Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 76    Accepted Submission(s): 47

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  

Source
2012 ACM/ICPC Asia Regional Chengdu Online  

Recommend liuyiding      
很裸的最大流的题。和POJ 3182 很相似。
用SAP算法不会超时，比较高效。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=11000;
const int MAXM=405000;
const int INF=0x3f3f3f3f;
struct Node
{
    int from,to,next;
    int cap;
}edge[MAXM];
int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];
int n;
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
    edge[tol].from=u;
    edge[tol].to=v;
    edge[tol].cap=w;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].from=v;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}

void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    int que[MAXN];
    int front,rear;
    front=rear=0;
    dep[end]=0;
    que[rear++]=end;
    while(front!=rear)
    {
        int u=que[front++];
        if(front==MAXN)front=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap!=0||dep[v]!=-1)continue;
            que[rear++]=v;
            if(rear>=MAXN)rear=0;
            dep[v]=dep[u]+1;
            ++gap[dep[v]];
        }
    }
}
int SAP(int start,int end)
{
    int res=0;
    BFS(start,end);
    int cur[MAXN];
    int S[MAXN];
    int top=0;
    memcpy(cur,head,sizeof(head));
    int u=start;
    int i;
    while(dep[start]<n)
    {
        if(u==end)
        {
            int temp=INF;
            int inser;
            for(i=0;i<top;i++)
              if(temp>edge[S[i]].cap)
              {
                  temp=edge[S[i]].cap;
                  inser=i;
              }
            for(i=0;i<top;i++)
            {
                edge[S[i]].cap-=temp;
                edge[S[i]^1].cap+=temp;
            }
            res+=temp;
            top=inser;
            u=edge[S[top]].from;
        }
        if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
           break;
        for(i=cur[u];i!=-1;i=edge[i].next)
           if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
             break;
        if(i!=-1)
        {
            cur[u]=i;
            S[top++]=i;
            u=edge[i].to;
        }
        else
        {
            int min=n;
            for(i=head[u];i!=-1;i=edge[i].next)
            {
                if(edge[i].cap==0)continue;
                if(min>dep[edge[i].to])
                {
                    min=dep[edge[i].to];
                    cur[u]=i;
                }
            }
            --gap[dep[u]];
            dep[u]=min+1;
            ++gap[dep[u]];
            if(u!=start)u=edge[S[--top]].from;
        }
    }
    return res;
}

int g[2000][2000];


char str[1200];
int main()
{
    int start,end;

    int N,F,D;

    int u;
    int i;
    while(scanf("%d%d%d",&N,&F,&D)!=EOF)
    {
        memset(g,0,sizeof(g));
        init();
        n=F+D+2*N;
        start=0;
        end=n+1;
        for(i=1;i<=F;i++)
        {
            scanf("%d",&g[0][i]);
            addedge(0,i,g[0][i]);
        }

        for(i=F+2*N+1;i<=F+2*N+D;i++)
        {
            scanf("%d",&g[i][end]);
            addedge(i,end,g[i][end]);
        }

        for(i=1;i<=N;i++)
           addedge(F+2*i-1,F+2*i,1);


        for(i=1;i<=N;i++)
        {
            scanf("%s",&str);
            for(int j=0;j<F;j++)
            {
                if(str[j]=='Y')
                {
                    addedge(j+1,F+2*i-1,1);
                }
            }
        }
        for(i=1;i<=N;i++)
        {
            scanf("%s",&str);
            for(int j=0;j<D;j++)
            {
                if(str[j]=='Y')
                {
                  addedge(F+2*i,F+2*N+j+1,1);
                }
            }
        }
        start=0;
        end=n+1;
        n+=2;
        printf("%d\n",SAP(start,end));
    }
    return 0;
}