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傳送門：

http://acm.hdu.edu.cn/showproblem.php?pid=1078

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13356 Accepted Submission(s): 5598

Problem Description 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

Source Zhejiang University Training Contest 2001 題目意思： 老鼠初始時在nXn的矩陣的（0.0），每次最多向一個方向移動k格，每次移動過去的那個格子裏面的數值必須比當前所在格子裏面的大，求出路徑上所有數值總和最大值 分析： 記憶化搜索 就是搜的時候搜過的存起來，下次就不用搜了 解析全在代碼： code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 105
int a[max_v][max_v];
int dp[max_v][max_v];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};//方向數組
int n,k,ans;
int dfs(int x,int y)
{
    if(dp[x][y]!=0)//已經搜過的直接返回 記憶化搜索核心
        return dp[x][y];
    int xx,yy;
    int sum=0;
    for(int i=1;i<=k;i++)//k步
    {
        for(int j=0;j<4;j++)//四個方向
        {
            xx=x+dir[j][0]*i;
            yy=y+dir[j][1]*i;//下一個到達的點的坐標
            if(xx<=0||xx>n||yy<=0||yy>n)//越界
                continue;
            if(a[xx][yy]>a[x][y])//要求下一個點奶酪大於本點
                sum=max(sum,dfs(xx,yy));//找max
        }
        dp[x][y]=a[x][y]+sum;//能找到的最大值加上當前點的奶酪等於從x,y出發最多能吃到的奶酪值
    }
    return dp[x][y];
}
int main()
{
    while(cin>>n>>k)
    {
        if(n==-1&&k==-1)
            break;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>a[i][j];
            }
        }
        memset(dp,0,sizeof(dp));//初始化，記憶化搜索核心 判斷這點有沒有搜過
        ans=dfs(1,1);//從1，1出發開始搜
        cout<<ans<<endl;
    }
    return 0;
}

HDU 1078 FatMouse and Cheese（記憶化搜索）