记忆化深搜+dp

hdu  1078

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2840    Accepted Submission(s): 1105

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 11 2 510 11 612 12 7-1 -1

 

Sample Output

37

 

每次能朝一个方向跑最多k格，然后格子里面的数必须递增，样例分析：1+2+5+6+11+12 = 37 

刚开始用DFS做TLE了，上网搜了代码，看了半天才弄懂，DP太弱，

因为是要求下一步必须比前一步的值大，所以吃过的不用做标记，下一次迟到的肯定不是上一次的

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=105;int n,m,ans,k;int map[maxn][maxn];int dp[maxn][maxn];// 存图int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};int dfs(int x,int y)// 状态就是当前点的坐标{int max1=0;if(dp[x][y]>0)return dp[x][y];for(int i=0;i<4;i++){    int num=1;for(num=1;num<=k;num++))//一次可以跳1-k步，k不中只吃到一次{    int nx=x+num*dir[i][0];// 状态的扩展int ny=y+num*dir[i][1];if(nx>=1&&nx<=n&&ny>=1&&ny<=n&&map[x][y]<map[nx][ny])//条件要求下一步必须比前一步的值大{    int ret=dfs(nx,ny);     nx=nx+dir[i][0];     ny=ny+dir[i][1];     max1=max(max1,ret);}}}dp[x][y]=max1+map[x][y];//本身也要加上//状态转移方程 DP[p][q]=max(dp[i][q],dp[p][j])+map[p][q]  其中i就是左右，j是上下从四周找到最大的数，return dp[x][y];}int main(){while(scanf("%d%d",&n,&k)!=EOF){    if(n==-1&&k==-1)    break;int sx,sy;for(int i=1;i<=n;i++)for(int j=1;j<=n;j++){cin>>map[i][j];}ans=0;memset(dp,0,sizeof(dp));ans=dfs(1,1);printf("%d\n",ans);//输出答案}return 0;}