zoj 1107 FatMouse and Cheese（动态规划 记忆化搜索）

FatMouse and Cheese

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Time Limit: 10 Seconds       Memory Limit: 32768 KB

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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

记忆化搜索，动态规划的实质是记忆化搜索

老鼠从地图左上角开始，走的下一个位置必须比当前位置的奶酪多，求能走的最长距离

View Code

#include <iostream>
# include<stdio.h>

using namespace std;

int map[105][105];
int vis[105][105];
int n,k;

bool judge(int x,int y)
{
    if(x>=0 && x<n && y>=0 &&y<n)
        return true;
    return false;
}
int dx[4]= {0,1,0,-1};
int dy[4]= {-1,0,1,0};
int dp(int x,int y)
{
    if(vis[x][y] != -1) return vis[x][y];
    int i,j;
    vis[x][y] = map[x][y];
    int max=-1,temp;
    for(i=1; i<=k; i++)
    {
        for(j=0; j<4; j++)
        {
            int a=x+i*dx[j];
            int b=y+i*dy[j];
            if(judge(a,b) && map[a][b]>map[x][y])
                if(max<(temp = map[x][y] + dp(a,b)))
                    max = temp;
        }
    }
    if(max != -1)
        vis[x][y] = max;
    return vis[x][y];
}
int main()
{
    int i,j;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        if(n==-1 && k==-1)
            break;
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
            {
                scanf("%d",&map[i][j]);
                vis[i][j] = -1;
            }
        printf("%d\n",dp(0,0));
    }
    return 0;
}