zoukankan      html  css  js  c++  java
  • Seeding(dfs)

    Seeding

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 101   Accepted Submission(s) : 52
    Problem Description
    It is spring time and farmers have to plant seeds in the field. Tom has a nice field, which is a rectangle with n * m squares. There are big stones in some of the squares. Tom has a seeding-machine. At the beginning, the machine lies in the top left corner of the field. After the machine finishes one square, Tom drives it into an adjacent square, and continues seeding. In order to protect the machine, Tom will not drive it into a square that contains stones. It is not allowed to drive the machine into a square that been seeded before, either.

    Tom wants to seed all the squares that do not contain stones. Is it possible?


    Input

    The first line of each test case contains two integers n and m that denote the size of the field. (1 < n, m < 7) The next n lines give the field, each of which contains m characters. 'S' is a square with stones, and '.' is a square without stones.

    Input is terminated with two 0's. This case is not to be processed.


    Output

    For each test case, print "YES" if Tom can make it, or "NO" otherwise.


    Sample Input

    4 4
    .S..
    .S..
    ....
    ....
    4 4
    ....
    ...S
    ....
    ...S
    0 0


    Sample Output

    YES
    NO

     题解:跟南阳最小步数差不多,这个是耕过的地就不再耕了,所以带个回溯,记录步数,如果走的步数等于‘.’的总个数就符合找不到就不对;

    代码:

     1 #include<stdio.h>
     2 int m,n,tot,nos;
     3 bool flag;
     4 char table[7][7];
     5 void dfs(int x,int y){
     6     //max=tot>max?tot:max;
     7     if(table[x][y]=='S'||x<0||x>=n||y>=m||y<0)return;
     8     table[x][y]='S';
     9     tot++;
    10     if(tot==nos){
    11         flag=true;
    12         return;
    13     }
    14     dfs(x+1,y);
    15     dfs(x,y+1);
    16     dfs(x-1,y);
    17     dfs(x,y-1);
    18     tot--;//回溯;
    19     table[x][y]='.';//回溯;
    20     return ;
    21 }
    22 int main(){
    23     int x,y,i,j;
    24     while(~scanf("%d%d",&n,&m),n||m){//max=0;
    25         for(x=0;x<n;x++)scanf("%s",table[x]);
    26         nos=0;tot=0;flag=false;
    27         for(x=0;x<n;x++){
    28             for(y=0;y<m;y++){
    29                 if(table[x][y]=='.'){
    30                     nos++;
    31                     if(nos==1)i=x,j=y;
    32                 }
    33             }
    34         }
    35         dfs(i,j);
    36     //    printf("%d
    ",max);
    37         //for(x=0;x<n;x++)printf("%s
    ",table[x]);
    38         if(flag)puts("YES");
    39         else puts("NO");
    40     }
    41     return 0;
    42 } 
  • 相关阅读:
    数据结构(2)-链表
    数据结构(1)-数组
    SpringMVC学习总结(一)--Hello World入门
    基本数据类型对象的包装类
    关于String的相关常见方法
    常见的集合容器应当避免的坑
    再一次生产 CPU 高负载排查实践
    分表后需要注意的二三事
    线程池没你想的那么简单(续)
    线程池没你想的那么简单
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4703409.html
Copyright © 2011-2022 走看看