HDU1078 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): 1635    Accepted Submission(s): 594

Problem Description

FatMouse has stored some cheese in a city. The city canbe 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 locationFatmouse has hid between 0 and 100 blocks of cheese in a hole. Now he's goingto enjoy his favorite food.

FatMouse begins by standing at location (0,0). He eats up the cheese where hestands and then runs either horizontally or vertically to another location. Theproblem is that there is a super Cat named Top Killer sitting near his hole, soeach time he can run at most k locations to get into the hole before beingcaught by Top Killer. What is worse -- after eating up the cheese at onelocation, FatMouse gets fatter. So in order to gain enough energy for his nextrun, he has to run to a location which have more blocks of cheese than thosethat were at the current hole.

Given n, k, and the number of blocks of cheese at each grid location, computethe 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 ofcheese at locations (0,0) (0,1) ... (0,n-1); the next line contains the numberof 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 integergiving the number of blocks of cheese collected.

 

 

Sample Input

3 1

1 2 5

10 11 6

12 12 7

-1 -1

 

 

简单记忆化搜索,刚开始没搞懂题意,以为走k步是可以转弯的,原来是不可以,WA了几次.

代码如下:

 

#include<stdio.h>#include<string.h>#include<cstdlib>#include<cmath>#define N 105int to[5][2]={{0,0},{1,0},{-1,0},{0,1},{0,-1}};int flag[3]={0,1,-1};int dp[N][N],g[N][N];bool mark[N][N],vis[N][N];int n,k;inline int max (int a,int b){ return a>b?a:b;}inline int min(int a,int b){ return a<b?a:b;}inline bool fix(int x,int y){ if (x>=0 && x<n && y>=0 && y<n) return true; return false;}void DP(int a,int b){ int i,j,x,y; mark[a][b]=1; vis[a][b]=1; for (i=1;i<=4;++i) { for (j=1;j<=k;++j) { x=a+to[i][0]*j; y=b+to[i][1]*j; if (fix(x,y) && !vis[x][y] && g[x][y]>g[a][b]) { if (mark[x][y]) { dp[a][b]=max(dp[a][b],dp[x][y]+g[a][b]); } else { DP(x,y); dp[a][b]=max(dp[a][b],dp[x][y]+g[a][b]); } } } } if (dp[a][b]==0) dp[a][b]=g[a][b]; vis[a][b]=0;}int main(){ int i,j; while (scanf("%d%d",&n,&k)) { if (n==-1 && k==-1) break; memset(mark,0,sizeof(mark)); memset(dp,0,sizeof(dp)); memset(vis,0,sizeof(vis)); for (i=0; i<n; ++i) { for (j=0; j<n; ++j) { scanf("%d",&g[i][j]); } } mark[0][0]=1; DP(0,0); printf("%d/n",dp[0][0]); } return 0;}