杭电1078（仓鼠吃奶酪，DFS+动态规划）

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/32768 K (Java/Others)

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

思路：

题意：有一个n * n的网格，每个洞里藏有0~100块奶酪；仓鼠从（0，0）开始走，由于一只猫的存在，所以仓鼠每次只能走1~k步；由于吃了奶酪仓鼠会变胖，所以每一次走到的洞里的奶酪都要比上一次多；求仓鼠最多能吃多少块奶酪。

方法：动态规划 + 深度优先搜索

AC代码：

#include 
#include 
#include 
using namespace std;

#define INF 0xfffffff
#define maxn 110
int dp[maxn][maxn];//dp[i][j]:从(i,j)开始能获得的最多奶酪 
int cheese[maxn][maxn];//记录每个位置有几块奶酪 
int n,k;//n个洞,每次最多走k步 

int dfs(int a,int b)
{
    if(dp[a][b]) return dp[a][b];
    dp[a][b]=cheese[a][b];
    for(int i=1;i<=k;i++)
    {
        //横坐标正向走i步,且下一个洞奶酪比当前的多 
        if(a+icheese[a][b])
        {
            dp[a][b]=max(dp[a][b],dfs(a+i,b)+cheese[a][b]);
        }
        //横坐标负向走i步,且下一个洞奶酪比当前的多
        if(a-i>=0&&cheese[a-i][b]>cheese[a][b])
        {
            dp[a][b]=max(dp[a][b],dfs(a-i,b)+cheese[a][b]);
        }
        //纵坐标负向走i步,且下一个洞奶酪比当前的多
        if(b-i>=0&&cheese[a][b-i]>cheese[a][b])
        {
            dp[a][b]=max(dp[a][b],dfs(a,b-i)+cheese[a][b]);
        }
        //纵坐标正向走i步,且下一个洞奶酪比当前的多
        if(b+icheese[a][b])
        {
            dp[a][b]=max(dp[a][b],dfs(a,b+i)+cheese[a][b]);
        }
    }
    return dp[a][b];
}

int main()
{
    while(cin>>n)
    {
        cin>>k;
        if(n==-1&&k==-1) break;
        memset(dp,0,sizeof(dp));//初始化
        for(int i=0;ifor(int j=0;jcin>>cheese[i][j];
            }
        }
        cout<0,0)<return 0;
}