UVa10259 - Hippity Hopscotch(动态规划)

The game of hopscotch involves chalk, sidewalks, jumping, and picking thingsup. Our variant of hopscotch involves money as well. The game is playedon a square grid of dimension n: each grid location is labelled (p,q) where0 <= p < n and 0 <= q < n. Each grid location has on it a stack of between0 and 100 pennies.

A contestant begins by standing at location (0,0). The contestant collectsthe money where he or she stands and then jumps either horizontally orvertically to another square. That square must be within the jumpingcapability of the contestant (say, k locations) and must have morepennies than those that were on the current square.

Given n, k, and the number of pennies at each grid location, compute themaximum number of pennies that the contestant can pick up before beingunable to move.

Input Specification

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

• a line containing two integers between 1 and 100: n and k
• n lines, each with n numbers: the first line contains the number of pennies at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of pennies at locations (1,0), (1,1), ... (1,n-1), and so on.

Output Specification

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

• a single integer giving the number of pennies collected

Sample Input

1

3 1
1 2 5
10 11 6
12 12 7

Output for Sample Input

37

题意：从(0,0)位置开始，可以水平或者竖直跳转不大于k的步数，跳的下一步中的值要大小当前步的值，而结果是经过的方格的数的和，求其最大值

思路：f(x,y) = grid[x][y] + max{f[x][y - 1], ...,f[x][y - k], f[x][y + 1],...,f[x][y + k], f[x - 1][y],......,f[x - k][y], f[x + 1][y], ...., f[x + k][y]})

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 110;

int grid[MAXN][MAXN];
int f[MAXN][MAXN];
int n, k;

void input()
{
	scanf("%d%d", &n, &k);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			scanf("%d", &grid[i][j]);
		}
	}
}

int dp(int x, int y)
{
	if (f[x][y] != -1) return f[x][y];

	int tmp = grid[x][y];
	for (int i = 1; i <= k; i++) {
		if (x + i < n && grid[x + i][y] > grid[x][y]) {
			tmp = max(tmp, grid[x][y] + dp(x + i, y));
		}

		if (x - i >= 0 && grid[x - i][y] > grid[x][y]) {
			tmp = max(tmp, grid[x][y] + dp(x - i, y));
		}

		if (y + i < n && grid[x][y + i] > grid[x][y]) {
			tmp = max(tmp, grid[x][y] + dp(x, y + i));
		}

		if (y - i >= 0 && grid[x][y - i] > grid[x][y]) {
			tmp = max(tmp, grid[x][y] + dp(x, y - i));
		}
	}

	return f[x][y] = tmp;

}

void solve()
{
	memset(f, -1, sizeof(f));
	dp(0, 0);
	printf("%d\n", f[0][0]);
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("d:\\OJ\\uva_in.txt", "r", stdin);
#endif

	int cas;
	scanf("%d", &cas);
	while (cas--) {
		input();
		solve();
		if (cas) printf("\n");
	}
	return 0;
}