HDU 4279 Food

Food

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 475 Accepted Submission(s): 222

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

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <list>

#define INT_INF 0x3fffffff
#define LL_INF 0x3fffffffffffffff
#define EPS 1e-12
#define MOD 1000000007
#define PI 3.141592653579798
#define N 5010
#define E 800080

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;

struct Edge
{
    int en,cap,flow,next;
} edge[E];
int head[N] , tot , now[N];
int source,sink,tot_num;
int pre[N] , dis[N] , gap[N];
queue<int> q;

void add_edge(int st,int en,int cap)
{
    edge[tot].en=en;
    edge[tot].cap=cap;
    edge[tot].flow=0;
    edge[tot].next=head[st];
    head[st]=tot++;

    edge[tot].en=st;
    edge[tot].cap=0;
    edge[tot].flow=0;
    edge[tot].next=head[en];
    head[en]=tot++;
}

void bfs()
{
    memset(dis,-1,sizeof(dis));
    memset(gap,0,sizeof(gap));
    while(!q.empty()) q.pop();
    gap[dis[sink]=0]=1; q.push(sink);
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        for(int i=head[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].en;
            if(edge[i].cap!=0 || dis[v]!=-1) continue;
            q.push(v);
            ++gap[dis[v]=dis[u]+1];
        }
    }
}

int sap()
{
    memset(pre,-1,sizeof(pre));
    bfs();
    memcpy(now,head,sizeof(head));
    int point=source , flow=0 , min_flow=INT_INF;
    while(dis[source]<tot_num)
    {
        bool fg=false;
        for(int i=now[point]; i!=-1; i=edge[i].next)
            if(edge[i].cap-edge[i].flow>0 && dis[point]==dis[edge[i].en]+1)
            {
                min_flow=min(min_flow,edge[i].cap-edge[i].flow);
                now[point]=i;
                pre[edge[i].en]=point;
                point=edge[i].en;
                if(point==sink)
                {
                    flow+=min_flow;
                    for(int u=source; u!=sink; u=edge[now[u]].en)
                    {
                        edge[now[u]].flow+=min_flow;
                        edge[now[u]^1].flow-=min_flow;
                    }
                    point=source;
                    min_flow=INT_INF;
                }
                fg=true;
                break;
            }
        if(fg) continue;
        if(--gap[dis[point]]==0) break;
        int Min=tot_num;
        for(int i=head[point]; i!=-1; i=edge[i].next)
            if(edge[i].cap-edge[i].flow>0 && Min>dis[edge[i].en])
            {
                Min=dis[edge[i].en];
                now[point]=i;
            }
        gap[dis[point]=Min+1]++;
        if(point!=source) point=pre[point];
    }
    return flow;
}

char s[400];
void build(int n,int f,int d)
{
    memset(head,-1,sizeof(head));
    tot=0;
    tot_num=f+d+n+n+2; source=0; sink=f+d+n+n+1;
    for(int i=1; i<=n; i++)
        add_edge(f+i,f+n+d+i,1);
    for(int i=1,val; i<=f; i++)  //讀入food
    {
        scanf("%d",&val);
        add_edge(source,i,val);
    }
    for(int i=1,val; i<=d; i++)  //讀入drink
    {
        scanf("%d",&val);
        add_edge(f+n+i,sink,val);
    }
    for(int i=1; i<=n; i++)
    {
        scanf("%s",s);
        for(int j=0; j<(int)strlen(s); j++)
        {
            if(s[j]=='N') continue;
            add_edge(j+1,f+i,1);
        }
    }
    for(int i=1; i<=n; i++)
    {
        scanf("%s",s);
        for(int j=0; j<(int)strlen(s); j++)
        {
            if(s[j]=='N') continue;
            add_edge(f+n+d+i,f+n+j+1,1);
        }
    }
}

int main()
{
    int n,f,d;
    while(scanf("%d%d%d",&n,&f,&d)!=EOF)
    {
        build(n,f,d);
        int ans=sap();
        printf("%d\n",ans);
    }
    return 0;
}