hdu1078 记忆化dfs

FatMouse and Cheese

 

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

---

题意：老鼠每次只能去LIS的hole，每次可以在col或者row方向走k步，求最大能吃到多大的cheese

思路：很简单的dp，深搜时注意记忆化dfs即可

理解错了，以为k步里面，每一小步都可以col或者row，其实每个k都确定了单一的方向了，所以tle了，但是要哭了，明明昨天做过类似的记忆化的，怎么会tle:cold_sweat:

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <algorithm>

const int inf = 0x3f3f3f;
const int MAXN = 1e3+10;

using namespace std;
int a[MAXN][MAXN];
int dp[MAXN][MAXN];
/*int mov[4][2] = {{1,1},{1,-1},{-1,1},{-1,-1}};*/
int n,k;

void init(){
    memset(dp,-1,sizeof(dp));
}

int check(int x,int y){
    if(x<0||x>=n||y<0||y>=n)return 0;
    else return 1;
}

int dfs(int x,int y){
    if(dp[x][y]!=-1)return dp[x][y];
    dp[x][y] = a[x][y];
    int nx,ny;
    /*for(int i=k;i>=0;i--){
        for(int j=k-i;j>=0;j--){
            if(i+j==0)continue;
            for(int w=0;w<4;w++){
                nx = x+mov[w][0]*i;
                ny = y+mov[w][1]*j;
                if(check(nx,ny)&&a[x][y]<a[nx][ny]){
                    //cout<<"ok"<<endl;
                    if(dfs(nx,ny)+a[x][y]>dp[x][y]){
                        dp[x][y] = dp[nx][ny]+a[x][y];
                    }
                }
            }
        }
    }*/
    //每次都是单向的
    for(int i=0;i<4;i++){
       // nx = x+move[j][0];
       // ny = y+move[j][1];
       nx = x;
       ny = y;
       for(int j=1;j<=k;j++){
            if(i==0)nx = x+j;
            else if(i==1)nx = x-j;
            else if(i==2)ny = y+j;
            else ny = y-j;
            if(check(nx,ny)&&a[x][y]<a[nx][ny]){
                //dfs(nx,ny,t+1);
                if(dfs(nx,ny)+a[x][y]>dp[x][y]){
                    dp[x][y] = dp[nx][ny]+a[x][y];
                }
            }
       }
    }

    return dp[x][y];
}

int main()
{
    while(scanf("%d%d",&n,&k)!=EOF&&n!=-1){
        init();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                scanf("%d",&a[i][j]);
            }
        }
        dfs(0,0);
        /*for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }*/
        cout<<dp[0][0]<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

View Code

 

 

 

转载于:https://www.cnblogs.com/EdsonLin/p/5479122.html