題意：

n * n的正方形格子（每個格子均放了奶酪），老鼠從(0,0)開始，每次最多移動k步，可以選擇上下左右四個方向移動，下一個移動點奶酪塊數量必須要大于當前點。

整理模板ing…

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.

Input

There 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.

Output

For 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

分析：

記憶化搜索入門，當采用記憶化搜索時，不必事先確定各狀態的計算順序，但需要記錄每個狀態“是否已經計算過”。
總結一下記憶化搜索是啥:
1.不依賴任何 外部變量(調用函數內的變量也沒有變化)
2.答案以返回值的形式存在, 而不能以參數的形式存在(就是不能將 dfs 定義成 dfs(pos ,tleft , nowans ) , 這里面的 nowans 不符合要求).
3.對于相同一組參數, dfs 返回值總是相同的（參見第一條）
就這道題復雜度為：O($n^2$)：每個點只遍歷了一次，從O($2^n$)~O($n^2$)這是記憶化搜索算法在時間復雜度上的優越性

AC代碼

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int n,m; int dp[110][110],a[110][110]; int c[4][2]= {0,1,0,-1,1,0,-1,0}; int dfs(int u,int v) {if(dp[u][v]!=-1)return dp[u][v];dp[u][v]=a[u][v];for(int i=0; i<4; i++)for(int j=1; j<=m; j++){int x=u+c[i][0]*j;int y=v+c[i][1]*j;if(x<0||y<0||x>=n||y>=n||a[u][v]>=a[x][y])continue;dp[u][v]=max(dp[u][v],dfs(x,y)+a[u][v]);}return dp[u][v]; } int main() {while(~scanf("%d%d",&n,&m)){if(n==-1&&m==-1)break;for(int i=0; i<n; i++)for(int j=0; j<n; j++)scanf("%d",&a[i][j]);memset(dp,-1,sizeof(dp));printf("%d\n",dfs(0,0));}return 0; }

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