HDU 1078 FatMouse and Cheese（记忆化搜索）

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=1078

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13356    Accepted Submission(s): 5598

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.

 

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

 

Source

Zhejiang University Training Contest 2001

 

题目意思：

老鼠初始时在nXn的矩阵的（0.0），每次最多向一个方向移动k格，每次移动过去的那个格子里面的数值必须比当前所在格子里面的大，求出路径上所有数值总和最大值

分析：

记忆化搜索

就是搜的时候搜过的存起来，下次就不用搜了

解析全在代码：

code：

#include<bits/stdc++.h>
using namespace std;
#define max_v 105
int a[max_v][max_v];
int dp[max_v][max_v];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};//方向数组
int n,k,ans;
int dfs(int x,int y)
{
    if(dp[x][y]!=0)//已经搜过的直接返回 记忆化搜索核心
        return dp[x][y];
    int xx,yy;
    int sum=0;
    for(int i=1;i<=k;i++)//k步
    {
        for(int j=0;j<4;j++)//四个方向
        {
            xx=x+dir[j][0]*i;
            yy=y+dir[j][1]*i;//下一个到达的点的坐标
            if(xx<=0||xx>n||yy<=0||yy>n)//越界
                continue;
            if(a[xx][yy]>a[x][y])//要求下一个点奶酪大于本点
                sum=max(sum,dfs(xx,yy));//找max
        }
        dp[x][y]=a[x][y]+sum;//能找到的最大值加上当前点的奶酪等于从x,y出发最多能吃到的奶酪值
    }
    return dp[x][y];
}
int main()
{
    while(cin>>n>>k)
    {
        if(n==-1&&k==-1)
            break;
        memset(a,0,sizeof(a));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
            {
                cin>>a[i][j];
            }
        }
        memset(dp,0,sizeof(dp));//初始化，记忆化搜索核心 判断这点有没有搜过
        ans=dfs(1,1);//从1，1出发开始搜
        cout<<ans<<endl;
    }
    return 0;
}