HDOJ2647拓扑排序模板程序连接矩阵实现

最新推荐文章于 2023-11-14 23:28:35 发布

原创最新推荐文章于 2023-11-14 23:28:35 发布 · 270 阅读

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本文探讨了一个工厂老板在春节前如何合理分配奖励给员工的问题。通过建立一个简单的图模型，并使用拓扑排序的方法来检查是否有环存在，进而确定能否满足所有员工的需求同时使用最少的资金。文章提供了一个完整的C++代码示例。

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9880 Accepted Submission(s): 3152

Problem Description
Dandelion’s uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a’s reward should more than b’s.Dandelion’s unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work’s reward will be at least 888 , because it’s a lucky number.

Input
One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a’s reward should be more than b’s.

Output
For every case ,print the least money dandelion ‘s uncle needs to distribute .If it’s impossible to fulfill all the works’ demands ,print -1.

Sample Input
2 1
1 2
2 2
1 2
2 1

Sample Output
1777
-1

Author
dandelion

Source
曾是惊鸿照影来

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问题就是判断是否构成环。
因为数据太小，用一张二维表存储实现起来比较方便，而且很直观。

#include <iostream>
#include <ctime>
#include <cstring>
#include <queue>
using namespace std;

const int maxn = 105;
int i,j,k,m,n,a,b;
int MAP[maxn][maxn],du[maxn];

void init(){
   memset(MAP,0,sizeof(MAP));
   memset(du,0,sizeof(du));
   for (i=1; i<=m; i++) {
      cin >> a >> b;
      if (!MAP[a][b]) {
          MAP[a][b] = 1;
          du[b]++;
      }
   }
}

bool tuopu(){
    int sum = 0,u,v;
    queue<int> que;
    for (i=0; i<n; i++) if (du[i]==0) que.push(i);
    while (!que.empty()){
        u = que.front();
        que.pop();
        sum++;
        for (i=0; i<n; i++) if (MAP[u][i]) {
             du[i]--;
             if (du[i]==0) que.push(i);
        }
    }
    return sum==n;
}

int main(){
    std::ios::sync_with_stdio(false);
    while (cin >> n >> m && n+m){
        init();
        if (tuopu()) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
    return 0;
}