HDU_1078_FatMouse and Cheese

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11199    Accepted Submission(s): 4748

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 11 2 510 11 612 12 7-1 -1

Sample Output

37

 

Source

Zhejiang University Training Contest 2001

• 注意落脚点奶酪数量单调递增
• 则从坐标A到坐标B后不可能回到A
• 对于当前状态S，挑选所有可能下一状态所可能达到的最大奶酪数，和当前状态奶酪数相加即可
  
• 那么就可以对于每个状态做dp，记录从此状态出发可以达到的最大奶酪数，记忆化搜索

#include <iostream>#include <string>#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>using namespace std;typedef long long           LL ;typedef unsigned long long ULL ;const int    maxn = 1000 + 10  ;const int    inf  = 0x3f3f3f3f ;const int    npos = -1         ;const int    mod  = 1e9 + 7    ;const int    mxx  = 100 + 5    ;const double eps  = 1e-6       ;const double PI   = acos(-1.0) ;int n, k;int a[maxn][maxn], dp[maxn][maxn];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};bool bound(int x, int y){return 0<=x && x<n && 0<=y && y<n;}int solve(int x, int y){if(dp[x][y]+1){return dp[x][y];}else{int ans=0;for(int i=0;i<4;i++){for(int j=1;j<=k;j++){int nx=x+j*dx[i];int ny=y+j*dy[i];if(bound(nx,ny) && a[nx][ny]>a[x][y]){ans=max(ans,solve(nx,ny));}}}return dp[x][y]=ans+a[x][y];}}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(~scanf("%d %d",&n,&k)){if(-1==n && -1==k){break;}for(int i=0;i<n;i++){for(int j=0;j<n;j++){scanf("%d",&a[i][j]);dp[i][j]=-1;}}printf("%d\n",solve(0,0));}return 0;}