hdu 4311 Meeting point-1(3级)
2013-06-01 09:10
190 查看
Meeting point-1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2313 Accepted Submission(s): 731
[align=left]Problem Description[/align]
It has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth anniversary. Because the retired TJU-ACMers may live in different places around the world, it
may be hard to find out where to celebrate this meeting in order to minimize the sum travel time of all the retired TJU-ACMers.
There is an infinite integer grid at which N retired TJU-ACMers have their houses on. They decide to unite at a common meeting place, which is someone's house. From any given cell, only 4 adjacent cells are reachable in 1 unit of time.
Eg: (x,y) can be reached from (x-1,y), (x+1,y), (x, y-1), (x, y+1).
Finding a common meeting place which minimizes the sum of the travel time of all the retired TJU-ACMers.
[align=left]Input[/align]
The first line is an integer T represents there are T test cases. (0<T <=10)
For each test case, the first line is an integer n represents there are n retired TJU-ACMers. (0<n<=100000), the following n lines each contains two integers x, y coordinate of the i-th TJU-ACMer. (-10^9 <= x,y <= 10^9)
[align=left]Output[/align]
For each test case, output the minimal sum of travel times.
[align=left]Sample Input[/align]
4 6 -4 -1 -1 -2 2 -4 0 2 0 3 5 -2 6 0 0 2 0 -5 -2 2 -2 -1 2 4 0 5 -5 1 -1 3 3 1 3 -1 1 -1 10 -1 -1 -3 2 -4 4 5 2 5 -4 3 -1 4 3 -1 -2 3 4 -2 2
[align=left]Sample Output[/align]
26 20 20 56 Hint In the first case, the meeting point is (-1,-2); the second is (0,0), the third is (3,1) and the last is (-2,2)
[align=left]Author[/align]
TJU
[align=left]Source[/align]
2012 Multi-University Training Contest 2
[align=left]Recommend[/align]
zhuyuanchen520
思路:x,y到某点的曼哈顿距离可以分开算,因此,分开枚举以某点为中心的值球最小就好了。
预处理,kx[x]以某点 x为中心的所有x距离和,ky[]同理。kx[i]=kx[i-1]+(i-n+i)*(f[i].x-f[i-1].x);
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; const int mm=2e5+9; class node { public: __int64 x,y; int id; } f[mm]; bool cmpx(node a,node b) { return a.x<b.x; } bool cmpy(node a,node b) { return a.y<b.y; } __int64 kx[mm],ky[mm]; __int64 sum; __int64 aabs(__int64 x) { if(x<0)return -x; return x; } int main() { int cas;__int64 n; while(~scanf("%d",&cas)) { while(cas--) { scanf("%I64d",&n); for(int i=0; i<n; ++i) scanf("%I64d%I64d",&f[i].x,&f[i].y); sort(f,f+n,cmpx); kx[0]=0; for(int i=0; i<n; ++i) { kx[0]+=aabs(f[i].x-f[0].x); f[i].id=i; } for(__int64 i=1; i<n; ++i) kx[i]=kx[i-1]+(i-n+i)*(f[i].x-f[i-1].x); //for(int i=0;i<n;++i) //cout<<" "<<kx[i]<<" ";puts(""); sort(f,f+n,cmpy); ky[0]=0; for(int i=0;i<n;++i) ky[0]+=aabs(f[i].y-f[0].y); for(__int64 i=1;i<n;++i) ky[i]=ky[i-1]+(i+i-n)*(f[i].y-f[i-1].y); //for(int i=0;i<n ;++i) //cout<<" "<<ky[i]<<" ";puts(""); sum=6e18; for(int i=0;i<n;++i) { //cout<<f[i].id<<" "<<i<<" "<<kx[f[i].id]<<" "<<ky[i]<<" "<<ky[i]+kx[f[i].id]<<endl; if(sum>ky[i]+kx[f[i].id])sum=ky[i]+kx[f[i].id]; } printf("%I64d\n",sum); } } return 0; }
相关文章推荐
- hdu 4311 Meeting point-1(3级)
- 【HDU 4311】Meeting point-1(前缀和求曼哈顿距离和)
- hdu 4311 - Meeting point-1(预处理)
- hdu 4311 Meeting point-1
- hdu - 4311 - Meeting point-1 - 想法题
- HDU 4311 Meeting point-1 简单几何
- hdu 4311 Meeting point-1 #manhattan距离
- HDU 4311 ——Meeting point-1(暴力)
- HDU 4311 Meeting point-1
- HDU 4311 Meeting point-1(曼哈顿距离最小)
- HDU-4311 Meeting point-1 曼哈顿距离快速计算
- HDU 4311 Meeting point-1
- HDU 4311 Meeting point-1(曼哈顿距离优化枚举)
- HDU 4311 Meeting point-1
- HDU 4311 Meeting point-1 求一个点到其它点的曼哈顿距离之和
- HDU 4311,4312 Meeting point(曼哈顿距离,切比雪夫距离)
- HDU 4311 Meeting point-1
- Meeting point-1 HDU - 4311
- Hdu 4311-Meeting point-1 曼哈顿距离,前缀和
- HDU 4311 4312 Meeting point 平面上的Manhattan距离和Chebyshev距离