kuangbin带你飞 并查集 题解

做这套题之前一直以为并查集是很简单的数据结构。

做了才发现自己理解太不深刻。只看重片面的合并集合。。

重要的时发现每个集合的点与这个根的关系，这个关系可以做太多事情了。

题解：

POJ 2236 Wireless Network

10S时限，尽量做到最优吧。一开始还怕会T

预处理出能到达的点。简单题看代码就行了

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <stack>
#include <queue>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define PI 3.1415926535897932626
using namespace std;
int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
const int MAXN = 1010;
struct point
{
    double x,y;
}src[MAXN];
int fa[MAXN];
int Find(int x) {return x == fa[x] ? x : fa[x] = Find(fa[x]);}
bool repair[MAXN];
double dis[MAXN][MAXN];
vector<int>go[MAXN];

int main()
{
    int N;
    double D;
    scanf("%d%lf",&N,&D);
    for (int i = 1 ; i <= N ; i++) scanf("%lf%lf",&src[i].x,&src[i].y);
    for (int i = 1 ; i <= N ; i++) go[i].clear();
    for (int i = 1 ; i <= N ; i++) fa[i] = i;
    for (int i = 1 ; i <= N ; i++)
    {
        for (int j = i + 1 ; j <= N ; j++)
        {
            dis[i][j] = sqrt((src[i].x - src[j].x) * (src[i].x - src[j].x) + (src[i].y - src[j].y) * (src[i].y - src[j].y));
            dis[j][i] = dis[i][j];
            if (dis[i][j] <= D) go[i].push_back(j);
            if (dis[i][j] <= D) go[j].push_back(i);
        }
    }

    memset(repair,false,sizeof(repair));
    char op[5];
    while (scanf("%s",op) != EOF)
    {
        if(op[0] == 'O')
        {
            int x;
            scanf("%d",&x);
            repair[x] = true;
            for (int j = 0 ; j < (int)go[x].size() ; j++)
            {
                if (repair[go[x][j]])
                {
                    int fx = Find(x);
                    int fy = Find(go[x][j]);
                    if (fx != fy) fa[fx] = fy;
                }
            }
        }
        else
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int fx = Find(x);
            int fy = Find(y);
            if (fx == fy) puts("SUCCESS");
            else puts("FAIL");
        }
    }
    return 0;
}

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