HDOJ 1078 FatMouse and Cheese （DFS+DP）

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

题目大意：输入n，k，一个n*n的地图。从起点（1，1）开始，每次可以横向或纵向移动k步。移动时下一步地图的权值要比当前大，求一条拥有最大权值和的路线的权值。

题解：DFS+记忆当前点的最优答案即可。

AC代码：

#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};int dp[105][105];int mp[105][105];int n, k,ans;int dfs(int sx, int sy){    int kk = 0;    if(!dp[sx][sy])//如果已经得到当前点的最优答案，直接返回    {        for(int i = 1; i <= k; i++)        {            for(int s = 0; s < 4; s++)            {                int nx = sx + i * dir[s][0];                int ny = sy + i * dir[s][1];                if(nx >= 0 && nx < n && ny >= 0 && ny < n && mp[sx][sy] < mp[nx][ny])                {                    kk = max(kk, dfs(nx, ny));//当前点的最优答案                }            }        }        dp[sx][sy]=kk+mp[sx][sy];    }    return dp[sx][sy];}int main(){    while(scanf("%d%d", &n, &k),n!=-1&&k!=-1)    {        ans = 0;        for(int i = 0; i < n; i++)        {            for(int j = 0; j < n; j++)            {                scanf("%d", &mp[i][j]);            }        }        ans=dfs(0, 0);        printf("%d\n", ans);        memset(dp,0,sizeof(dp));    }    return 0;}