搜索与减枝：FatMouse and Cheese HDU - 1078

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本文介绍了一个有趣的问题：一只名为FatMouse的老鼠在一个n×n的城市网格中寻找并吃掉尽可能多的奶酪块。该问题涉及算法设计，特别是递归搜索和动态规划策略的应用。文章提供了详细的解决方案和代码实现。

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
问题描述
在第一行输入n和k。表示有n*n个老鼠洞，k表示每一次老鼠可以移动多长距离。
接下来的n行n列存储每一个老鼠洞中存储的奶酪的多少。每一次胖老鼠只能移动到存有更多奶酪的老鼠洞中。问老鼠从（0，0）点开始最多可以吃到多少奶酪。
问题分析
假设有一个函数int find(int x,int y);返回胖老鼠从当前节点开始最多能吃到多少奶酪。
那么我们只需要遍历从当前节点开始能够到达的点(x,y)计算find(x,y)取最大值加上当前节点的奶酪数量即为最大值。
注：要记得存储当前计算过的值避免重复计算
代码实现

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int array[102][102];
int array_max[102][102];
int find(int x,int y,const int k,const int N){
    if(array_max[x][y]!=-1)return array_max[x][y];
    int x_max=min(x+k+1,N);
    int y_max=min(y+k+1,N);
    int x_min=max(x-k,0);
    int y_min=max(y-k,0);
    int maxi=0;
    for(int i=x+1;i<x_max;i++)
        if(array[i][y]>array[x][y])
        maxi=max(maxi,array_max[i][y]=find(i,y,k,N));
    for(int i=x-1;i>=x_min;i--)
        if(array[i][y]>array[x][y])
        maxi=max(maxi,array_max[i][y]=find(i,y,k,N));
    for(int i=y+1;i<y_max;i++)
        if(array[x][i]>array[x][y])
        maxi=max(maxi,array_max[x][i]=find(x,i,k,N));
    for(int i=y-1;i>=y_min;i--)
        if(array[x][i]>array[x][y])
        maxi=max(maxi,array_max[x][i]=find(x,i,k,N));
    array_max[x][y]=maxi+array[x][y];
    return array_max[x][y];
}
int main(){
    int N,k;
    while(scanf("%d %d",&N,&k)==2){
        if(N==-1&&k==-1)break;
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++){
                scanf("%d",&array[i][j]);
                array_max[i][j]=-1;
            }
            int max=find(0,0,k,N);
            printf("%d\n",max);
    }
    return 0;
}