(h,0,sizeof(h));
        s=0;
        t=n*2+F+D+1;
        cnt=1;
        for(int j=1;j<=F;j++)
        {
            int u;
            scanf("%d",&u);
            add(0,j,u);
            add(j,0,0);
        }
        for(int j=1;j<=D;j++)
        {
            int u;
            scanf("%d",&u);
            add(j+F+n*2,t,u);
            add(t,j+F+n*2,0);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",x);
            int len=strlen(x);
            for(int j=0;j<len;j++)
            {
                if(x[j]=='Y')
                {
                    add(j+1,i+F,1);
                    add(i+F,j+1,0);
                }
            }

            add(i+F,i+F+n,1);
            add(i+F+n,i+F,0);
        }
        for(int i=1;i<=n;i++)
        {
            scanf("%s",x);
            int len=strlen(x);
            for(int j=0;j<len;j++)
            {
                if(x[j]=='Y')
                {
                    add(i+F+n,F+n*2+j+1,1);
                    add(F+n*2+j+1,i+F+n,0);
                }
            }
        }
        ll ans=0;
        while(!q.empty())
            q.pop();
        while(bfs())
        {
            ans+=dfs(s,0x7fffffff);
            //cout<<ans<<endl;
        }
        printf("%lld\n",ans);
    }

    return 0;
}