hdu1078 FatMouse and Cheese(記憶化搜尋)

標籤：動態規劃   hdu   dfs   

轉載請註明出處：http://blog.csdn.net/u012860063

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

Problem DescriptionFatMouse 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.
 InputThere 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. 
OutputFor each test case output in a line the single integer giving the number of blocks of cheese collected. 
Sample Input

3 11 2 510 11 612 12 7-1 -1

 Sample Output

37

SourceZhejiang University Training Contest 2001 

題目大意：

　　　　有一個地圖，上面有正整數值，然後要求從（0, 0）點出發，每次只能走向比當前點值更大的點，並且每次可以向四個方向走最多k步，要求走到最後所能得到的最大值。

解題思路：

              由於題目要求有四個方向可以走，所以想到搜尋，而要求走得最優解。所以可以用記憶化搜尋。動態規劃基礎題，不過相當經典。

代碼如下：（93ms）

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
10907677	2014-06-26 10:55:49	Accepted	1078	93MS	392K	910 B	G++	天資4747tym

#include <cstdio>#include <cstring>int dp[147][147],map[147][147];int n,k;int xx[4]={0,0,1,-1};int yy[4]={1,-1,0,0};int DFS(int x,int y){int ans = 0,MAX = 0;if(!dp[x][y]){for(int i = 1; i <= k; i++){for(int j = 0; j < 4; j++){int dx = x+xx[j]*i;int dy = y+yy[j]*i;if(dx>=0 && dx<n && dy>=0 && dy<n && map[x][y]<map[dx][dy]){ans = DFS(dx,dy);//每條路徑能吃到的乳酪量if(ans > MAX)MAX = ans;}}dp[x][y] = MAX+map[x][y];//從dp[x][y]開始能吃到的最大乳酪量}}return dp[x][y];}int main(){while(~scanf("%d%d",&n,&k)){if(n == -1 && k == -1)break;memset(dp,0,sizeof(dp));memset(map,0,sizeof(map));for(int i = 0; i < n; i++){for(int j = 0; j < n; j++){scanf("%d",&map[i][j]);}}int TT = DFS(0,0);printf("%d\n",TT);}return 0;}