zoukankan      html  css  js  c++  java
  • 5.1.4 Dragon Balls

    Dragon Balls

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 324 Accepted Submission(s): 154

    Problem Description
    Five hundred years later, the number of dragon balls will increase unexpectedly, so it\\\\\\\'s too difficult for Monkey King(WuKong) to gather all of the dragon balls together.

    His country has N cities and there are exactly N dragon balls in the world. At first, for the ith dragon ball, the sacred dragon will puts it in the ith city. Through long years, some cities\\\\\\\' dragon ball(s) would be transported to other cities. To save physical strength WuKong plans to take Flying Nimbus Cloud, a magical flying cloud to gather dragon balls.
    Every time WuKong will collect the information of one dragon ball, he will ask you the information of that ball. You must tell him which city the ball is located and how many dragon balls are there in that city, you also need to tell him how many times the ball has been transported so far.
     

    Input
    The first line of the input is a single positive integer T(0 < T <= 100).
    For each case, the first line contains two integers: N and Q (2 < N <= 10000 , 2 < Q <= 10000).
    Each of the following Q lines contains either a fact or a question as the follow format:
      T A B : All the dragon balls which are in the same city with A have been transported to the city the Bth ball in. You can assume that the two cities are different.
      Q A : WuKong want to know X (the id of the city Ath ball is in), Y (the count of balls in Xth city) and Z (the tranporting times of the Ath ball). (1 <= A, B <= N)
     

    Output

                For each test case, output the test case number formated as sample output. Then for each query, output a line with three integers X Y Z saparated by a blank space.
     

    Sample Input
    2
    3 3
    T 1 2
    T 3 2
    Q 2
    3 4
    T 1 2
    Q 1
    T 1 3
    Q 1
     

    Sample Output
    Case 1:
    2 3 0
    Case 2:
    2 2 1
    3 3 2

    思路:这题居然1A了,太神奇了。并查集嘛,对于个数这个很好求,但是对于转移次数需动下脑筋。我们可以用一个tran表示此节点距离它自己现在的祖先(f数组)的转移次数,有点难理解,也就是说如图所示

     1 #include <cstdio>
     2 #include <iostream>
     3 #include <cstdlib>
     4 #include <algorithm>
     5 #include <cstring>
     6 #include <string>
     7 using namespace std;
     8 
     9 const int maxn=100100;
    10 int f[maxn],tran[maxn],sum[maxn];
    11 int af,bf,a,b,T,n,m;
    12 char s[2];
    13 
    14 void close()
    15 {
    16     fclose(stdin);
    17     fclose(stdout);
    18     exit(0);
    19 }
    20 
    21 int getfather(int k)
    22 {
    23     if (k==f[k])
    24         return k;
    25     return getfather(f[k]);
    26 }
    27 
    28 
    29 int find(int k)
    30 {
    31     if (f[k]==k)
    32         return k;
    33     int t=f[k];
    34     f[k]=find(f[k]);
    35     if (t!=f[t])
    36         tran[k]+=tran[t];
    37     return f[k];
    38 }
    39 
    40 void work()
    41 {
    42 }
    43 
    44 void init ()
    45 {
    46 freopen("dragon.in","r",stdin);
    47 freopen("dragon.out","w",stdout);
    48     scanf("%d",&T);
    49      int cnt=0;
    50      while (T--)
    51      {
    52          cnt++;
    53          printf("Case %d:\n",cnt);
    54          memset(tran,0,sizeof(tran));
    55          memset(f,0,sizeof(f));
    56          scanf("%d %d",&n,&m);
    57          for (int i=1;i<=n;i++)
    58          {
    59              f[i]=i;
    60              sum[i]=1;
    61          }
    62          for (int i=1;i<=m;i++)
    63          {
    64              scanf("%s",s);
    65              if (s[0]=='T')
    66              {
    67                  scanf("%d %d",&a,&b);
    68                  af=getfather(a);
    69                  bf=getfather(b);
    70                  f[af]=bf;
    71                  tran[af]++;
    72                  sum[bf]+=sum[af];
    73              }
    74              else
    75              {
    76                  scanf("%d",&a);
    77                  af=find(a);
    78                  printf("%d %d %d\n",af,sum[af],tran[a]);
    79              }
    80          }
    81      }
    82 }
    83 
    84 int main ()
    85 {
    86     init();
    87     work();
    88     close();
    89     return 0;
    90 }

    把freopen那些删掉

  • 相关阅读:
    sqlmap使用和X-Forwarded-For注入漏洞测试
    Spring 配置文件中将common.properties文件外置
    git rebase 和 reset的区别
    fatal: Not a valid object name: 'master'.
    Git error on commit after merge
    git 常用命令使用
    git常用命令学习
    关于idea 修改jsp文件后不能生效
    mysql 表被锁处理方案
    Redis 基本操作
  • 原文地址:https://www.cnblogs.com/cssystem/p/2911196.html
Copyright © 2011-2022 走看看