1349 - Aladdin and the Optimal Invitation

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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.

Note

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

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

链接：http://lightoj.com/volume_showproblem.php?problem=1349

题意： 有n*m的 格子。 然后输入每个格子的人数。 最后决定所有人到达一个格子，要求每个人走的路都最少。

做法：横坐标为i 的 所有格子人数和 放在numr【i】里 。 纵坐标同理。 这个格子的 横坐标和纵坐标的选择是不会相互影响的。  我们只用找到 所有格子的总人数和，取一个中位数mid=sum/2，然后判断第mid个人在第几行，那行就是答案的横坐标。

因为，如果这个横坐标向下移，因为它本来是中位数，所以上面的人数肯定变得大于下面的人数。而结果会导致，这个横坐标上面的人  要多走一步，下面的人少走一步，最后就是总步数增加了。

int rnum[50010];
int lnum[50010];

int main()
{
	int t,n;
	int r,l,c,q,m;
	cin>>t;
	int cas=1;
	while(t--)
	{
		cin>>n>>m>>q;
		//50000
		for(int i=1;i<=n;i++)
			rnum[i]=0;
		for(int i=1;i<=m;i++)
			lnum[i]=0;
		int sum=0;
		while(q--)
		{
			scanf("%d%d%d",&r,&l,&c);
			sum+=c;
			rnum[r]+=c;
			lnum[l]+=c;
		}
		int ansr,ansc;
		int num=0;
		int mid=(sum+1)/2;
		for(int i=1;i<=n;i++)
		{
			num+=rnum[i];
			if(num>=mid)
			{
				ansr=i;
				break;
			}
		}
		num=0;
		for(int i=1;i<=m;i++)
		{
			num+=lnum[i];
			if(num>=mid)
			{
				ansc=i;
				break;
			}
		}
		printf("Case %d: %d %d\n",cas++,ansr,ansc);
	}
	return 0;
}