hdu 1078【記憶化搜尋入門題】FatMouse and Cheese

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
           

題意  

從（0,0）這個點出發，每次最多走K步，要求走到的格子數大于原來的，求問總那麼多步後總和最大是多少。

記憶化搜尋，很多路徑有重複的，跟DP有些相似。

【思考】

Select Code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <cmath>
using namespace std;
 
const int  maxn =105;
int a[maxn][maxn],dp[maxn][maxn];
int n,k;
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
 
int dfs(int x, int y){
    int tmp=0;//執行一次dfs就要化為0 ，不是很懂！！！！ 
    for (int j=1;j<=k;j++){
      for (int i=0;i<4;i++){
        int xx=x+dx[i]*j; //k步是水準或豎直的
        int yy=y+dy[i]*j;
        if (xx<0 || yy<0 || xx>=n || yy>=n) continue;
        if (a[x][y]>=a[xx][yy]) continue;
        if (dp[xx][yy]) { //如果已經探索過了，就沒必要繼續深搜了
            tmp=max(tmp,dp[xx][yy]); //取最大值
            continue;
        }
        tmp=max(tmp,dfs(xx,yy)); //取所有方案中的最大值
      }
    }
    return dp[x][y]=a[x][y]+tmp;///不懂遞歸過程！！！！！ ，了解在思考區 
}
 
int main(){
    std::ios::sync_with_stdio(false); //提高cin輸入效率
    while (cin >> n >> k){
        if (n==-1 && k == -1) break;
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++) cin>>a[i][j];
        }
        memset(dp,0,sizeof(dp));
        cout<<dfs(0,0)<<endl; //最優解在出發點上
    }
    return 0;
}