Top k path--动态规划

Description

A robot is located at the top-left corner of a m×nm×n grid. The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid. There is a non-negative integer in each small grid which means the score that the robot get when it passes this grid. Now you need to calculate the top kk total scores when the robot reach the destination.

Note: The top k total scores may contain same total score got from different paths.

Input

The first line of input are m and n,which denote the size of the grid. The second line of input is the kk. The next m lines are the score matrix. (1≤m,n,k≤1001≤m,n,k≤100,and 0≤score≤1000≤score≤100)

Output

There is only one line in the Output,which is a descending sequence consist of the top k total scores.

Sample1

Input

Output

13 13

Sample2

Input

Output

30 29 27

#include <iostream>
using namespace std;
const int maxn = 200;

int m, n, K;//矩阵m*n,输出TOPK的值

int dp[maxn][maxn][maxn];
int a[maxn][maxn];

void solve()
{

	dp[1][1][1] = a[1][1];
	for (int k = 2; k <= K; k++) dp[1][1][k] = -1;

	for (int j = 2; j <= n; j++)
	{
		dp[1][j][1] = dp[1][j - 1][1] + a[1][j];
		for (int k = 2; k <= K; k++) dp[1][j][k] = -1;
	}

	for (int i = 2; i <= m; i++)
	{
		for (int k = 1; k <= K; k++)
		{
			if (~dp[i - 1][1][k])  dp[i][1][k] = a[i][1] + dp[i - 1][1][k];
			else          dp[i][1][k] = -1;
		}


		for (int j = 2; j <= n; j++)
		{
			int ple = 1, pri = 1;
			int nex = 1;
			while (nex <= K)
			{
				int & x = dp[i - 1][j][ple];
				int & y = dp[i][j - 1][pri];

				if (x == -1 && y == -1) { dp[i][j][nex++] = -1; continue; }

				if (x >= y) { dp[i][j][nex++] = x + a[i][j]; ple++; }
				else { dp[i][j][nex++] = y + a[i][j]; pri++; }
			}

		}

	}


}


int main()
{
	
		cin >> m >> n >> K;
		for (int i = 1; i <= m; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				cin >> a[i][j];
			}
		}	

		solve();

		for (int k = 1; k <= K; k++)
		{
			cout << dp[m][n][k] << (k == K ? '\n' : ' ');//控制输出格式
		}
	 

	return 0;
}