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2013
11-27

HDU 1078 FatMouse and Cheese-DFS-[解题报告] C++

FatMouse and Cheese

问题描述 :

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.

输入:

There are several test cases. Each test case consists ofa 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.

输出:

For each test case output in a line the single integer giving the number of blocks of cheese collected.

样例输入:

3 1
1 2 5
10 11 6
12 12 7
-1 -1

样例输出:

37

题目链接:Click
here~~

 

题意:在n*n的格子上,每个点各有若干块奶酪,胖老鼠从左上角出发,每次最多走k步(只能直走),且下一点必须比这一点的奶酪多,问最多能吃到多少块奶酪。

 

以前做过一道记忆搜索的题,叫skiing。好久不写深搜,犯了个很2的错误,把mmax声明成全局的了,WA了n次,rbl。

#include <stdio.h>
#include <string.h>
#include <iostream>

using namespace std;

#define N 105
#define CLR(a,v) memset(a,v,sizeof(a))

int map[N][N],dp[N][N];
int dir[]={1,0,0,1,-1,0,0,-1};
int n,k;
int Dfs(int x,int y)
{
    if(dp[x][y]+1)
        return dp[x][y];
    int _x,_y,mmax = 0;         //不要把这里的变量声明成全局变量
    for(int t=1;t<=k;t++)
    {
        for(int i=0;i<8;)
        {
            _x = x + t*dir[i++];
            _y = y + t*dir[i++];
            if(!(_x<1 || _y<1 || _x>n || _y>n) && map[_x][_y] > map[x][y])
                mmax = max(mmax,Dfs(_x,_y));
        }
    }
    return dp[x][y] = map[x][y]+mmax;
}
int main()
{
    while(scanf("%d%d",&n,&k),n+1)
    {
        CLR(dp,-1);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&map[i][j]);
        printf("%d\n",Dfs(1,1));
    }
    return 0;
}

  1. 漂亮。佩服。
    P.S. unsigned 应该去掉。换行符是n 不是/n
    还可以稍微优化一下,
    int main() {
    int m,n,ai,aj,bi,bj,ak,bk;
    while (scanf("%d%d",&m,&n)!=EOF) {
    ai = sqrt(m-1);
    bi = sqrt(n-1);
    aj = (m-ai*ai-1)>>1;
    bj = (n-bi*bi-1)>>1;
    ak = ((ai+1)*(ai+1)-m)>>1;
    bk = ((bi+1)*(bi+1)-n)>>1;
    printf("%dn",abs(ai-bi)+abs(aj-bj)+abs(ak-bk));
    }
    }

  2. A猴子认识的所有猴子和B猴子认识的所有猴子都能认识,这句话用《爱屋及乌》描述比较容易理解……

  3. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
    O(n) = O(n-1) + O(n-2) + ….
    O(n-1) = O(n-2) + O(n-3)+ …
    O(n) – O(n-1) = O(n-1)
    O(n) = 2O(n-1)