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  • (记忆化搜索) FatMouse and Cheese(hdu 1078)

    题目大意:
     
    给n*n地图,老鼠初始位置在(0,0),它每次行走要么横着走要么竖着走,每次最多可以走出k个单位长度,且落脚点的权值必须比上一个落脚点的权值大,求最终可以获得的最大权值 
     
    (题目很容易会理解错题意,道友小心)

     

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    
    #define met(a,b) (memset(a,b,sizeof(a)))
    #define N 110
    #define INF 0xffffff
    
    int a[N][N], dp[N][N];
    int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
    
    int DFS(int n, int k, int x, int y)
    {
        if(!dp[x][y])
        {
            int ans=0;
    
            for(int i=1; i<=k; i++)
            {
                int temp=0;
    
                for(int j=0; j<4; j++)
                {
                    int nx = x + dir[j][0]*i;
                    int ny = y + dir[j][1]*i;
    
                    if(nx>=1 && nx<=n && ny>=1 && ny<=n && a[nx][ny]>a[x][y])
                    {
                        temp = max(temp, DFS(n, k, nx, ny));
                    }
                }
                ans = max(ans, temp);
            }
            dp[x][y] = ans + a[x][y];
        }
        return dp[x][y];
    }
    
    int main()
    {
        int n, k;
    
        while(scanf("%d%d", &n, &k), n!=-1||k!=-1)
        {
            int i, j;
    
            met(a, 0);
            met(dp, 0);
    
            for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
               scanf("%d", &a[i][j]);
    
            printf("%d
    ", DFS(n, k, 1, 1));
        }
    
        return 0;
    }

    http://acm.hdu.edu.cn/showproblem.php?pid=1078

     
     

     

    FatMouse and Cheese

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7774    Accepted Submission(s): 3221


    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
     

    Sample Output
    37
     

    Source
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  • 原文地址:https://www.cnblogs.com/YY56/p/5448139.html
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