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 Specification
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 Specification
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
Output for Sample Input
37【分析】
题意：老鼠从（0,0）开始每次可以往一个方向连续走[1,k]中任意的步数，但是每次要求所到达的位置的value要大于当前位置的value，求走到不能走为止，老鼠能吃到最大的value和
矩阵不大，直接考虑搜索....搜索很简单，按题意搜就完事了
但是显然复杂度超时了 粗略估计是O(4k^100)当然不可能真的这么大，那么显然dfs做的大量操作都是重复的操作...每个点只需要遍历一次找四个方向能去的4k个点之后能到达的最大value和就可以了
那么考虑记忆化搜索的话复杂度估计是O(100 * 4k)...每个点往上下左右四个方向各走k步
那也就没啥了....直接爆搜就完了...每个点记录下上下左右各k步返回的最大值加上当前value[x][y]记录即可
当然既然可以记忆化搜索，那么dp也是可以的...dp就比较麻烦一点，需要对所有点进行处理，因为显然要从value小的值开始搜索....当然难也不是特别难，但是肯定没有记忆化搜索那么容易理解了...
【代码】
#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <stack>
#include <set>
#include <deque>
#include <queue>
#include <algorithm>
#include <assert.h>
using namespace std;
#define clr(a,b) memset(a, b, sizeof(a))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define lson(x) (x << 1)
#define rson(x) (x << 1 | 1)
typedef long long LL;
typedef vector<LL> VI;
typedef pair<int,int> PII;
const LL mod=1000000007;
LL powmod(LL a,LL b) {LL res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
LL gcd(LL a,LL b) { return b?gcd(b,a%b):a;}
LL mod_inv(LL a, LL k) {return powmod(powmod(a, k), mod - 2) % mod;}
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
//std::ios::sync_with_stdio(false);
// head
int n,k;
int dirx[] = {0,-1,0,1};
int diry[] = {1,0,-1,0};
int vis[110][110];
int a[110][110];
int dfs(int x,int y){
if (vis[x][y]) return vis[x][y];
else{
int ans = 0 ;
for (int i = 1 ; i <= k ; ++i){
for (int j = 0 ; j < 4 ; ++j){
int xx = x + dirx[j] * i;
int yy = y + diry[j] * i;
if (a[xx][yy] <= a[x][y] || xx < 0 || xx >= n || yy < 0 || yy >= n) continue;
ans = max(ans , dfs(xx,yy));
}
}
return vis[x][y] = ans + a[x][y];
}
}
int main()
{
while (~scanf("%d%d",&n,&k) && n > 0 && k > 0 ){
memset(vis,0,sizeof(vis));
for (int i = 0 ; i < n ; ++i)
for (int j = 0 ; j < n ; ++j)
scanf("%d",&a[i][j]);
printf("%d\n", dfs(0,0));
}
return 0;
}