3170: [Tjoi 2013]松鼠聚会
Time Limit: 10 Sec Memory Limit: 128 MBDescription
有N个小松鼠,它们的家用一个点x,y表示,两个点的距离定义为:点(x,y)和它周围的8个点即上下左右四个点和对角的四个点,距离为1。现在N个松鼠要走到一个松鼠家去,求走过的最短距离。
Input
第一行给出数字N,表示有多少只小松鼠。0<=N<=10^5
下面N行,每行给出x,y表示其家的坐标。
-10^9<=x,y<=10^9
Output
表示为了聚会走的路程和最小为多少。
Sample Input
6
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2
-4 -1
-1 -2
2 -4
0 2
0 3
5 -2
Sample Output
20
HINT
Source
转换一下就A了,,,
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<cstdlib> 7 #include<vector> 8 using namespace std; 9 typedef long long ll; 10 typedef long double ld; 11 typedef pair<int,int> pr; 12 const double pi=acos(-1); 13 #define rep(i,a,n) for(int i=a;i<=n;i++) 14 #define per(i,n,a) for(int i=n;i>=a;i--) 15 #define Rep(i,u) for(int i=head[u];i;i=Next[i]) 16 #define clr(a) memset(a,0,sizeof(a)) 17 #define pb push_back 18 #define mp make_pair 19 #define fi first 20 #define sc second 21 #define pq priority_queue 22 #define pqb priority_queue <int, vector<int>, less<int> > 23 #define pqs priority_queue <int, vector<int>, greater<int> > 24 #define vec vector 25 ld eps=1e-9; 26 ll pp=1000000007; 27 ll mo(ll a,ll pp){if(a>=0 && a<pp)return a;a%=pp;if(a<0)a+=pp;return a;} 28 ll powmod(ll a,ll b,ll pp){ll ans=1;for(;b;b>>=1,a=mo(a*a,pp))if(b&1)ans=mo(ans*a,pp);return ans;} 29 void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); } 30 //void add(int x,int y,int z){ v[++e]=y; next[e]=head[x]; head[x]=e; cost[e]=z; } 31 int dx[5]={0,-1,1,0,0},dy[5]={0,0,0,-1,1}; 32 ll read(){ ll ans=0; char last=' ',ch=getchar(); 33 while(ch<'0' || ch>'9')last=ch,ch=getchar(); 34 while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar(); 35 if(last=='-')ans=-ans; return ans; 36 } 37 const int N=100005; 38 ll X[N],Y[N],A[N],B[N],sA[N],sB[N],ans=1e15; 39 int main(){ 40 int n=read(),a,b; 41 for (int i=1;i<=n;i++){ 42 a=read(),b=read(); 43 X[i]=a+b; Y[i]=a-b; 44 A[i]=a+b; B[i]=a-b; 45 } 46 sort(A+1,A+n+1); 47 sort(B+1,B+n+1); 48 for (int i=1;i<=n;i++){ 49 sA[i]=sA[i-1]+A[i]; sB[i]=sB[i-1]+B[i]; 50 } 51 int p,q; 52 for (int i=1;i<=n;i++){ 53 p=lower_bound(A+1,A+n+1,X[i])-A; q=lower_bound(B+1,B+n+1,Y[i])-B; 54 ans=min(ans,(X[i]*p-sA[p])+((sA[n]-sA[p])-X[i]*(n-p))+(Y[i]*q-sB[q])+((sB[n]-sB[q])-Y[i]*(n-q))); 55 } 56 cout<<ans/2; 57 return 0; 58 }