HDU 4292 Food 第37届ACM/ICPC 成都赛区网络赛1005题 （最大流）

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;

}