FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14142 Accepted Submission(s): 5978
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
题意：给n,k两个整数，在n*n矩阵中，可以垂直或水平移动到达一个格点，最远移动距离为k,当前到达格点的权值一定要大于上一个起点的权值，一直到无法移动（在可移动范围内，权值没用比当前格点大的），输出权值之和。
思路：记忆化搜索
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int mat[105][105];//原权值
int dp[105][105];//保存以i,j为起点，k步内的最大值
int f[4][2]={0,1,1,0,-1,0,0,-1};
int n,k;
int dfs(int x,int y) {
int maxn=0;
if(dp[x][y]>=0) return dp[x][y];//已经找到以x,y为起点的最大值，直接返回
for(int i=1;i<=k;i++) {
for(int j=0;j<4;j++) {
int xx=x+i*f[j][0];
int yy=y+i*f[j][1];
if(xx>=0&&yy>=0&&yy<n&&xx<n&&mat[x][y]<mat[xx][yy]) {
maxn=max(maxn,dfs(xx,yy));
}
}
}
return dp[x][y]=maxn+mat[x][y];//以x,y为起点，k步内的最大值。
}
int main()
{
while(~scanf("%d%d",&n,&k)&&n!=-1&&k!=-1) {
for(int i=0;i<n;i++) {
for(int j=0;j<n;j++) {
scanf("%d",&mat[i][j]);
}
}
memset(dp,-1,sizeof(dp));
printf("%d\n",dfs(0,0));
}
return 0;
}