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.

　　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, please print a single line with one integer, the maximum number of people to be satisfied.

2012 ACM/ICPC Asia Regional Chengdu Online

liuyiding

 

题解：比赛的时候没看 现在看了一下很简单的网络流问题 开一个F+D+2*N+2的一个网络 源点与各个食物点F相连 权值为各个食物的权值 再将食物点按照字符条件 将食物点与前N个顾客点对应条件一一相连 权值为1 为了保证顾客不会获得两种或者两种以上的食物或饮料 将前N个顾客点与后N个顾客点相连 权值为1 再将后N个顾客点与饮料点相连 权值为1 再将饮料点与汇点相连 就可以找最大值了 这题起初我还考虑有的顾客只要食物 或者有的顾客只要饮料 但是这种测试数据里好像没有 所以我程序也没加

 

#include <iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int oo=1e9;
/**oo表示无穷大*/
const int mm=111111111;
/**mm表示边的最大数量，记住要是原图的两倍，在加边的时候都是双向的*/
const int mn=9999999;
/**mn表示点的最大数量*/
int node,src,dest,edge;
/**node表示节点数，src表示源点，dest表示汇点，edge统计边数*/
int ver[mm],flow[mm],next[mm];
/**ver 边指向的节点，flow 边的容量 ，next 链表的下一条边*/
int head[mn],work[mn],dis[mn],q[mn];
/**head 节点的链表头，work 用于算法中的临时链表头，dis 计算距离*/

/**初始化链表及图的信息*/
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;
}
/**增加一条u到v容量为c的边*/
void addedge(int u,int v,int c)
{
    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
/**广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束*/
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=next[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                /**这条边必须有剩余容量*/
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)return 1;
            }
    return 0;
}
/**寻找可行流的增广路算法，按节点的距离来找，加快速度*/
int Dinic_dfs(int u,int exp)
{
    if(u==dest)return exp;
    /**work是临时链表头，这里用i引用它，这样寻找过的边不再寻找*/
    for(int &i=work[u],v,tmp; i>=0; i=next[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            /**正反向边容量改变*/
            return tmp;
        }
    return 0;
}
/**求最大流，直到没有可行流*/
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()
{
    int n,f,d,ef,ed,ans;
    string sf,sd;
    while(cin>>n>>f>>d)
    {
        prepare(f+d+2*n+2,0,f+d+2*n+1);
        for(int i=1; i<=f; i++)
        {
            cin>>ef;
            addedge(src,i,ef);
        }
        for(int i=1; i<=d; i++)
        {
            cin>>ed;
            addedge(f+n+n+i,dest,ed);
        }
        for(int i=1; i<=n; i++)
        {
            addedge(f+i,f+n+i,1);
            cin>>sf;
            for(int j=0; j<f; j++)
                if(sf[j]=='Y')
                    addedge(j+1,f+i,1);
        }
        for(int i=1; i<=n; i++)
        {
            cin>>sd;
            for(int j=0; j<d; j++)
                if(sd[j]=='Y')
                    addedge(f+n+i,f+n+n+j+1,1);
        }
        ans=Dinic_flow();
        cout<<ans<<endl;
    }

    return 0;
}