K - FatMouse and Cheese

最新推荐文章于 2019-07-28 18:44:15 发布

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最新推荐文章于 2019-07-28 18:44:15 发布

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分类专栏：记忆法搜索

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博客讲述了FatMouse吃奶酪问题，城市是n*n的方格，老鼠在(0,0)位置，每次水平或垂直最多走k步，且下一个位置奶酪数要更多。问题是求老鼠最多能吃多少奶酪，题解采用从(1,1)开始的记忆法搜索，搜到(n,n)记录各点最多能吃的奶酪数。

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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

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

Sample Output

题意：n*n的格子里都是奶酪，一只老鼠在(0,0)位置，每次只能水平或垂直走最多k步，且下一次到达位置格子的奶酪数必须大于本次格子奶酪数，问这只老鼠最多能吃多少奶酪。

题解：从（1，1）开始搜索，搜到最后（n，n)记下每一个点的最多能吃多少奶酪，所以是记忆法搜索。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<string>
#include<map>
#include<queue>
using namespace std;
int n,k;
int dp[110][110],s[110][110];
int nextt[4][2]={1,0,-1,0,0,1,0,-1};
int dfs(int x,int y)
{
    int maxx=0;
    if(dp[x][y])
        return dp[x][y];
    for(int i=1;i<=k;i++)
    {
        for(int j=0;j<4;j++)
        {
            int tx=x+nextt[j][0]*i;
            int ty=y+nextt[j][1]*i;
            if(tx<1||ty<1||tx>n||ty>n)
                continue;
            if(s[x][y]<s[tx][ty])
            {
                maxx=max(maxx,dfs(tx,ty));
            }
        }
    }
    dp[x][y]=maxx+s[x][y];
    return dp[x][y];
}
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        if(n==-1&&k==-1)
            break;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&s[i][j]);
            }
        }
        memset(dp,0,sizeof(dp));
        int ans;
        ans=dfs(1,1);
        printf("%d\n",ans);
    }
    return 0;
}