题意:平面上有许多点,过原点,作一条线把它们分为两个部分,求两边权重和的乘的最大值
思路:求每个点对于x轴的角度,以y轴为初始线分为两边(可能有点在y轴,加个精度判断),排序,每次扫一个点,看这点去到了哪一边,更新ans
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<stdlib.h>
#include<math.h>
#include<vector>
#include<list>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#include<numeric>
#include<functional>
using namespace std;
typedef long long ll;
const int maxn = 5e4+5;
const double PI = acos(-1.0);
struct data
{
int x,y,val;
double jiao;
}node[maxn];
int cmp(const data &a,const data &b)
{
return a.jiao < b.jiao;
}
int main(void)
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i = 1; i <= n; i++)
{
scanf("%d%d%d",&node[i].x,&node[i].y,&node[i].val);
if(node[i].x == 0)
node[i].jiao = PI/2.0;
else
node[i].jiao = atan((double)node[i].y / node[i].x);
}
sort(node+1,node+1+n,cmp);
ll ans;
int sum1 = 0,sum2 = 0;
for(int i = 1; i <= n; i++)
{
if((double)node[i].x < 1e-6)
sum1 += node[i].val;
else
sum2 += node[i].val;
}
ans = (ll)sum1 * sum2;
for(int i = 1; i <= n; i++)
{
if((double)node[i].x > 1e-6)
{
sum1 += node[i].val;
sum2 -= node[i].val;
}
else
{
sum1 -= node[i].val;
sum2 += node[i].val;
}
ans = max(ans,(ll)sum1*sum2);
}
printf("%lld\n",ans);
}
return 0;
}