HDU4292 Food —— 最大流 + 拆点

题目链接：https://vjudge.net/problem/HDU-4292

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6457    Accepted Submission(s): 2197

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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题解：

此题（POJ3281 Dining）的变形，只不过是把超级汇点连向食物的边改为库存，饮料连向超级汇点的边改为库存。

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int mod = 1e9+;
 const int MAXN = 1e3+;

 int maze[MAXN][MAXN];
 int gap[MAXN], dis[MAXN], pre[MAXN], cur[MAXN];
 int flow[MAXN][MAXN];

 int sap(int start, int end, int nodenum)
 {
     memset(cur, , sizeof(cur));
     memset(dis, , sizeof(dis));
     memset(gap, , sizeof(gap));
     memset(flow, , sizeof(flow));
     int u = pre[start] = start, maxflow = , aug = INF;
     gap[] = nodenum;

     while(dis[start]<nodenum)
     {
         loop:
         for(int v = cur[u]; v<nodenum; v++)
         if(maze[u][v]-flow[u][v]> && dis[u] == dis[v]+)
         {
             aug = min(aug, maze[u][v]-flow[u][v]);
             pre[v] = u;
             u = cur[u] = v;
             if(v==end)
             {
                 maxflow += aug;
                 for(u = pre[u]; v!=start; v = u, u = pre[u])
                 {
                     flow[u][v] += aug;
                     flow[v][u] -= aug;
                 }
                 aug = INF;
             }
             goto loop;
         }

         int mindis = nodenum-;
         for(int v = ; v<nodenum; v++)
             if(maze[u][v]-flow[u][v]> && mindis>dis[v])
             {
                 cur[u] = v;
                 mindis = dis[v];
             }
         if((--gap[dis[u]])==) break;
         gap[dis[u]=mindis+]++;
         u = pre[u];
     }
     return maxflow;
 }

 char str[MAXN];
 int main()
 {
     int N, F, D;
     while(scanf("%d%d%d",&N,&F,&D)!=EOF)
     {
         int start = , end = F+D+*N+;
         memset(maze, , sizeof(maze));
         for(int i = ; i<=F; i++)
         {
             int Food_supply;
             scanf("%d", &Food_supply);
             maze[start][i] = Food_supply;
         }
         for(int i = ; i<=D; i++)
         {
             int Drink_supply;
             scanf("%d", &Drink_supply);
             maze[F+i][end] = Drink_supply;
         }
         for(int i = ; i<=N; i++)
         {
             scanf("%s", str+);
             for(int j = ; j<=F; j++)
                 if(str[j]=='Y')
                     maze[j][F+D+i] = ;
         }
         for(int i = ; i<=N; i++)
         {
             scanf("%s", str+);
             for(int j = ; j<=D; j++)
                 if(str[j]=='Y')
                     maze[F+D+N+i][F+j] = ;
         }
         for(int i = ; i<=N; i++)
             maze[F+D+i][F+D+N+i] = ;

         cout<< sap(start, end, F+D+*N+) <<endl;
     }
 }