贪心也能过？

wangtairan114 @ 2023-10-19 20:01:15

by wangtairan114 @ 2023-10-19 20:02:19

我拿这个写一篇题解应该没人会打我吧

by tehllra @ 2023-10-19 20:41:40

智慧法挺多的吧

by wangtairan114 @ 2023-10-23 19:56:40

@tehllra 我专门设置了一个变量，用时间换数据点，调高一点就骗过去了

by arthurrr @ 2023-11-08 19:32:10

康康代码

by wangtairan114 @ 2023-11-11 08:57:39

@arthurrr

#include <cstring>
#include <string>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <stack>
#include <queue>
#include <limits.h>
#include <list>
#include <set>
#include <map>
using namespace std;
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)>(b)?(b):(a))
#define INF 0x3f3f3f3f
#define ll long long
#define sc scanf
#define pr printf
#define v1 first
#define v2 second
struct node{
    double x,y;
}a[200005];
int main()
{
    int n;
    sc("%d",&n);
    for(int i=1; i <= n; i++)
    {
        sc("%lf%lf",&a[i].x,&a[i].y);
    }
    sort(a+1,a+n+1,[](node x,node y){if(x.x!=y.x)return x.x<y.x;return x.y<y.y;});
    double minloc;
    double minn=1919810.00;
    for(int i=1; i <= n; i++)
    {
        minloc=1919810.00;
        int ok=30;
        for(int j=i-1; j >= 1; j--)
        {
            if(i!=j){
                minloc=min(minloc,sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y)));
                if(sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y))>minloc&&!ok)
                    break;
                else
                {
                    if(sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y))>minloc)
                        ok--;
                }
            }
        }
        for(int j=i+1; j <= n; j++)
        {
            if(i!=j){
                minloc=min(minloc,sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y)));
                if(sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y))>minloc&&!ok)
                    break;
                else
                {
                    if(sqrt(abs(a[i].x-a[j].x)*abs(a[i].x-a[j].x)+abs(a[i].y-a[j].y)*abs(a[i].y-a[j].y))>minloc)
                        ok--;
                }
            }
        }
        minn=min(minn,minloc);
    }
    pr("%.4lf",minn);
}

by 20230332fuben @ 2024-01-04 21:37:32

@wangtairan114 P7883的点能骗过去不

by wangtairan114 @ 2024-01-13 15:00:31

@20230332fuben 过不了 试了，要是过了帖子也不会发在这里