FatMouse and Cheese 【dp思想的dfs】

                            emmmmm 遞歸的dp？ 就是在搜索的時候像dp一樣找到最優子狀態並儲存。

題目：

FatMouse has stored some cheese in a city. The city can be considered as a square grid of dimension n: each grid location is labelled (p,q) where 0 <= p < n and 0 <= q < n. At each grid location Fatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he's going to enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where he stands and then runs either horizontally or vertically to another location. The problem is that there is a super Cat named Top Killer sitting near his hole, so each time he can run at most k locations to get into the hole before being caught by Top Killer. What is worse -- after eating up the cheese at one location, FatMouse gets fatter. So in order to gain enough energy for his next run, he has to run to a location which have more blocks of cheese than those that were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, compute the maximum amount of cheese FatMouse can eat before being unable to move.

InputThere are several test cases. Each test case consists of

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 blocks of cheese at locations (0,0) (0,1) ... (0,n-1); the next line contains the number of blocks of cheese at locations (1,0), (1,1), ... (1,n-1), and so on.
The input ends with a pair of -1's.
OutputFor each test case output in a line the single integer giving the number of blocks of cheese collected.
Sample Input

3 1
1 2 5
10 11 6
12 12 7
-1 -1

Sample Output

37

題意：應該很好理解。。。就懶得翻譯了

思路：

我們要找到由1，1走出來的最大解。那我們就以dp[1][1]爲最終狀態，而得到這個狀態就要知道它能到達位置的最大解。用遞歸的思想求解。dp[i][j]就是從i,j出發達到的最大解。代碼很容易看懂                                                                                                             第一次做，還是思想有堵塞。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#define ll long long
#define ld long double
#define INF 0x3f3f3f3f
using namespace std;
int map[102][102],dp[102][102];
int n,k;
int dfs(int x,int y){
	if(dp[x][y]>0)
	  return dp[x][y];
	int Max=0;
	for(int i=1;i<=k;i++){
		if((x+i)<=n&&map[x][y]<map[x+i][y])       //向下 
		    Max=max(Max,dfs(x+i,y));
		if((x-i)>0&&map[x][y]<map[x-i][y])        //向上 
		    Max=max(Max,dfs(x-i,y));
		if((y+i)<=n&&map[x][y]<map[x][y+i])       //向右 
		    Max=max(Max,dfs(x,y+i));
		if((y-i)>0&&map[x][y]<map[x][y-i])        //向左 
		    Max=max(Max,dfs(x,y-i));
	}
	dp[x][y]=Max+map[x][y];
	return  dp[x][y];
}
int main(){
	std::ios::sync_with_stdio(false);

	while(cin>>n>>k&&n!=-1){
		for(int i=1;i<=n;i++)
		  for(int j=1;j<=n;j++)
		     cin>>map[i][j];
	    memset(dp,0,sizeof dp);
		cout<<dfs(1,1)<<'\n';
	}
	
	return 0;
}