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:
    根据中国气象局提供的API接口实现天气查询
    小程序——云函数发送请求
    apifm-wxapi API工厂
    首次使用 linux 阿里云服务器,入门及使用
    Android立体旋转动画实现与封装(支持以X、Y、Z三个轴为轴心旋转)
    Android来电监听和去电监听
    Android 源码下载方法(Git 方式clone)
    HandlerThread 创建一个异步的后台线程
    Android Toast cancel和show 不踩中不会知道的坑
    PopupWindow 点击外部和返回键无法消失背后的真相(setBackgroundDrawable(Drawable background))
  • 原文地址:https://www.cnblogs.com/homura/p/4730729.html
Copyright © 2011-2022 走看看