lightoj 1349 - Aladdin and the Optimal Invitation (中位数的运用)

1349 - Aladdin and the Optimal Invitation

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Time Limit: 4 second(s)Memory Limit: 32 MB

Finally Aladdin reached home, with the great magical lamp. He was happier than ever. As he was a nice boy, he wanted to share the happiness with all people in the town. So, he wanted to invite all people in town in some place such that they can meet there easily. As Aladdin became really wealthy, so, number of people was not an issue. Here you are given a similar problem.

Assume that the town can be modeled as an m x n 2D grid. People live in the cells. Aladdin wants to select a cell such that all people can gather here with optimal overall cost. Here, cost for a person is the distance he has to travel to reach the selected cell. If a person lives in cell (x, y) and he wants to go to cell (p, q), then the cost is |x-p|+|y-q|. So, distance between (5, 2) and (1, 3) is |5-1|+|2-3| which is 5. And the overall cost is the summation of costs for all people.

So, you are given the information of the town and the people, your task to report a cell which should be selected by Aladdin as the gathering point and the overall cost should be as low as possible.

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with a blank line. Next line contains three integers: m, n and q (1 ≤ m, n, q ≤ 50000), m and n denote the number of rows and columns of the grid respectively. Each of the next q lines contains three integers u v w (1 ≤ u ≤ m, 1 ≤ v ≤ n, 1 ≤ w ≤ 10000), meaning that there are w persons who live in cell (u, v). You can assume that there are no people in the cells which are not listed. You can also assume that each of the q lines contains a distinct cell.

Output

For each case, print the case number and the row and column position of the cell where the people should be invited. There can be multiple solutions, any valid one will do.

Sample Input

Output for Sample Input

2

 

5 1 1

2 1 10

 

5 5 4

1 1 1

2 2 1

4 4 1

5 5 1

Case 1: 2 1

Case 2: 3 3

Note

1.      This is a special judge problem; wrong output format may cause 'Wrong Answer'.

2.      Dataset is huge, use faster I/O methods.

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PROBLEM SETTER: JANE ALAM JAN

嗯，题目求曼哈顿距离，给出n*m的表格，p个位置各有w人，问在那个坐标集合使所有人走得距离最小，

思路：算出总人数，x y排序，位于中位数的那个人的xy 坐标

#include<cstdio>#include<algorithm>#include<cstring> using namespace std;struct T{int x,y,w;}a[50010];bool cmp1(T u,T v){return u.x<v.x;}bool cmp2(T u,T v){return u.y<v.y;}int main(){int t,n,m,q,k=1;int xx,yy;long long num,cnt;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&m,&q);memset(a,0,sizeof(a));num=0;for(int i=0;i<q;i++){scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].w);num+=a[i].w;}num=(num+1)/2;//所有的人 中位数 sort(a,a+q,cmp1);cnt=0;for(int i=0;i<q;i++)//中位数所在横坐标 {cnt+=a[i].w;if(cnt>=num){//printf("%d %d==\n",cnt,num);xx=a[i].x;break;}}sort(a,a+q,cmp2);cnt=0;for(int i=0;i<q;i++)//中位数所在纵坐标 {cnt+=a[i].w;if(cnt>=num){yy=a[i].y;break;}}printf("Case %d: %d %d\n",k++,xx,yy);}return 0;}