并查集之旅

1、POJ 2524 Ubiquitous Religions

最基础的并查集。

#include <iostream>

using namespace std;
int s[50005], hight[50005];
void init(int n)
{
    for (int i = 1; i <= n + 5; i++)
    {
        s[i] = i;
        hight[i] = 1;
    }
}

int getroot(int x)
{
    if (s[x] != x)
        x = getroot(s[x]);
    return s[x];
}

void merge(int x, int y)
{
    x = getroot(x);
    y = getroot(y);
    if (hight[x] == hight[y])
    {
        hight[x]++;
        s[y] = x;
    }
    else
    {
        if (hight[x] > hight[y])
            s[y] = x;
        else
            s[x] = y;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    int n, m, x, y;
    int tmp = 0;
    while (cin >> n >> m)
    {
        if (!n && !m)
            break;
        int ans = 0;
        init(n);
        while (m--)
        {
            cin >> x >> y;
            merge(x, y);
        }
        for (int i = 1; i <= n; i++)
        {
            if (i == s[i])
                ans++;
        }
        cout << "Case " << ++tmp << ": ";
        cout << ans << endl;
    }
    return 0;
}

---

2、POJ1611 The Suspects

类似于并查集的连通问题。

#include <iostream>
using namespace std;

int s[30005], sum[30005]; //sum用来记录当前集合的元素个数，s为并查集

void init(int n) //初始话
{
    for (int i = 0; i < n; i++)
    {
        s[i] = i;
        sum[i] = 1;
    }
}

int getroot(int x) //压缩状态
{
    if (x != s[x])
        s[x] = getroot(s[x]);
    return s[x];
}

int merge(int x, int y) //集合的交
{
    x = getroot(x);
    y = getroot(y);
    if (x != y)
    {
        s[y] = x;
        sum[x] += sum[y]; //加到x上
    }
}
int main()
{
    ios::sync_with_stdio(false);
    int n, m, t, x, y;
    while (cin >> n >> m)
    {
        if (n == 0 && m == 0)
            break;
        init(n);
        while (m--)
        {
            cin >> t;
            cin >> x; //多组数据的处理
            t--;
            while (t--)
            {
                cin >> y;
                merge(x, y);
            }
        }
        cout << sum[getroot(0)] << endl;
    }
    return 0;
}

3、POJ 1703 Find them, Catch them

并查集提高题，并查集维护不同集合的关系。

#include <iostream>
using namespace std;
const int MAXN = (1e5 + 5) * 2;
int s[MAXN], high[MAXN]; //存根
void itin(int n)         //初始化
{
    for (int i = 1; i <= 2 * n; i++)
    {
        s[i] = i;
        high[i] = 1;
    }
}

int getroot(int x) //压缩状态
{
    if (x != s[x])
        s[x] = getroot(s[x]);
    return s[x];
}
void merge(int x, int y) //合并优化
{
    x = getroot(x);
    y = getroot(y);
    if (x == y)
        return;
    if (high[x] == high[y])
    {
        s[y] = x;
        high[x]++;
    }
    else
    {
        if (high[x] > high[y])
            s[y] = x;
        else
            s[x] = y;
    }
}
bool same(int x, int y) //判断函数
{
    x = getroot(x);
    y = getroot(y);
    return x == y;
}
int main()
{
    int t, n, m, x, y;
    char ch;
    cin >> t;
    while (t--)
    {
        cin >> n >> m;
        itin(n);
        while (m--)
        {
            cin >> ch;
            cin >> x >> y;
            if (ch == 'A') //put
            {
                if (same(x, y) || same(x + n, y + n))
                    cout << "In the same gang." << endl;
                else if (same(x + n, y) || same(x, y + n))
                    cout << "In different gangs." << endl;
                else
                    cout << "Not sure yet." << endl;
            }
            else //merge
            {
                merge(x + n, y);
                merge(x, y + n);
            }
        }
    }
    return 0;
}

4、POJ 2236 Wireless Network

与坐标有关的并查集。
处理方法，结构体存放坐标，其他与普通并查集相同。

#include <iostream>
#include <math.h>
using namespace std;
int s[1005], high[1005];
struct zb
{
    int x, y, flag; //flag：1代表修好，0代表没修好
} num[1005];
double fx(int x1, int y1, int x2, int y2)
{
    return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int find_set(int x)
{
    if (x != s[x])
        s[x] = find_set(s[x]);
    return s[x];
}
bool some(int x, int y) //check
{
    return find_set(x) == find_set(y);
}
void union_set(int x, int y) //merge
{
    x = find_set(x);
    y = find_set(y);
    if (x == y)
        return;
    if (high[x] == high[y])
    {
        s[y] = x;
        high[x]++;
    }
    else if (high[x] > high[y])
        s[y] = x;
    else
        s[x] = y;
}
int main()
{
    ios::sync_with_stdio(false);
    int n, D;
    char ch;
    cin >> n >> D;
    for (int i = 1; i <= n; i++) //在存入坐标的同时进行初始化
    {
        cin >> num[i].x >> num[i].y;
        num[i].flag = 0;
        s[i] = i;
        high[i] = 0;
    }
    int t1, t2;
    while (cin >> ch)
    {
        if (ch == 'O') //合并结构体
        {
            cin >> t1;
            num[t1].flag = 1; //只有修好的情况下才能合并
            for (int i = 1; i <= n; i++)
            {
                if (t1 == i)
                    continue;
                if (fx(num[i].x, num[i].y, num[t1].x, num[t1].y) <= D && num[i].flag)
                    union_set(i, t1);
            }
        }
        else //检查结构体
        {
            cin >> t1 >> t2;
            if (some(t1, t2))
                cout << "SUCCESS" << endl;
            else
                cout << "FAIL" << endl;
        }
    }
    return 0;
}