【2017 Multi-University Training Contest - Team 7】Hard challenge
2017-10-04 18:44
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【Link】:http://acm.hdu.edu.cn/showproblem.php?pid=6127
【Description】
平面上有n个点,每个点有一个价值,每两个点之间都有一条线段,定义线段的值为两个点价值的乘积,现在让你找一条过原点的直线(直线不经过任何一个节点),将这条直线所经过的所有线段的值求和,问最大的和是多少.
【Solution】
假设有一条线把x轴上方和x轴下方的点分开了;
这样这条线的答案就为(val上1+val上2+…+val上n)*(val下1+val下2+…+val下n);
把上边的点的权值加起来,下边的点的权值也加起来.然后做下乘法就好.
之后,我们只要一点一点地逆时针旋转这条直线就好了;
每次遇到的第一个点,就改变上半部分删掉它之后权值的改变量;
在所有里面取最大值即可;
在转的时候,不管是直线的哪一个地方,只要遇到了一个点就停下来;
然后计算改变量.
如果是在x轴的下方的点的话,就是从直线的下方变成上方.
如果是在x轴的上方…
遇到的是哪一点并不好判断!
于是,我们考虑把x轴下方的点按原点对称到上方来.
(记录它原来是下方的);
这样,我们只要按照角升序排一下.
然后顺序处理,就能知道下一个会遇到的点是哪一个点了.
因为不存在两点经过原点,所以不会出现重复点.
虽然我们把它翻到了x轴上方,但我们在处理的时候,还是在原图上基础上处理的,即每个点转到之后和直线的位置关系
只不过能更清楚的知道下一个遇到的点是什么
【NumberOf WA】
0
【Reviw】
老是重新算很麻烦,就尝试一步一步地改变.
【Code】
#include <bits/stdc++.h> 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 ri(x) scanf("%d",&x) #define rl(x) scanf("%lld",&x) #define rs(x) scanf("%s",x+1) #define oi(x) printf("%d",x) #define ol(x) printf("%lld",x) #define oc putchar(' ') #define os(x) printf(x) #define all(x) x.begin(),x.end() #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 = 5e4; struct abc{ LL x,y,val; int tag; double c; friend bool operator < (const abc &a,const abc &b){ return a.c > b.c; } }; int n; abc a[N+10]; LL sqr(LL x){ return x*x; } int main(){ //Open(); //Close(); int T; ri(T); while (T--){ ri(n); rep1(i,1,n){ rl(a[i].x),rl(a[i].y),rl(a[i].val); if (a[i].y < 0){ a[i].tag = 0; a[i].x = - a[i].x; a[i].y = - a[i].y; }else a[i].tag = 1; a[i].c = 1.0*a[i].x/(1.0*sqrt((double)(sqr(a[i].x)+sqr(a[i].y)))); } sort(a+1,a+1+n); LL s0 = 0,s1 = 0,temp = 0,ans; rep1(i,1,n){ if (a[i].tag==0) s0 += a[i].val; else s1 += a[i].val; } rep1(i,1,n) if (a[i].tag == 1) temp += s0*a[i].val; ans = temp; rep1(i,1,n){ if (a[i].tag == 1){ temp = temp - a[i].val*s0 + a[i].val*(s1-a[i].val); s1 -= a[i].val; s0 += a[i].val; }else{ temp = temp - a[i].val*s1 + a[i].val*(s0-a[i].val); s1 += a[i].val; s0 -= a[i].val; } ans = max(ans,temp); } ol(ans);puts(""); } return 0; }
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