HDU 4292 FOOD 2012 ACM/ICPC Asia Regional Chengdu Online

　　　　　　　　　　　　　　　　　　　　　　Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3399    Accepted Submission(s): 1141

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

 

题意：

一个餐馆有F种食物和D种饮料，每种食物和饮料都有固定的量，和N个顾客。

每个顾客会同时点饮料和食物，而且只会点他们喜欢的食物和饮料。

若顾客没有得到他喜欢的饮料和食物，则这位顾客不满意，会马上离开。

问要怎么分配食物和饮料给顾客，使得尽量多的顾客满意。

 

网络流

建图：S-食物F—顾客N—顾客N—饮料D—T

S和F，边的容量为食物的量

F和N，若顾客喜欢该种食物，则他们之间有一条容量为1的边

N和N，相应的点加边，容量为1（保证了每个顾客只点一份饮料+食物）

N和D，若顾客喜欢该种饮料，则加边，容量为1

D和T，边的容量为饮料的量

然后跑一遍dinic，即可。

问题：

1.刚开始，数组edge忘记初始化了

2.初始化的时候从下标1开始，而下标0（即源点）忘记初始化了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

const int MAXN=810;
const int INF=0x3f3f3f3f;

struct Edge
{
    int to,cap,rev;
};
vector<Edge>edge[MAXN];
int level[MAXN];
int iter[MAXN];
const int S=0;
int T;
char str[205];

void addedge(int u,int v,int cap)
{
    edge[u].push_back((Edge){v,cap,edge[v].size()});
    edge[v].push_back((Edge){u,0,edge[u].size()-1});
}

void bfs()
{
    memset(level,-1,sizeof(level));
    queue<int>que;
    while(!que.empty())
        que.pop();

    level[S]=0;
    que.push(S);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        for(int i=0;i<edge[u].size();i++)
        {
            Edge &e=edge[u][i];
            if(e.cap>0&&level[e.to]<0)
            {
                level[e.to]=level[u]+1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int u,int f)
{
    if(u==T)
        return f;
    for(int &i=iter[u];i<edge[u].size();i++)
    {
        Edge &e=edge[u][i];
        if(e.cap>0&&level[e.to]>level[u])
        {
            int d=dfs(e.to,min(f,e.cap));
            if(d>0)
            {
                e.cap-=d;
                edge[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}

int max_flow()
{
    int flow=0;
    while(true)
    {
        bfs();
        if(level[T]<0)
            return flow;
        memset(iter,0,sizeof(iter));
        int f;
        while(f=dfs(S,INF)>0)
        {
            flow+=f;
        }
    }
}

int main()
{
    int N,F,D;
    while(~scanf("%d%d%d",&N,&F,&D))
    {
        for(int i=0;i<MAXN;i++)
            edge[i].clear();

        T=F+2*N+D+1;
        for(int i=1;i<=F;i++)
        {
            int w;
            scanf("%d",&w);
            addedge(S,i,w);
        }
        for(int i=1;i<=D;i++)
        {
            int w;
            scanf("%d",&w);
            addedge(F+2*N+i,T,w);
        }
        for(int i=1;i<=N;i++)
        {
            addedge(i+F,i+N+F,1);
        }
        for(int i=1;i<=N;i++)
        {
            scanf("%s",str);
            for(int j=1;j<=F;j++)
            {
                if(str[j-1]=='Y')
                    addedge(j,F+i,1);
            }
        }
        for(int i=1;i<=N;i++)
        {
            scanf("%s",str);
            for(int j=1;j<=D;j++)
            {
                if(str[j-1]=='Y')
                    addedge(F+N+i,F+2*N+j,1);
            }
        }

        int flow=max_flow();

        printf("%d
",flow);
    }

    return 0;
}