FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3521 Accepted Submission(s): 1394
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.
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.
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
1 2 5
10 11 6
12 12 7
-1 -1
Sample Output
37
/* 记忆化搜索 当访问到一个点的时候，将从这个点出发遍历到其他的点，但是由于一个的被访问的次数 是很多次的，所以造成了很多的重叠。 用记忆化数组，保存最优的解。 val[102][102]代表该点所能到达的“最优值”，这样在当前的点，我们可以从所能遍历 的所有点中选择出最大的 val[][] + map[x][y] ==> 就是当前点的“最优值”. */ #include<stdio.h> #include<stdlib.h> #include<string.h> int n,m; int map[102][102]; int val[102][102]; int max(int x,int y) { if(x>y) return x; else return y; } int dfs(int x,int y,int num) { int i,l,r,sum=0; if(val[x][y]>0) return val[x][y]; //最值已知，返回。避免重复。 if(x+m>=n) r=n; else r=x+m; if(x-m<1) l=1; else l=x-m; for(i=l;i<=r;i++) { if(i!=x && num<map[i][y]) { sum=max(dfs(i,y,map[i][y]),sum); } } if(y+m>=n) r=n; else r=y+m; if(y-m<1) l=1; else l=y-m; for(i=l;i<=r;i++) { if(i!=y && num<map[x][i]) { sum=max(sum,dfs(x,i,map[x][i])); } } //选择出 “最优值”。 val[x][y]=sum+map[x][y]; return val[x][y]; } int main() { int i,j,k; while(scanf("%d%d",&n,&m)>0) { if(n==-1&&m==-1)break; for(i=1;i<=n;i++) for(j=1;j<=n;j++) scanf("%d",&map[i][j]); memset(val,0,sizeof(val)); k=dfs(1,1,map[1][1]); printf("%d ",k); } return 0; }