POJ.2236 Wireless Network (并查集)

POJ.2236 Wireless Network (并查集)

题意分析

查询点直接O(n)搞就行了，我想的是画个内切正方形，然后二分，或者kdtree。10sec其实不用了。

代码总览

#include <cstdio>#include <cstring>#include <algorithm>#define nmax 1200using namespace std;struct point{    int x, y;    bool isok;}p[nmax];int father[nmax];int rnk[nmax];int n,d;void makeset(int x){    father[x] = x;    rnk[x] = 0;}int findset(int x){    int r  = x,temp;    while(father[r] != r) r = father[r];    while(x != r){        temp = father[x];        father[x] = r;        x = temp;    }    return r;}void unionset(int x, int y){    x = findset(x);    y = findset(y);    if(x == y) return;    else{        if(rnk[x] > rnk[y])            father[y] = x;        else{            father[x] = y;            if(rnk[x] == rnk[y]) rnk[x]++;        }    }}int main(){//    freopen("in.txt","r",stdin);    scanf("%d %d",&n,&d);    for(int i = 1;i<=n;++i) scanf("%d %d",&p[i].x,&p[i].y),makeset(i);    char op;    while(scanf(" %c",&op) != EOF){        int a,b;        if(op  == 'O'){            scanf("%d",&a);            p[a].isok = true;            int temp = 0;            for(int i = 1;i<=n;++i){                if(i == a) continue;                temp = (p[i].x - p[a].x) * (p[i].x - p[a].x) + (p[i].y - p[a].y) *(p[i].y - p[a].y);                if(temp <= d * d && p[i].isok) unionset(a,i);            }        }else{            scanf("%d %d",&a,&b);            a = findset(a);            b = findset(b);            if(a == b) printf("SUCCESS\n");            else printf("FAIL\n");        }    }    return 0;}