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  • poj 3687 拓扑排序

          

    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:有n个球重量分别1到n;对每个球进行编号,给出编号之间的重量大小,按重量排序从小到大(让编号小的尽可能重量小)。
    这题意有点绕,wrong了好几次,以5—>6->1,4->3->2为例,正确的顺序应该是5 6 1 4 3 2,而不是4 3 2 5 6 1;难点在于让编号小的排在前面;
    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    using namespace std;
    const int N=210;
    int n,deep[N],topo[N];
    bool map[N][N];
    int toposort()
    {
       int temp[N],c=0,k,m,i,j;
       for(i=0;i<=n;i++)
       {
           temp[i]=deep[i];
       }
       for(i=n;i>0;i--)
       {
           m=0;
           for(j=n;j>0;j--)
           {
               if(temp[j]==0){m++;k=j;break;}
           }
           if(m==0) return 0;
           topo[k]=i;
           temp[k]=-1;
           for(j=1;j<=n;j++)
           {
               if(map[k][j]==1)
               {
                   temp[j]--;
               }
           }
       }
       return 1;
    }
    int main()
    {
        int t,m,i,x,y;
        scanf("%d",&t);
        while(t--)
        {
            memset(map,0,sizeof(map));
            memset(deep,0,sizeof(deep));
            scanf("%d%d",&n,&m);
            for(i=1;i<=m;i++)
            {
                scanf("%d%d",&x,&y);
                if(!map[y][x])
                {
                    deep[x]++;
                }
                map[y][x]=1;
            }
            int sign=toposort();
            if(!sign)
               printf("-1");
            else
            for(i=1;i<=n;i++)
            {
                if(i==1)
                   printf("%d",topo[i]);
                 else
                    printf(" %d",topo[i]);
            }
            printf("
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yuanbo123/p/4840738.html
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