zoukankan      html  css  js  c++  java
  • POJ1988 Cube Stacking

    Time Limit: 2000MS   Memory Limit: 30000KB   64bit IO Format: %lld & %llu

    Description

    Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) operation. There are two types of operations: 
    moves and counts. 
    * In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y. 
    * In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with cube X that are under the cube X and report that value. 

    Write a program that can verify the results of the game. 

    Input

    * Line 1: A single integer, P 

    * Lines 2..P+1: Each of these lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a 'M' for a move operation or a 'C' for a count operation. For move operations, the line also contains two integers: X and Y.For count operations, the line also contains a single integer: X. 

    Note that the value for N does not appear in the input file. No move operation will request a move a stack onto itself. 

    Output

    Print the output from each of the count operations in the same order as the input file. 

    Sample Input

    6
    M 1 6
    C 1
    M 2 4
    M 2 6
    C 3
    C 4
    

    Sample Output

    1
    0
    2
    

    Source

    偏移量并查集。

    维护每个块所在栈的块数,每个块下面的块数。

     1 /*by SilverN*/
     2 #include<algorithm>
     3 #include<iostream>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 #include<vector>
     8 using namespace std;
     9 const int mxn=100010;
    10 int fa[mxn],und[mxn],cnt[mxn];
    11 int find(int x){
    12     if(fa[x]==x)return x;
    13     int ff=fa[x];
    14     fa[x]=find(fa[x]);
    15     und[x]+=und[ff];
    16     return fa[x];
    17 }
    18 int n,m;
    19 int main(){
    20     scanf("%d",&m);
    21     char op[3];int x;
    22     for(x=1;x<=30000;x++)fa[x]=x,cnt[x]=1,und[x]=0;
    23     int u,v;
    24     while(m--){
    25         scanf("%s",op);
    26         if(op[0]=='M'){
    27             scanf("%d%d",&u,&v);
    28             u=find(u);
    29             v=find(v);
    30             if(u!=v){
    31                 fa[u]=v;
    32                 und[u]+=cnt[v];
    33                 cnt[v]+=cnt[u];
    34                 cnt[u]=0;
    35             }
    36         }
    37         else{
    38             scanf("%d",&u);
    39             find(u);
    40             printf("%d
    ",und[u]);
    41         }
    42     }
    43     return 0;
    44 }
  • 相关阅读:
    使用Java发送qq邮件
    docker部署nacos1.4
    职责链模式
    策略模式
    状态模式
    解释器模式
    备忘录模式
    js中数组常用方法总结
    微信小程序生成二维码工具
    小程序登录过程
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/5998057.html
Copyright © 2011-2022 走看看