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

 

 

Recommend

liuyiding

 

 

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+7;
const int MAXN = 1e3+10;

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, 0, sizeof(cur));
    memset(dis, 0, sizeof(dis));
    memset(gap, 0, sizeof(gap));
    memset(flow, 0, sizeof(flow));
    int u = pre[start] = start, maxflow = 0, aug = INF;
    gap[0] = nodenum;

    while(dis[start]<nodenum)
    {
        loop:
        for(int v = cur[u]; v<nodenum; v++)
        if(maze[u][v]-flow[u][v]>0 && dis[u] == dis[v]+1)
        {
            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-1;
        for(int v = 0; v<nodenum; v++)
            if(maze[u][v]-flow[u][v]>0 && mindis>dis[v])
            {
                cur[u] = v;
                mindis = dis[v];
            }
        if((--gap[dis[u]])==0) break;
        gap[dis[u]=mindis+1]++;
        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 = 0, end = F+D+2*N+1;
        memset(maze, 0, sizeof(maze));
        for(int i = 1; i<=F; i++)
        {
            int Food_supply;
            scanf("%d", &Food_supply);
            maze[start][i] = Food_supply;
        }
        for(int i = 1; i<=D; i++)
        {
            int Drink_supply;
            scanf("%d", &Drink_supply);
            maze[F+i][end] = Drink_supply;
        }
        for(int i = 1; i<=N; i++)
        {
            scanf("%s", str+1);
            for(int j = 1; j<=F; j++)
                if(str[j]=='Y')
                    maze[j][F+D+i] = 1;
        }
        for(int i = 1; i<=N; i++)
        {
            scanf("%s", str+1);
            for(int j = 1; j<=D; j++)
                if(str[j]=='Y')
                    maze[F+D+N+i][F+j] = 1;
        }
        for(int i = 1; i<=N; i++)
            maze[F+D+i][F+D+N+i] = 1;

        cout<< sap(start, end, F+D+2*N+2) <<endl;
    }
}