HDU4292 Food(基础最大流)

Food

Time Limit: 1000 MS Memory Limit: 32768 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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 31 1 11 1 1YYNNYYYNYYNYYNYYYNYYNNNY

Sample Output

3

Source

2012 ACM/ICPC Asia Regional Chengdu Online

题意：有n个人，F种食物（各Fi个），D种饮料（各Di个）。各人都有喜爱的食物和饮料，每个人想要一个食物和一个饮料，问最大匹配是多少。最大流模板，注意让人与人连一次就行。G：yuan-food-ren-ren-drink-hui。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>#include<queue>using namespace std;struct edge{   int to,cap,rev;   edge(int a=0,int b=0,int c=0):to(a),cap(b),rev(c){}};const int maxn = 802;const int inf = 2147483647;vector<edge> G[maxn];int iter[maxn];int level[maxn];void Add(int f,int t,int c){    G[f].push_back(edge{t,c,G[t].size()  } );    G[t].push_back(edge{f,0,G[f].size()-1} );}void bfs(int s){    memset(level,-1,sizeof(level));    queue<int> que;    level[s] = 0;    que.push(s);    while(!que.empty()){        int v = que.front();        que.pop();        for(int i = 0;i < G[v].size();i++){            edge &e = G[v][i];            if(e.cap > 0 && level[e.to] == -1){                level[e.to] = level[v]+1;                que.push(e.to);            }        }    }}int dfs(int v,int t,int f){    if(v == t) return f;    for(int &i = iter[v];i < G[v].size();i++){        edge &e = G[v][i];        if(e.cap > 0 && level[v] < level[e.to]){            int d = dfs(e.to,t,min(e.cap,f));            if(d > 0){                e.cap -= d;                G[e.to][e.rev].cap += d;                return d;            }        }    }    return 0;}int mflow(int s,int t,int fmax){    int flow = 0;    while(true){        bfs(s);        if(level[t] == -1) return flow;        memset(iter,0,sizeof(iter));        int f = 0;        while((f = dfs(s,t,inf)) > 0)            flow += f;    }}void init(){    for(int i = 0;i < maxn;i++){        G[i].clear();    }}int main(){    int n,food,drink;    while(scanf("%d%d%d",&n,&food,&drink) != EOF){        init();        int yuan = 0;        int hui = n*2+food+drink+1;        int num;        for(int i = 1;i <= n;i++){            Add(food+i,food+n+i,1);        }        for(int i = 1;i <= food;i++){            scanf("%d",&num);            Add(yuan,i,num);        }        for(int i = 1;i <= drink;i++){            scanf("%d",&num);            Add(n*2+food+i,hui,num);        }        char chi;        for(int i = 1;i <= n;i++){            getchar();            for(int j = 1;j <= food;j++){                scanf("%c",&chi);                if(chi == 'Y'){                    Add(j,food+i,1);                }            }        }        for(int i = 1;i <= n;i++){            getchar();            for(int j = 1;j <= drink;j++){                scanf("%c",&chi);                if(chi == 'Y'){                    Add(n+food+i,n*2+food+j,1);                }            }        }        int ans = mflow(yuan,hui,inf);        printf("%d\n",ans);    }    return 0;}