題目連結

FatMouse and Cheese

Time Limit: 10 Seconds      Memory Limit: 32768 KB

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

【題意】

小倉鼠吃乳酪，有一個n*n的矩陣，每個單元格內是乳酪的數量，倉鼠可以往上下左右任意方向走1-k步，每一次獲得的乳酪數量一定要比上一次大，計算倉鼠能吃到的最大乳酪數量。

【解題思路】

嗯...第一次接觸記憶化搜尋，其實和普通的搜尋也是差不多的，就是記錄掉了狀態。因為當前點的最優解是航下左右四個方向各走k步，也就是所有可能到達的狀態的下一步中最大的值，也就是說是這個點出去能到達的最大的答案。所以對於每次的(x,y)都已經遍歷過所有能去的點，然後從這些返回值中找到最大值並記錄給vis[x][y]，也就是說(x,y)這個點往後能到達的最大的答案是已知的狀態，如果下次有另一條路也到達了(x,y)，就可以直接返回答案了。

【程式碼】

#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
int dirx[]={-1,1,0,0};
int diry[]={0,0,1,-1};
int n,k;
int a[maxn][maxn],vis[maxn][maxn];
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+i*dirx[j];
                int yy=y+i*diry[j];
                if(a[xx][yy]>a[x][y] && xx>=0 && xx<n && yy>=0 && yy<n)
                    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;
}