
给n*n地图,老鼠初始位置在(0,0),它每次行走要么横着走要么竖着走,每次最多可以走出k个单位长度,且落脚点的权值必须比上一个落脚点的权值大,求最终可以获得的最大权值
(题目很容易会理解错题意,道友小心)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std; #define met(a,b) (memset(a,b,sizeof(a)))
#define N 110
#define INF 0xffffff int a[N][N], dp[N][N];
int dir[][]={{-,},{,},{,-},{,}}; int DFS(int n, int k, int x, int y)
{
if(!dp[x][y])
{
int ans=; for(int i=; i<=k; i++)
{
int temp=; for(int j=; j<; j++)
{
int nx = x + dir[j][]*i;
int ny = y + dir[j][]*i; if(nx>= && nx<=n && ny>= && ny<=n && a[nx][ny]>a[x][y])
{
temp = max(temp, DFS(n, k, nx, ny));
}
}
ans = max(ans, temp);
}
dp[x][y] = ans + a[x][y];
}
return dp[x][y];
} int main()
{
int n, k; while(scanf("%d%d", &n, &k), n!=-||k!=-)
{
int i, j; met(a, );
met(dp, ); for(i=; i<=n; i++)
for(j=; j<=n; j++)
scanf("%d", &a[i][j]); printf("%d\n", DFS(n, k, , ));
} return ;
}
http://acm.hdu.edu.cn/showproblem.php?pid=1078
FatMouse and Cheese
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7774 Accepted Submission(s): 3221
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.
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.
1 2 5
10 11 6
12 12 7
-1 -1