hdu4292 Food 网络最大流 拆点

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 31 1 11 1 1YYNNYYYNYYNYYNYYYNYYNNNY

 

Sample Output

3

      这道题比较简单只需将每个人拆分成两点即可，但数据量很大，最开始我用熟悉的Ford-Fulkerson 算法提交 ，结果为TLE，改为Dinic ，运行时间为375MS。

Dinic算法

#include <cstdio>#include <cstring>#define inf 0x3f3f3f3f#define maxn 810int s, t, n, N, F, D;char temp[210];int c[maxn][maxn], dep[maxn]; int queue[maxn];void init(){    s = 0; t = N + N + F + D + 1; n = t + 1;    memset(c, 0, sizeof(c));    for(int i = 1; i <= F; i++)        scanf("%d", &c[s][i]);    for(int i = 1; i <= D; i++)        scanf("%d", &c[F+N+N+i][t]);    for(int i = 1; i <= N; i++)    c[F+i][F+N+i] = 1;    for(int i = 1; i <= N; i++){        scanf("%s", temp + 1);        for(int j = 1; j <= F; j++)            if(temp[j] == 'Y')                  c[j][F+i] = 1;    }       for(int i = 1; i <= N; i++){        scanf("%s", temp + 1);        for(int j = 1; j <= D; j++)            if(temp[j] == 'Y')                  c[F+N+i][F+N+N+j] = 1;    }}bool bfs(){    int f = 0, r = 0, u, v;      memset(dep, -1, sizeof(dep));      queue[r++] = s;      dep[s] = 0;       while(f < r){          u = queue[f++];          for(v = 0; v < n; v++)              if(c[u][v] && dep[v] < 0){                  dep[v] = dep[u] + 1;                  queue[r++] = v;              }              if(u == t) break;        }      return dep[t] < 0; }int Dinic(){    int k, u, v, num, res = 0, top, tag;    while(1){        if( bfs() ) break;         top = 0;        u = s;        while(1){            if(u == t){                  queue[top++] = t;                for(k = 0, num = inf; k < top - 1; k++)                    if(c[ queue[k] ][ queue[k+1] ] < num){                        tag = k;                         num = c[ queue[k] ][ queue[k+1] ];                    }                for(k = 0; k < top - 1; k++){                    c[ queue[k] ][ queue[k+1] ] -= num;                    c[ queue[k+1] ][ queue[k] ] += num;                }                res += num;                top = tag;                 u = tag;            }            for(v = 0; v < n; v++)                 if(c[u][v] && dep[u] + 1 == dep[v]) break;             if(v<n) {queue[top++] = u; u = v;}             else{                if(top == 0) break;                dep[u] = -1;                u = queue[--top];             }        }    }    printf("%d\n", res);}int main(){while(~scanf("%d %d %d", &N, &F, &D)){        init();        Dinic();}    return 0;}