FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14070 Accepted Submission(s): 5928
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
Source
Zhejiang University Training Contest 2001
Recommend
小老鼠 每次 走的步数<=k，而且必须跑到一个比现在的洞更多的奶酪块。
dp[i][j]表示从i，j出发能获取的最大奶酪
一只名叫Top Killer的超级猫坐在他的洞附近，所以每次他都可以跑到最多的k个位置进入洞，然后被Top Killer抓住
计算FatMouse在无法移动之前可以吃的最大奶酪量。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
int mapp[110][110],dp[110][110],n,k;
int dxy[2][4]={1,0,-1,0,0,1,0,-1};
int dfs(int x,int y)
{
int maxx=0;
if(!dp[x][y])
{
for(int i=0;i<4;i++)
{
for(int j=1;j<=k;j++)
{
int dx=x+dxy[0][i]*j;
int dy=y+dxy[1][i]*j;
if(dx>=0&&dx<n&&dy>=0&&dy<n&&mapp[dx][dy]>mapp[x][y])
{
int temp=dfs(dx,dy);
if(temp>maxx)
{
maxx=temp;
}
}
}
}
dp[x][y]=maxx+mapp[x][y];
}return dp[x][y];
}
int main()
{
while(scanf("%d%d",&n,&k))
{
if(n==-1&&k==-1) break;
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
scanf("%d",&mapp[i][j]);
dp[i][j]=0;
}
}
int ans=dfs(0,0);
printf("%d\n",ans);
}return 0;
}
