POJ 3714 Raid

Raid

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 9636 Accepted: 2941

Description

After successive failures in the battles against the Union, the Empire retreated to its last stronghold. Depending on its powerful defense system, the Empire repelled the six waves of Union’s attack. After several sleepless nights of thinking, Arthur, General of the Union, noticed that the only weakness of the defense system was its energy supply. The system was charged by N nuclear power stations and breaking down any of them would disable the system.

The general soon started a raid to the stations by N special agents who were paradroped into the stronghold. Unfortunately they failed to land at the expected positions due to the attack by the Empire Air Force. As an experienced general, Arthur soon realized that he needed to rearrange the plan. The first thing he wants to know now is that which agent is the nearest to any power station. Could you, the chief officer, help the general to calculate the minimum distance between an agent and a station?

Input

The first line is a integer T representing the number of test cases.
Each test case begins with an integer N (1 ≤ N ≤ 100000).
The next N lines describe the positions of the stations. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the station.
The next following N lines describe the positions of the agents. Each line consists of two integers X (0 ≤ X ≤ 1000000000) and Y (0 ≤ Y ≤ 1000000000) indicating the positions of the agent. 　

Output

For each test case output the minimum distance with precision of three decimal placed in a separate line.

Sample Input

2
4
0 0
0 1
1 0
1 1
2 2
2 3
3 2
3 3
4
0 0
0 0
0 0
0 0
0 0
0 0
0 0
0 0

Sample Output

1.414
0.000

Source
POJ Founder Monthly Contest – 2008.12.28, Dagger

这道题就是套用的BZOJ2648(http://blog.csdn.net/javays1/article/details/50432064)的解题思路，而且更简单，因为不用新增点。 直接上代码：

#include <stdio.h>#include <algorithm>#include<math.h>using namespace std;#define N 100000short currentDim;int nextIndex = 0;struct point{    int pos[2];    int left;    int right;    int Min[2], Max[2];    bool operator < (const point &u) const    {        return pos[currentDim] < u.pos[currentDim];    }} point_set[N];point kd_tree[N];pair<double, point> nearest_point;double square(int x){    return (double)x * (double)x;}inline void construct_kdTree(int p, int r, int index, short depth){    if(index == -1)        return;    if(p <= r)    {        currentDim = depth % 2;        int mid = (p + r) / 2;        nth_element(point_set + p, point_set + mid, point_set+r+1);        kd_tree[index] = point_set[mid];        nextIndex = index + 1;        kd_tree[index].Min[0] = kd_tree[index].Max[0] = kd_tree[index].pos[0];        kd_tree[index].Min[1] = kd_tree[index].Max[1] = kd_tree[index].pos[1];        if(p > mid - 1)            kd_tree[index].left = -1;        else if(nextIndex < N)            kd_tree[index].left = nextIndex++;        else            kd_tree[index].left = -1;        construct_kdTree(p, mid-1, kd_tree[index].left, depth+1);        if(mid + 1 > r)            kd_tree[index].right = -1;        else if(nextIndex < N)            kd_tree[index].right = nextIndex++;        else            kd_tree[index].right = -1;        construct_kdTree(mid+1, r, kd_tree[index].right, depth+1);        if(kd_tree[index].left != -1)        {            int leftChild = kd_tree[index].left;            if(kd_tree[index].Max[0] < kd_tree[leftChild].Max[0])                kd_tree[index].Max[0] = kd_tree[leftChild].Max[0];            if(kd_tree[index].Max[1] < kd_tree[leftChild].Max[1])                kd_tree[index].Max[1] = kd_tree[leftChild].Max[1];            if(kd_tree[index].Min[0] > kd_tree[leftChild].Min[0])                kd_tree[index].Min[0] = kd_tree[leftChild].Min[0];            if(kd_tree[index].Min[1] > kd_tree[leftChild].Min[1])                kd_tree[index].Min[1] = kd_tree[leftChild].Min[1];        }        if(kd_tree[index].right != -1)        {            int rightChild = kd_tree[index].right;            if(kd_tree[index].Max[0] < kd_tree[rightChild].Max[0])                kd_tree[index].Max[0] = kd_tree[rightChild].Max[0];            if(kd_tree[index].Max[1] < kd_tree[rightChild].Max[1])                kd_tree[index].Max[1] = kd_tree[rightChild].Max[1];            if(kd_tree[index].Min[0] > kd_tree[rightChild].Min[0])                kd_tree[index].Min[0] = kd_tree[rightChild].Min[0];            if(kd_tree[index].Min[1] > kd_tree[rightChild].Min[1])                kd_tree[index].Min[1] = kd_tree[rightChild].Min[1];        }    }}inline double dist(int index, point p){    int x = 0, y = 0;    if(p.pos[0] < kd_tree[index].Min[0])        x = kd_tree[index].Min[0] - p.pos[0];    if(p.pos[0] > kd_tree[index].Max[0])        x = p.pos[0] - kd_tree[index].Max[0];    if(p.pos[1] < kd_tree[index].Min[1])        y = kd_tree[index].Min[1] - p.pos[1];    if(p.pos[1] > kd_tree[index].Max[1])        y = p.pos[1] - kd_tree[index].Max[1];    return square(x) + square(y);}inline void query_kdTree(point p, int index, short depth){    if(index == -1)        return;    pair<double, point> current_node(0, kd_tree[index]);    current_node.first = square(p.pos[0] - current_node.second.pos[0])        + square(p.pos[1] - current_node.second.pos[1]);    //printf("%lf %d %d %d %d\n", current_node.first, p.pos[0], p.pos[1],        //current_node.second.pos[0], current_node.second.pos[1]);    if(current_node.first < nearest_point.first)    {        nearest_point.first = current_node.first;        nearest_point.second = current_node.second;    }    int lchild = kd_tree[index].left;    int rchild = kd_tree[index].right;    //printf("left:%d %d %d\n",lchild, kd_tree[lchild].pos[0], kd_tree[lchild].pos[1]);    //printf("right:%d %d %d\n",rchild, kd_tree[rchild].pos[0], kd_tree[rchild].pos[1]);    if(lchild != -1 && rchild != -1)    {        double ldist = dist(lchild, p);        double rdist = dist(rchild, p);        //printf("ldist:%lf\n", ldist);        //printf("rdist:%lf\n", rdist);        if(ldist < rdist)        {            if(ldist < nearest_point.first)                query_kdTree(p, lchild, depth+1);            if(rdist < nearest_point.first)                query_kdTree(p, rchild, depth+1);        }        else        {            if(rdist < nearest_point.first)                query_kdTree(p, rchild, depth+1);            if(ldist < nearest_point.first)                query_kdTree(p, lchild, depth+1);        }    }    else if(lchild != -1)    {        double ldist = dist(lchild, p);        if(ldist < nearest_point.first)            query_kdTree(p, lchild, depth+1);    }    else if(rchild != -1)    {        double rdist = dist(rchild, p);        if(rdist < nearest_point.first)            query_kdTree(p, rchild, depth+1);    }}int main(){    int T,n;    scanf("%d", &T);    for(int t = 0; t < T; t++)    {        nearest_point.first = 9999999999999999999999999999.99999;        scanf("%d", &n);        for(int i = 0; i < n; i++)        {            scanf("%d%d", &point_set[i].pos[0], &point_set[i].pos[1]);        }        memset(kd_tree, 0, N*sizeof(point));        construct_kdTree(0, n-1, 1, 0);        point tmpv;        for(int i = 0; i < n; i++)        {            scanf("%d%d", &tmpv.pos[0], &tmpv.pos[1]);            query_kdTree(tmpv, 1, 0);        }        printf("%.3lf\n", sqrt(nearest_point.first));    }    return 0;}

值得提醒的一点是如果提交使用g++，那么最后printf的输出格式需要使用”%3.f\n”， 而非”%3.lf\n”，这是g++编译器的一个问题。