zoukankan      html  css  js  c++  java
  • HDU 3687 Labeling Balls(逆向拓扑)

    Labeling Balls

     

    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, bN) 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

     1 #include<cstdio>
     2 #include<vector>
     3 #include<cstring>
     4 #include<queue>
     5 #include<algorithm>
     6 using namespace std;
     7 
     8 int indegree[205];
     9 int topo[205];
    10 int G[205][205];
    11 int m,n;
    12 
    13 void init()
    14 {
    15     memset(indegree,0,sizeof(indegree));
    16     memset(G,0,sizeof(G));
    17     //memset(topo,0,sizeof(topo));
    18 }
    19 
    20 bool toposort()
    21 {
    22     int t=n;
    23     priority_queue<int,vector<int>,less<int> >q;
    24     for(int i=n;i>=1;i--)
    25         if(indegree[i]==0)
    26         q.push(i);
    27     while(!q.empty())
    28     {
    29         int u=q.top();
    30         q.pop();
    31         topo[u]=t--;
    32         for(int v=1;v<=n;v++)
    33         {
    34             if(G[u][v])
    35             {
    36                 indegree[v]--;
    37                 if(indegree[v]==0)
    38                 q.push(v);
    39             }
    40         }
    41     }
    42     if(t==0)
    43     return true;
    44     return false;
    45 }
    46 
    47 int main()
    48 {
    49     int i,a,b,t;
    50     scanf("%d",&t);
    51     while(t--)
    52     {
    53         init();
    54         scanf("%d%d",&n,&m);
    55         for(i=0;i<m;i++)
    56         {
    57             scanf("%d%d",&a,&b);
    58             if(!G[b][a])
    59             indegree[a]++;
    60             G[b][a]=1;
    61         }
    62         bool flag=toposort();
    63         if(flag)
    64         {
    65             for(i=1;i<=n;i++)
    66             {
    67                 if(i!=n)
    68                 printf("%d ",topo[i]);
    69                 else
    70                 printf("%d
    ",topo[i]);
    71             }
    72         }
    73         else
    74             printf("-1
    ");
    75     }
    76     return 0;
    77 }
  • 相关阅读:
    mongodb(2)
    mongodb(1)
    分片集群
    副本集 集群
    java连接腾讯云上的redis
    maven打包
    生成ftp文件的目录树
    从ftp获取文件并生成压缩包
    android 连续点击退出程序
    Android之完美退出方法
  • 原文地址:https://www.cnblogs.com/homura/p/4730729.html
Copyright © 2011-2022 走看看