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  • poj 1988 Cube Stacking

    Cube Stacking
    Time Limit: 2000MS   Memory Limit: 30000K
    Total Submissions: 23540   Accepted: 8247
    Case Time Limit: 1000MS

    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

    
    
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 const int MAX = 30005;
     6 int parent[MAX];
     7 int sum[MAX];//若parent[i]=i,sum[i]表示砖块i所在堆的砖块数
     8 int under[MAX];//under[i]表示砖块i下面有多少砖块
     9 
    10 void init(){
    11     for(int i = 0; i < MAX; i++){
    12         parent[i] = i;
    13         sum[i] = 1;
    14         under[i] = 0;
    15     }
    16 }
    17 
    18 int GetParent(int a){//获取a的根,并把a的父节点改为根
    19     if(parent[a] == a)
    20         return a;
    21     int p = GetParent(parent[a]);
    22     under[a] += under[parent[a]];
    23     parent[a] = p;
    24     return parent[a];
    25 }
    26 
    27 void merge(int a, int b){
    28     //把b所在的堆,叠放到a所在的堆。
    29     int pa = GetParent(a);
    30     int pb = GetParent(b);
    31     if(pa == pb)
    32         return ;
    33     parent[pb] = pa;
    34     under[pb] = sum[pa];//under[pb]赋值前一定是0,因为parent[pb] = pb,pb一定是原b所在堆最底下的
    35     sum[pa] += sum[pb];
    36 }
    37 
    38 int main(){
    39     int p;
    40     init();
    41     scanf("%d", &p);
    42     for(int i = 0; i < p; i++){
    43         char s[20];
    44         int a, b;
    45         scanf("%s", s);
    46         if(s[0] == 'M'){
    47             scanf("%d%d", &a, &b);
    48             merge(b, a);
    49         }
    50         else {
    51             scanf("%d", &a);
    52             GetParent(a);
    53             printf("%d
    ", under[a]);
    54         }
    55     }        
    56     return 0;
    57 }
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  • 原文地址:https://www.cnblogs.com/qinduanyinghua/p/5703987.html
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