BZOJ3170: [Tjoi 2013]松鼠聚会

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000000000ll
 24 
 25 #define maxn 100000+5
 26 
 27 #define maxm 500+100
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define mod 1000000007
 44 
 45 using namespace std;
 46 
 47 inline int read()
 48 
 49 {
 50 
 51     int x=0,f=1;char ch=getchar();
 52 
 53     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 54 
 55     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 56 
 57     return x*f;
 58 
 59 }
 60 struct rec{int x,y;ll z;}a[maxn];
 61 int n;
 62 inline bool cmp1(rec a,rec b){return a.x<b.x;}
 63 inline bool cmp2(rec a,rec b){return a.y<b.y;}
 64 
 65 int main()
 66 
 67 {
 68 
 69     freopen("input.txt","r",stdin);
 70 
 71     freopen("output.txt","w",stdout);
 72 
 73     n=read();
 74     for1(i,n)
 75     {
 76         int x=read(),y=read();
 77         a[i].x=x+y;a[i].y=x-y;
 78     }
 79     sort(a+1,a+n+1,cmp1);
 80     ll sum=0;
 81     for1(i,n)
 82     {
 83         a[i].z+=(ll)a[i].x*(ll)(i-1)-sum;
 84         sum+=(ll)a[i].x;
 85     }
 86     sum=0;
 87     for3(i,n,1)
 88     {
 89         a[i].z+=sum-(ll)a[i].x*(ll)(n-i);
 90         sum+=(ll)a[i].x;
 91     }
 92     sort(a+1,a+n+1,cmp2);
 93     sum=0;
 94     for1(i,n)
 95     {
 96         a[i].z+=(ll)a[i].y*(ll)(i-1)-sum;
 97         sum+=(ll)a[i].y;
 98     }
 99     sum=0;
100     for3(i,n,1)
101     {
102         a[i].z+=sum-(ll)a[i].y*(ll)(n-i);
103         sum+=(ll)a[i].y;
104     }
105     ll ans=inf;
106     for1(i,n)ans=min(ans,a[i].z);
107     printf("%lld\n",ans>>1);
108 
109     return 0;
110 
111 }