FatMouse and Cheese---老鼠和奶酪 dfs+记忆化搜索+回溯~~ 当你非常肯定思路对，还一直wa

本文已参与「新人创作礼」活动，一起开启掘金创作之路。

编辑

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 Specification

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 Specification

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
Output for Sample Input
37

dfs+记忆化搜索+回溯~~

当你非常肯定思路对，还一直wa的时候，首先看看自己定义的变量....是不是有的ambiguous。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,k,a[110][110];
int book[110][110];
int f[4][2]= {-1,0,1,0,0,1,0,-1};
int dfs(int p,int q)        ///注意！！习惯于一直定义i，j的小可爱们，下面定义的变量要避开i，j啊~~~~坑死我了哟。。。。
{
    if(book[p][q])return book[p][q];   ///记忆化搜索
    int ans=0;
    for(int i=0; i<4; i++)    ///4个方向
    {
        for(int j=1; j<=k; j++) ///移动步数
        {
            int dp=p+f[i][0]*j;
            int dq=q+f[i][1]*j;
            if(dp<0||dq<0||dp>=n||dq>=n||a[dp][dq]<=a[p][q])continue;
                                                    ///下一个值必须大于上一个值
            ans=max(ans,dfs(dp,dq));   ///dp，dq这一步选与不选
        }
        book[p][q]=ans+a[p][q];///加上此刻位置的奶酪数
    }
    return book[p][q];
}
int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        if(n==-1&&k==-1)break;
        memset(book,0,sizeof(book));
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                scanf("%d",&a[i][j]);
        printf("%d\n",dfs(0,0));
    }
    return 0;
}