Problem Description
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
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
Sample Input
3 1 1 2 5 10 11 6 12 12 7 -1 -1
Sample Output
37
题意:一个矩形的分割成很多网格,每个格子里都有食物,老鼠初始位置在(0,0),每次最多走k步,而且必须满足下一步到达的格子里面的食物大于当前,求最多能得到多少食物。
分析:用dp[i][j]记录从(i,j)开始能得到的最多的食物,则
dp[i][j]=max(dp[i+move[][0]*c][j+move[][1]*c])+map[i][j],其中move表示移动方向
代码:
#include<stdio.h>
#include<string.h>
#define N 101
int max(int a,int b)
{
if(a>b)
return a;
return b;
}
int n,k,map[N][N],dp[N][N];
int move[][2]={0,1,1,0,0,-1,-1,0};
int dfs(int x,int y)
{
if(dp[x][y])
return dp[x][y];
int i,j,a,b;
int num=0;
for(i=1;i<=k;i++)
{
for(j=0;j<4;j++)
{
a=x+move[j][0]*i;
b=y+move[j][1]*i;
if(a>=0&&a<n&&b>=0&&b<n&&map[a][b]>map[x][y])
num=max(num,dfs(a,b));
}
}
return dp[x][y]=num+map[x][y];
}
int main()
{
while(scanf("%d%d",&n,&k)!=EOF)
{
int i,j;
if(n==-1&&k==-1)
break;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
scanf("%d",&map[i][j]);
memset(dp,0,sizeof(dp));
dfs(0,0);
int sum=0;
for(i=0;i<n;++i)
for(j=0;j<n;++j)
sum=max(sum,dp[i][j]);
printf("%d\n",sum);
}
return 0;
}