3170: [Tjoi 2013]松鼠聚会

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=3170

Description

Input

表示为了聚会走的路程和最小为多少.

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100010
#define eps 1e-9
int Num;
//const int inf=0x7fffffff;   //§&szlig;§é§à§é¨f§3
const int inf=0x3f3f3f3f;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//**************************************************************************************struct node
{double x,y;int id;
}p[maxn];
double ans[maxn];
bool cmp1(node a,node b)
{return a.x<b.x;
}
bool cmp2(node a,node b)
{return a.y<b.y;
}int main()
{int n=read();double sumx=0,sumy=0;for(int i=0;i<n;i++){double x,y;scanf("%lf%lf",&x,&y);p[i].x = (x+y)/2;p[i].y = (x-y)/2;sumx += p[i].x;sumy += p[i].y;p[i].id = i;}sort(p,p+n,cmp1);double tmp=0;for(int i=0;i<n;i++){ans[p[i].id]+=(i)*p[i].x - tmp;ans[p[i].id]+=(sumx-tmp)-(n-i)*p[i].x;tmp+=p[i].x;}sort(p,p+n,cmp2);tmp=0;for(int i=0;i<n;i++){ans[p[i].id]+=(i)*p[i].y - tmp;ans[p[i].id]+=(sumy-tmp)-(n-i)*p[i].y;tmp+=p[i].y;}double Ans = ans[0];for(int i=0;i<n;i++)Ans = min(Ans,ans[i]);printf("%.0lf\n",Ans);
}