第一次用两种数据结构解的题目，纪念一下

USE 并查集和线段树

The k-th Largest Group
Time Limit:2000MS Memory Limit:131072K
Total Submit:1222 Accepted:290

Description

Newman likes playing with cats. He possesses lots of cats in his home. Because the number of cats is really huge, Newman wants to group some of the cats. To do that, he first offers a number to each of the cat (1, 2, 3, …, n). Then he occasionally combines the group cat i is in and the group cat j is in, thus creating a new group. On top of that, Newman wants to know the size of the k-th biggest group at any time. So, being a friend of Newman, can you help him?

Input

1st line: Two numbers N and M (1 ≤ N, M ≤ 200,000), namely the number of cats and the number of operations.

2nd to (m + 1)-th line: In each line, there is number C specifying the kind of operation Newman wants to do. If C = 0, then there are two numbers i and j (1 ≤ i, j ≤ n) following indicating Newman wants to combine the group containing the two cats (in case these two cats are in the same group, just do nothing); If C = 1, then there is only one number k (1 ≤ k ≤ the current number of groups) following indicating Newman wants to know the size of the k-th largest group.

Output

For every operation “1” in the input, output one number per line, specifying the size of the kth largest group.

Sample Input

Sample Output

Hint

When there are three numbers 2 and 2 and 1, the 2nd largest number is 2 and the 3rd largest number is 1.

Source
POJ Monthly--2006.08.27, zcgzcgzcg

#include < iostream >
using namespace std;
const int MAXN = 200001 ;

class UFset
{
public:
    int parent[MAXN];
    UFset();
    int Find(int);
    void Union(int, int);
} ;

UFset::UFset()
{
    memset(parent, -1, sizeof(parent));
}

int UFset::Find( int x)
{
    if (parent[x] < 0)
        return x;
    else
    {
        parent[x] = Find(parent[x]);
        return parent[x];
    }// 压缩路径
}

void UFset::Union( int x, int y)
{
    int pX = Find(x);
    int pY = Find(y);
    int tmp;
    if (pX != pY)
    {
        tmp = parent[pX] + parent[pY]; // 加权合并
        if (parent[pX] > parent[pY])
        {
            parent[pX] = pY;
            parent[pY] = tmp;
        }
        else
        {
            parent[pY] = pX;
            parent[pX] = tmp;
        }
    }
}

int f[(MAXN + 1 ) * 3 ] = {0} ;
int n, m;

void initTree()
{
    int l = 1, r = n;
    int c = 1;
    while (l < r)
    {
        f[c] = n;
        c = c * 2;
        r = (l + r) / 2;
    }
    f[c] = n;//叶子初始化
}

void insertTree( int k)
{
    int l = 1, r = n;
    int c = 1;
    int mid;

    while (l < r)
    {
        f[c]++;
        mid = (r + l) / 2;
        if (k > mid)
        {
            l = mid + 1;
            c = c * 2 + 1;
        }
        else
        {
            r = mid;
            c = c * 2;
        }
    }
    f[c]++;//叶子增加1
}

void delTree( int k)
{
    int l = 1, r = n;
    int c = 1;
    int mid;

    while (l < r)
    {
        f[c]--;
        mid = (r + l) / 2;
        if (k > mid)
        {
            l = mid + 1;
            c = c * 2 + 1;
        }
        else
        {
            r = mid;
            c = c * 2;
        }
    }
    f[c]--;//叶子减少1
}

int searchTree( int k)
{
    int l = 1, r = n;
    int c = 1;
    int mid;

    while (l < r)
    {
        mid = (l + r) / 2;
        if (k <= f[2*c+1])
        {
            l = mid + 1;
            c = c * 2 + 1;
        }
        else
        {
            k -= f[2*c+1];
            r = mid;
            c = c * 2;
        }
    }
    return l;
}

int main()
{
    int i, j;
    int x, y;
    int k;
    int l, r;
    int cmd;
    int px, py;
    int tx, ty, tz;
    UFset UFS;


    scanf("%d%d", &n, &m);
    initTree();
    for (i=0; i<m; i++)
    {
        scanf("%d", &cmd);
        if (cmd == 0)
        {
            scanf("%d%d", &x, &y);
            px = UFS.Find(x);
            py = UFS.Find(y);
            if (px != py)
            {
                tx = -UFS.parent[px];
                ty = -UFS.parent[py];
                tz = tx + ty;
                UFS.Union(x, y);
                insertTree(tz);
                delTree(tx);
                delTree(ty);
            }
        }
        else
        {
            scanf("%d", &k);
            printf("%d\n", searchTree(k));
        }
    }
    return 0;
}

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