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:
    csu 1503: 点弧之间的距离-湖南省第十届大学生计算机程序设计大赛
    Android MediaPlayer 和 NativePlayer 播放格式控制
    国内互联网企业奇妙招数
    [Oracle] Insert All神奇
    代码杂记
    R.layout.main connot be resolved 和R.java消失
    计算机安全篇(1)
    深入浅出谈开窗函数(一)
    PHP JSON_ENCODE 不转义中文汉字的方法
    c#indexof使用方法
  • 原文地址:https://www.cnblogs.com/homura/p/4730729.html
Copyright © 2011-2022 走看看