HDU-1078 FatMouse and Cheese ( 记忆化搜索 )

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

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

 

题目大意 给一个方阵，一老鼠从(0,0)出发，每次只能横走或竖走最多k格，且到达的地方权值要比当前格大，求最大权值和

记忆化搜索 看完题目，很容易想到dfs，但是不加优化的话，肯定会超时，这时候就需要记忆化搜索。简单模拟一下dfs可以发现，有很多重复的搜索，如果我们将重复的搜索储存起来，就可以快许多，于是可以用dp[i][j]表示以(i,j)为起点进行dfs的结果，从而下次扫到(i,j)时就不需要再扫一遍了。

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cmath>
 4 #include<string>
 5 #include<algorithm>
 6 using namespace std;
 7 
 8 int map[105][105];
 9 int dp[105][105];
10 int n, k;
11 int d[2][4] ={ -1, 1, 0, 0, 0, 0, -1, 1 };
12 
13 int dfs( int x, int y ){
14     int ans = 0;
15     if( !dp[x][y] ){
16         for( int i = 1; i <= k; i++ )
17             for( int j = 0; j < 4; j++ ){
18                 int dx = x + d[0][j] * i;
19                 int dy = y + d[1][j] * i;
20                 if( dx < 0 || dx >= n || dy < 0 || dy >= n ) continue;
21                 if( map[dx][dy] > map[x][y] ){
22                     ans = max( ans, dfs( dx, dy ) );
23                 }
24             }
25         dp[x][y] = ans + map[x][y];
26     }
27     return dp[x][y];
28 }
29 
30 int main(){
31     ios::sync_with_stdio( false );
32 
33     while( cin >> n >> k, ( n + 1 ) || ( k + 1 ) ){
34         for( int i = 0; i < n; i++ )
35             for( int j = 0; j < n; j++ )
36                 cin >> map[i][j];
37         memset( dp, 0, sizeof( dp ) );
38         cout << dfs( 0, 0 ) << endl;
39     }
40 
41     return 0;
42 }

 

转载于:https://www.cnblogs.com/hollowstory/p/5452188.html