前往小程序,Get更优阅读体验!
立即前往
首页
学习
活动
专区
工具
TVP
发布
社区首页 >专栏 >hdoj 1078 FatMouse and Cheese(记忆化搜索)

hdoj 1078 FatMouse and Cheese(记忆化搜索)

作者头像
xindoo
发布2021-01-22 12:50:35
2780
发布2021-01-22 12:50:35
举报
文章被收录于专栏:XINDOO的专栏XINDOO的专栏

题目链接

Problem 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.

………………

简单的记忆化搜索,和其他不一样的地方就是这个一次可以走K步,其他没啥!!

代码语言:javascript
复制
#include <iostream>
#include <string.h>
#include <stdio.h>

using namespace std;

int n, k, dp[101][101];
int map[101][101];

int dfs(int a,int b)
{
    if(dp[a][b]) return dp[a][b];
    dp[a][b] = map[a][b];
    for(int i = 1;i <= k; i++)
    {
        if(a+i < n && map[a+i][b]>map[a][b])
            dp[a][b] = max(dp[a][b],dfs(a+i,b)+map[a][b]);
        if(a-i >= 0 && map[a-i][b]>map[a][b])
            dp[a][b] = max(dp[a][b],dfs(a-i,b)+map[a][b]);
        if(b-i >= 0 && map[a][b-i]>map[a][b])
            dp[a][b] = max(dp[a][b],dfs(a,b-i)+map[a][b]);
        if(b+i < n && map[a][b+i]>map[a][b])
            dp[a][b] = max(dp[a][b],dfs(a,b+i)+map[a][b]);
    }
    return dp[a][b];
}

int main()
{
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        if(n==-1&&k==-1)
            break;
        for(int i = 0;i<n;i++)
        {
            for(int j = 0;j<n;j++)
                scanf("%d",&map[i][j]);
        }
        memset(dp,0,sizeof(dp));
        printf("%d\n",dfs(0,0));
    }
}
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2013-04-29,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档