HDU 1078 FatMouse and Cheese

首先吐槽一下英语水平真心的不好啊，读了好几遍，又是百度又是谷歌的终于看明白了啊、、、

题目的大体意思是：一个老鼠他很厉害啊，他在一个最大100*100的数组中藏了很多的蛋糕，他每次(0，0)点开始吃，由于那只猫的原因他每次最多前进K个格子，但是走是有条件的就是每次走到的点所藏的蛋糕必须比之前的多、、

一开始以为是bfs但是因为他求的是吃到最多的蛋糕的数量所以会导致不是最长路的情况下也会吃到的最大值。

所以得用记忆化搜索、、然后dp一下、、公式是：DP[p][q]=max(dp[i][q],dp[p][j])+map[p][q]  其中i就是左右，j是上下

FatMouse and Cheese

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3824    Accepted Submission(s): 1531

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 11 2 510 11 612 12 7-1 -1

 

Sample Output

37

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;int dp[120][120], f[120][120];int k, n, t[4][2]={1,0,-1,0,0,1,0,-1};int dfs(int x, int y){    int i, j, xx, yy, _max = 0, ans;    if(!dp[x][y])    {        for(i = 1; i <= k; i++)        {            for(j = 0; j < 4; j++)            {                xx = x + i*t[j][0];                yy = y + i*t[j][1];                if(xx >= 0 && xx < n && yy >= 0 && yy < n && f[xx][yy] > f[x][y])                {                    ans = dfs(xx,yy);                    if(ans > _max)                        _max = ans;                }            }        }        dp[x][y] = _max + f[x][y];    }    return dp[x][y];}int main(){    int i, j;    while(~scanf("%d %d",&n, &k))    {        if(n == k && n == -1)            break;        for(i = 0; i < n; i++)            for(j = 0; j < n; j++)            scanf("%d",&f[i][j]);        memset(dp, 0 , sizeof(dp));        int sum;        sum = dfs(0,0);        printf("%d\n",sum);    }    return 0;}