HDU 1856 More is better

传送门

找最大连通分量，没有一对pairs就输出1。

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

const int MAXN = 1e7 + 2;
int pre[MAXN];
int pairs;
int opt;

int f(int x)
{
    int f0 = x, f1 = x;
    for (; pre[f0] > 0;)
        f0 = pre[f0];
    for (; pre[f1] > 0;)
    {
        int t = f1;
        f1 = pre[f1];
        pre[t] = f0;
    }
    return f0;
}

void u(int a, int b)
{
    int f1 = f(a);
    int f2 = f(b);
    if (f1 != f2)
    {
        if (pre[f1] <= pre[f2])
        {
            pre[f1] += pre[f2];
            if (-pre[f1] > opt) opt = -pre[f1];
            pre[f2] = f1;
        }
        else
        {
            pre[f2] += pre[f1];
            if (-pre[f2] > opt) opt = -pre[f2];
            pre[f1] = f2;
        }
    }
}

int main()
{
    for (; ~scanf("%d", &pairs);)
    {
        int a, b;
        memset(pre, -1, sizeof pre);
        opt = 1;                               // n=0 时直接输出 1     又是这种坑点，服
        for (int i = 0; i < pairs; i++)
        {
            scanf("%d%d", &a, &b);
            u(a, b);
        }
        printf("%d\n", opt);
    }

    return 0;
}