【POJ 3714】Raid

【题目链接】:http://poj.org/problem?id=3714

【题意】



给你两类的点;

各n个;

然后让你求出2*n个点中的最近点对的距离;

这里的距离定义为不同类型的点之间的距离;

【题解】



分治法;

不断划分小区间;

求出小区间内的最小值;

为后面的大区间剪枝;

只是要求点的类型不同才能剪枝了;



【Number Of WA】



0



【完整代码】

#include <iostream>

#include <cstdio>

#include <iomanip>

#include <algorithm>

#include <cmath>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define ms(x,y) memset(x,y,sizeof x)

#define Open() freopen("F:\\rush.txt","r",stdin)

#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

const int N = 2e5+100;

const LL oo = 5e18;

struct node{

    LL x,y;

    int p;

};

int n;

node a[N],b[N];

bool cmp1(node a,node b){

    return a.x < b.x;

}

bool cmp2(node a,node b){

    return a.y < b.y;

}

LL sqr(LL x){

    return x*x;

}

LL dis(node a,node b){

    LL temp = 0;

    temp+=sqr(a.x-b.x);

    temp+=sqr(a.y-b.y);

    return temp;

}

LL solve(int l,int r){

    LL ret = oo;

    if (l>=r) return ret;

    if (l+1==r) {

            if (a[l].p!=a[r].p)

                return dis(a[l],a[r]);

            else

                return ret;

    }

    int m = (l+r)>>1;

    LL ret1 = solve(l,m),ret2 = solve(m+1,r),temp;

    ret = min(ret1,ret2);

    int k = 0;

    rep2(i,m,l){

        temp =sqr(a[m].x-a[i].x);

        if (temp>ret && a[m].p != a[i].p) break;

        b[++k] = a[i];

    }

    rep1(i,m+1,r){

        temp = sqr(a[m].x-a[i].x);

        if (temp>ret && a[m].p != a[i].p) break;

        b[++k] = a[i];

    }

    sort(b+1,b+1+k,cmp2);

    rep1(i,1,k){

        rep1(j,i+1,k){

            if (b[i].p==b[j].p) continue;

            temp = sqr(b[i].y-b[j].y);

            if (temp > ret) break;

            temp = dis(b[i],b[j]);

            if (temp < ret) ret = temp;

        }

    }

    return ret;

}

int main(){

    //Open();

    Close();

    int T;

    cin >> T;

    while (T--){

        cin >> n;

        rep1(i,1,n){

            cin >> a[i].x >> a[i].y;

            a[i].p = 0;

        }

        rep1(i,n+1,2*n){

            cin >> a[i].x >> a[i].y;

            a[i].p = 1;

        }

        n<<=1;

        sort(a+1,a+1+n,cmp1);

        cout << fixed << setprecision(3) << double (sqrt(1.0*solve(1,n))) << endl;

    }

    return 0;

}