POJ 3687 Labeling Balls

最新推荐文章于 2022-08-03 12:41:58 发布

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最新推荐文章于 2022-08-03 12:41:58 发布

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本文介绍了一种使用拓扑排序算法解决特定约束条件下的球体重排序问题的方法，并提供了一个具体的C++实现示例。

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Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8264		Accepted: 2247

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:

No two balls share the same label.
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

Sample Output

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

刚看到这题的时候就有点发懵，虽然我知道这是拓扑排序但都头来还是没做出来，不断的变化思路，最终还是WA，无奈之下看了看解题报告，明白了思想，感觉特别简单。推荐大家看

http://www.cnblogs.com/rainydays/archive/2011/07/20/2112047.html

很容易懂

贴一下我的代码：

#include <iostream>
#include <string>
using namespace std;
class num
{
    public:
    int end,next;
}a[1000000];
int b[210];
int status[210];
int res[210];
int in[210],out[210],n;
int main()
{
    int deal();
    int i,j,m,s,t,k;
    int x,y;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        memset(b,-1,sizeof(b));
        memset(in,0,sizeof(in));
        memset(out,0,sizeof(out));
        k=0;
        for(i=1,j=0;i<=m;i++)
        {
            cin>>x>>y;
            in[x]+=1; out[y]+=1;
            a[j].end=x; a[j].next=b[y];
            b[y]=j; j++;
            if(x==y)
            {
                k=1;
            }
        }
        if(k)
        {
            cout<<-1<<endl;
            continue;
        }
        k=deal();
        if(!k)
        {
            cout<<-1<<endl;
        }else
        {
            for(i=1;i<=n;i++)
            {
                if(i==1)
                {
                    cout<<res[i];
                }else
                {
                    cout<<" "<<res[i];
                }
            }
            cout<<endl;
        }
    }
    return 0;
}
int deal()
{
    int i,j,k;
    memset(status,0,sizeof(status));
    for(i=n;i>=1;i--)
    {
        k=-1;
        for(j=n;j>=1;j--)
        {
            if(!in[j]&&!status[j])
            {
                k=j;
                break;
            }
        }
        if(k==-1)
        {
            return 0;
        }
        res[k]=i; status[k]=1;
        for(j=b[k];j!=-1;j=a[j].next)
        {
            in[a[j].end]-=1;
        }
    }
    return 1;
}