hdu 1078 FatMouse and Cheese【经典记忆化搜索】

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7299    Accepted Submission(s): 3010

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

记忆化搜索：算法上依然是搜索的流程，但是搜索到的一些解用动态规划的那种思想和模式作一些保存。

一般说来，动态规划总要遍历所有的状态，而搜索可以排除一些无效状态。

更重要的是搜索还可以剪枝，可能剪去大量不必要的状态，因此在空间开销上往往比动态规划要低很多。

记忆化算法在求解的时候还是按着自顶向下的顺序，但是每求解一个状态，就将它的解保存下来，

以后再次遇到这个状态的时候，就不必重新求解了。

这种方法综合了搜索和动态规划两方面的优点，因而还是很有实用价值的。

我的第一发记忆化搜索、经典题。

直接暴力敲的话，如果k比较大，图也在极限大小的时候，是会超时的，所以用记忆化搜索的方法来优化。

AC代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int a[105][105];int dp[105][105];int fx[4]={0,0,1,-1};int fy[4]={1,-1,0,0};int n,k;int dfs(int x,int y){    int ans=0;    if(dp[x][y]==0)    {        for(int i=1;i<=k;i++)        {            for(int j=0;j<4;j++)            {                int xx=x+fx[j]*i;                int yy=y+fy[j]*i;                if(xx>=1&&xx<=n&&yy>=1&&yy<=n)                {                    if(a[xx][yy]>a[x][y])                    {                        ans=max(ans,dfs(xx,yy));                    }                }            }        }        dp[x][y]=ans+a[x][y];    }    return dp[x][y];}int main(){    while(~scanf("%d%d",&n,&k))    {        if(n<0||k<0)break;        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&a[i][j]);            }        }        memset(dp,0,sizeof(dp));        printf("%d\n",dfs(1,1));    }}