FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9586    Accepted Submission(s): 4039

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  

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挺麻烦的一道dfs 拿到手不难看出是一道搜索题  但是！没有条件！那跟枚举有啥区别，所以1s肯定会超时。。。只能加个DP了。加了DP也不好写，卡顿了好久，好不容易测试没问题了交了下还是超时。。。无奈之下看了别人的题解  只要加上if(dp[x][y]) return dp[x][y];就不会超时了，还是太年轻 值得纪念的一道好题。。。

#include<bits/stdc++.h>
using namespace std;
int n,k;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int mp[1000][1000];
int ans=0;
int dp[1000][1000];
int dfs(int x,int y){
	if(dp[x][y]) return dp[x][y]; //没有记忆化是会超时的 关键！
    int sum=0;
	for(int i=0;i<4;i++){
		for(int j=1;j<=k;j++){
			int x1=x+j*dir[i][0];
			int y1=y+j*dir[i][1];	
			if(x1>=n||y1>=n||x1<0||y1<0) break;
			else if(mp[x1][y1]>mp[x][y]){			
			sum=max(sum,dfs(x1,y1));
			}	
		}
	}
     return dp[x][y]=sum+mp[x][y];
}
int main(){
	while(scanf("%d%d",&n,&k)){
		if(n==-1&&k==-1) break;
		for(int i=0;i<n;i++)
		    for(int j=0;j<n;j++) scanf("%d",&mp[i][j]);
		    memset(dp,0,sizeof(dp));
		    cout<<dfs(0,0)<<endl;
	}
}