POJ 3687 Labeling Balls (拓扑排序)

Labeling Balls

 

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14312 Accepted: 4179

 

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:

  1. No two balls share the same label.
  2. The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled with b".

Can you help windy to find a solution?

 

Input

The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.

 

Output

For each test case output on a single line the balls' weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.

 

Sample Input

5

4 0

4 1
1 1

4 2
1 2
2 1

4 1
2 1

4 1
3 2

 

Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4

 

题意

标号为 1~n 的 N 个球,满足给定的 M 个编号约束关系,输出最终满足关系的球的标号。

 

思路

每一个标号都有可能被其他标号所约束,而对于这样的题目我们可以联想到拓扑排序。

题目给出了球的轻重关系,最终需要输出的是小球的重量。

但是小球的序号与重量并没有什么关系,所以我们不能直接使用字典序最小来解决这道问题。

采用逆向建图的方法,然后从最大标号开始判断,因为这样可以保证重量大一点的球会在前面,最后反转之后也就到后面啦!

 

AC 代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
using namespace std;
#define M 210
int in[M],arr[M];
vector<int>G[M];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            G[i].clear();
        memset(in,0,sizeof(in));
        for(int i=0; i<m; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            G[b].push_back(a);      //反向建立邻接表
            in[a]++;                //点的入度
        }
        int w;
        for(w=n; w>0; w--)          //从最大点开始
        {
            int i;
            for(i=n; i>0; i--)      //寻找入度为0的点
                if(!in[i])break;
            if(i==0)break;          //没有找到
            arr[i]=w;
            in[i]=-1;               //删除该点
            for(int j=0; j<(int)G[i].size(); j++)
            {
                int v=G[i][j];      //临接点的入度-1
                if(in[v]>0)
                    in[v]--;
            }
        }
        if(w!=0)
            printf("-1\n");
        else
        {
            for(int i=1; i<=n; i++)
                printf(i!=n?"%d ":"%d\n",arr[i]);
        }
    }
    return 0;
}