TOJ 3294. Building Block 並查集

3294.   Building Block

Time Limit: 1.0 Seconds
Memory Limit: 65536K
Total Runs: 840   Accepted Runs:235

John are playing with blocks. There are N blocks (1 ≤
N ≤ 30000) numbered 0.. N-1. Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 ≤ P ≤ 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containingY. IfX and
Y are in the same pile, just ignore this command.
C X : Count the number of blocks under block X

You are request to find out the output for each C operation.

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

Output

Output the count for each C operations in one line.

Sample Input

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

Sample Output

102

Author:SUN, Chao

Source:Multi-School Training Contest
- TOJ Site #1

題意： 有N個木塊，放在N個平台上，M x，y移動有x的平台，全部放到y的平台上

c x 輸出x木塊下有幾個木塊

 

一道並查集的題；

但是由于思考的不夠細，一直wa，，再加上不懂c讀取字元，導致了幾次re。 一次ce

連水題都被難倒了，，貼一下。。。紀念又一個悲劇

 

設定三個數字： up [x]  x上有幾個木塊

down[x]  下有幾個木塊

pre[x]是下面的一個木塊是幾號

//一堆木塊，除了最底下的父親是自己外，其他的木塊父親最終指向最底下

#include<cstdio>
using namespace std;
const int maxn = 30005;
int up[maxn];//記錄木塊上面的木塊個數
int down[maxn];//記錄木塊下面的木塊個數
int pre[maxn];//記錄父親結點
int stack[maxn];//堆棧
int sp = 0;
int find_pre(int root){//尋找根
    int u = root;
    sp = 0;
    while(pre[root] != root)//找根，且把鏈上的結點都加入棧
    {
        stack[sp++] = root;
        root = pre[root];
    }
    for(int i = sp - 2 ;i > -1; i--)//更新鏈上的每個結點
    {// -2 的原因是如果他的父親是最底下的木塊，就不用更新了。因為在  合并的時候更新過 1.0處
        down[stack[i]] += down[stack[i+1]];
        pre[stack[i]] = root;
    }
    return root;
}

int main()
{
    int N,x,y;
    char a;
    while(scanf("%d",&N) != EOF)
    {
        for(int i = 0; i <= maxn; i++)
        {
            pre[i] = i;
            up[i] = 0;
            down[i] = 0;
        }
        for(int i = 0;i < N; i++)
        {
            getchar();
            a = getchar();//讀取字元
            if(a == 'M')
            {
                scanf("%d%d",&x,&y);  //1.0，更新點
                int fx = find_pre(x);//找x的根
                int fy = find_pre(y);//找y的根
                if(fx == fy) continue;
                down[fx] += up[fy] + 1;//更新結點資訊
                up[fy] += up[fx] + 1;
                pre[fx] = fy;
            }
            else
            {
                scanf("%d",&x);
                find_pre(x);
                printf("%d\n",down[x]);
            }
        }
    }
    return 0;
}