FatMouse and Cheese
Time Limit: 10 Seconds Memory Limit: 32768 KB
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 Specification
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 Specification
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
Output for Sample Input
37
【题意】
小仓鼠吃奶酪，有一个n*n的矩阵，每个单元格内是奶酪的数量，仓鼠可以往上下左右任意方向走1-k步，每一次获得的奶酪数量一定要比上一次大，计算仓鼠能吃到的最大奶酪数量。
【解题思路】
嗯...第一次接触记忆化搜索，其实和普通的搜索也是差不多的，就是记录掉了状态。因为当前点的最优解是航下左右四个方向各走k步，也就是所有可能到达的状态的下一步中最大的值，也就是说是这个点出去能到达的最大的答案。所以对于每次的(x,y)都已经遍历过所有能去的点，然后从这些返回值中找到最大值并记录给vis[x][y]，也就是说(x,y)这个点往后能到达的最大的答案是已知的状态，如果下次有另一条路也到达了(x,y)，就可以直接返回答案了。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
int dirx[]={-1,1,0,0};
int diry[]={0,0,1,-1};
int n,k;
int a[maxn][maxn],vis[maxn][maxn];
int dfs(int x,int y)
{
if(vis[x][y])return vis[x][y];
else
{
int ans=0;
for(int i=1;i<=k;i++)
{
for(int j=0;j<4;j++)
{
int xx=x+i*dirx[j];
int yy=y+i*diry[j];
if(a[xx][yy]>a[x][y] && xx>=0 && xx<n && yy>=0 && yy<n)
ans=max(ans,dfs(xx,yy));
}
}
return vis[x][y]=ans+a[x][y];
}
}
int main()
{
while(~scanf("%d%d",&n,&k) && n>0 &&k>0)
{
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
scanf("%d",&a[i][j]);
printf("%d\n",dfs(0,0));
}
return 0;
}