标签:bsp ble typedef false div ++ 题目 平面最近点对 abs
题目链接:https://www.luogu.com.cn/problem/P4423
此题给定n个点,从n个点中选取3个点组成三角形的周长最小。
暴力枚举不要考虑。
那回想平面最近点对,平面最近点对是求一对点对之间的最小距离。
我们是否也可以用此种想法呢。
平面最近点对的代码,是针对一个点,求与另外一个点的距离,并不断的更新minx。
同样,我们这个题,不就是针对一个点,求与另外两个点的距离和最小。并不断的更新minx。
按此思路,我们将平面最近点对的模板进行更改
#include"stdio.h"
#include"string.h"
#include"vector"
#include"math.h"
#include"algorithm"
using namespace std;
typedef struct Node{
double x,y;
int id;
}Node;
int n;
Node node[200100];
Node tran[200100];
double minx = 1e20;
bool same(double a, double b) { /// 1e-5精度意义下的浮点数相等
if(fabs(a-b) <= 1e-5) return true; return false;
}
int cmpx(Node a,Node b){
if(!same(a.x, b.x)) return a.x<b.x; return a.y<b.y;
}
int cmpy(Node a,Node b){
if(!same(a.y, b.y)) return a.y<b.y; return a.x<b.x;
}
void dist_minx(Node a,Node b,Node c){
double s1 = sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
double s2 = sqrt((a.x - c.x) * (a.x - c.x) + (a.y - c.y) * (a.y - c.y));
double s3 = sqrt((b.x - c.x) * (b.x - c.x) + (b.y - c.y) * (b.y - c.y));
minx = min(minx,s1 + s2 + s3);
}
void merge_node(int l,int mid,int r)
{
int t = 0;
int i = l,j = mid + 1;
while(i <= mid && j <= r)
{
if(node[i].y < node[j].y)
{
tran[++ t] = node[i]; i ++; continue;
} else {
tran[++ t] = node[j]; j ++; continue;
}
}
while(i <= mid) tran[++ t] = node[i ++];
while(j <= r) tran[++ t] = node[j ++];
for(int i = l; i <= r; i ++)
node[i] = tran[i - l + 1];
}
void Blocking(int l,int r)
{
if(r - l <= 4)
{
for(int i = l; i < r; i ++)
{
for(int j = i + 1; j <= r; j ++)
{
for(int k = j + 1; k <= r; k ++)
dist_minx(node[i],node[j],node[k]);
}
}
sort(node + l,node + r + 1,cmpy);
return ;
}
int mid = (l + r) >> 1;
double midx = node[mid].x;
Blocking(l,mid); Blocking(mid + 1,r);
merge_node(l,mid,r);
vector<Node> Q;
for(int i = l; i <= r; i ++)
{
if(fabs(node[i].x - midx) >= minx / 2) continue;
for(int j = Q.size() - 1; j >= 0; j --)
{
if(fabs(Q[j].y - node[i].y) >= minx) break;
for(int k = j - 1; k >= 0; k --)
dist_minx(Q[j],node[i],Q[k]);
}
Q.push_back(node[i]);
}
Q.clear();
return ;
}
int main()
{
scanf("%d",&n);
for(int i = 1; i <= n; i ++)
{
scanf("%lf%lf",&node[i].x,&node[i].y);
node[i].id = i;
}
sort(node + 1,node + n + 1,cmpx);
Blocking(1,n);
printf("%0.6lf\n",minx);
}
标签:bsp ble typedef false div ++ 题目 平面最近点对 abs
原文地址:https://www.cnblogs.com/yrz001030/p/12340019.html