zoukankan      html  css  js  c++  java
  • hdu 1078 FatMouse and Cheese

    FatMouse and Cheese

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 8274    Accepted Submission(s): 3466


    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
    #include<stdio.h>
    #include<iostream>
    #include<queue>
    using namespace std;
    int ma[105][105];
    int dx[]={
        0,0,1,-1
    };
    int dy[]={1,-1,0,0};
    int dp[105][105];
    int n,k;
    int dfs(int x,int y){
        if(dp[x][y]){
            return dp[x][y];
        }
        for(int i=1;i<=k;i++){
            for(int j=0;j<4;j++){
                int nx=dx[j]*i+x;
                int ny=dy[j]*i+y;
                if((nx>=0&&ny>=0)&&(nx<n&&ny<n)&&(ma[nx][ny]>ma[x][y])){
                    int z=dfs(nx,ny);
                    dp[x][y]=max(dp[x][y],z+ma[nx][ny]);
                }
     
            }
        }
        return dp[x][y];
    }
    
    int main(){
        while((scanf("%d%d",&n,&k))&&(n!=-1&&k!=-1)){
            for(int i=0;i<n;i++)
              for(int j=0;j<n;j++)
                dp[i][j]=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<n;j++){
                    scanf("%d",&ma[i][j]);
                }
            }
            printf("%d
    ",ma[0][0]+dfs(0,0));
        }
    }
    View Code
  • 相关阅读:
    java 查看 class文件编译时使用的编译器版本
    eclipse新建maven工程(web工程)
    U盘插入后电脑提示“使用驱动器F:中的光盘之前需要将其格式化”,该怎么办?(实测有用)
    随笔安装theano的那些故事(亲测有效,附安装包)
    Python学习4Python的交互
    Mysql热备份总结
    php中date()函数的使用
    在linux中用C语言实现ping命令的部分功能
    【转】/etc/sysconfig/目录详解
    python学习(一)
  • 原文地址:https://www.cnblogs.com/superxuezhazha/p/5710026.html
Copyright © 2011-2022 走看看