Ural 1397. Points Game

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假设玩家A取的石子集合是A，B的是B

val = Sum( distant(pi,pj) ) { i<j && i,j belong to A} - Sum( distant(pi,pj) ) {i<j && i,j belong to B}

= Sum( distant(pi,pj) ) {i,j belong to Q} - Sum( distant(pi,pj) ) { i belong to B && j belong to Q}

我们记录每个石子到其他石子的距离的总和，然后排序。

这样的方法就是贪心取。。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<vector>#include<set>#include<queue>#include<map>#include<sstream>#include<iostream>using namespace std;#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)#define bug puts("Fuck");#define LL long long#define pb push_back#define mp make_pair#define nMax 1010#define eps 1e-8#define inf 0x7fffffffdouble dir(double x) { return x*x; }class P{public:double x,y;void read() { scanf("%lf%lf",&x,&y);};double dis(P b) {return sqrt(dir(x-b.x) + dir(y-b.y));}};P p[nMax];int n;int input() {if(scanf("%d",&n)==EOF) return 0;for(int i=0;i<2*n;i++) p[i].read();n <<= 1;return 1;}double sum[nMax];void sovle(){double total=0.0;for(int i=0;i<n;i++) {sum[i] = 0;for(int j=0;j<n;j++) sum[i] += p[i].dis(p[j]);total += sum[i];}total /= 2;sort(sum,sum+n);for(int i=0;i<n;i+=2) total -= sum[i];printf("%.3lf\n",total+1e-9);}int main(){#ifndef ONLINE_JUDGE  freopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);#endifwhile(input())sovle();return 0;}