Building Block 并查集

Building Block

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                 Total Submission(s): 4548    Accepted Submission(s): 1408

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。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 containing Y. If X 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

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

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#include <cstdio>#include <iostream>using namespace std;const int MAXN = 30000 + 10;int Fa[MAXN], Under[MAXN], Num[MAXN];void Init(){    for (int i = 0; i < MAXN; i++)    {        Fa[i] = i;              //初始化并查集，自成一个连通分量        Num[i] = 1;             //i是根节点，初始化以它为根的树共有1个节点    }}int Find(int x){    if (x == Fa[x]) return x;    int fa = Fa[x];    Fa[x] = Find(Fa[x]);        //递归查找老祖宗    Under[x] += Under[fa];      //加上父亲的    return Fa[x];}void Union(int x, int y){    int tx = Find(x), ty = Find(y); //先查找两者的祖先    if (tx == ty) return;    Fa[tx] = ty;                //合并    Under[tx] = Num[ty];        //tx下面多了Num[ty]个    Num[ty] += Num[tx];         //以ty为根的树多了Num[tx]个节点    Num[tx] = 0;                //以tx为根的树没有结点（此时tx已经不是一个树的真正根节点）}int main(){    int t, x, y;    char cmd[10];    Init();    scanf("%d", &t);    while (t--)    {        scanf("%s", cmd);        if (cmd[0] == 'C')        {            scanf("%d", &x);            Find(x);            printf("%d\n", Under[x]);        }        else        {            scanf("%d%d", &x, &y);            Union(x, y);        }    }    return 0;}

//超时了的代码   不造怎么超了#include<iostream>#include<cstdio>using namespace std;int f[30005],vis[30005],down[30005],i,x,y,T;char op[2];int zbb(int x){    if(f[x]!=x)    {        down[x]+=down[f[x]];        return f[x]=zbb(f[x]);    }    else return x;}void hb(int x,int y){    int a=zbb(x),b=zbb(y);    if(a!=b)    {        f[a]=b;        down[a]=vis[b];   //x堆底下面的数量为y堆的数量        vis[b]+=vis[a];   //更新整个y堆加上合并x堆后的数量    }}void init(int n)             //初始化{    for(i=0;i<n;i++)    {        f[i]=i;        down[i]=0;        vis[i]=1;    }}    int main()    {        scanf("%d",&T);        init(T);        while(T--)        {            scanf("%s",op);            if(op[0]=='M')            {                scanf("%d%d",&x,&y);                hb(x,y);            }            if(op[0]=='C')            {                scanf("%d",&x);                zbb(x);                printf("%d\n",down[x]);            }        }        return 0;    }