zoukankan      html  css  js  c++  java
  • Codeforces Round #302(Div. 2)——B—— Sea and Islands

    A map of some object is a rectangular field consisting of n rows and n columns. Each cell is initially occupied by the sea but you can cover some some cells of the map with sand so that exactly k islands appear on the map. We will call a set of sand cells to be island if it is possible to get from each of them to each of them by moving only through sand cells and by moving from a cell only to a side-adjacent cell. The cells are called to be side-adjacent if they share a vertical or horizontal side. It is easy to see that islands do not share cells (otherwise they together form a bigger island).

    Find a way to cover some cells with sand so that exactly k islands appear on the n × n map, or determine that no such way exists.

    Input

    The single line contains two positive integers nk (1 ≤ n ≤ 100, 0 ≤ k ≤ n2) — the size of the map and the number of islands you should form.

    Output

    If the answer doesn't exist, print "NO" (without the quotes) in a single line.

    Otherwise, print "YES" in the first line. In the next n lines print the description of the map. Each of the lines of the description must consist only of characters 'S' and 'L', where 'S' is a cell that is occupied by the sea and 'L' is the cell covered with sand. The length of each line of the description must equal n.

    If there are multiple answers, you may print any of them.

    You should not maximize the sizes of islands.

    Sample test(s)
    input
    5 2
    output
    YES
    SSSSS
    LLLLL
    SSSSS
    LLLLL
    SSSSS
    input
    5 25
    output
    NO
    大意:岛和海,问你布局,只要行相邻为1,列相邻为1,花式模拟
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int a[110][110];
    int j;
    int main()
    {
        int n,k;
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&k);
        int temp = 0;
        if(n%2 == 1){
            if( k > (1+n)*(1+n)/4 + (n-1)*(n-1)/4){
                printf("NO
    ");
                return 0;
            }
            else {
                printf("YES
    ");
              int i = 1;
                while(k > temp){
                  if(i%2 == 1)  
                     j = 1;
                  else  j = 2;
                  if(j == 1){
                      for(j = 1; j <= n ;j += 2){
                    a[i][j] = 1;
                    temp++;
                    if(temp == k)
                        break;
                      }
                      i++;
                  }
                  else {
                      for(j = 2; j <= n-1 ; j+=2){
                    a[i][j] = 1;
                    temp++;
                    if(temp == k)
                        break;
                      }
                      i++;
                   }
                }
            }
       }
        else {
            if(k > n*n/2){
            printf("NO
    ");
            return 0;
            }
            else {
            printf("YES
    ");
            int i = 1;
            while(k > temp){
                if(i%2 == 1)
                 j = 1;
                else j = 2;
                if(j == 1){
                    for(j = 1; j <= n - 1; j+=2){
                        a[i][j] = 1;
                        temp++;
                        if(temp == k) break;
                    }
                    i++;
                   }
                else {
                    for(j = 2; j <= n ; j+=2){
                        a[i][j] = 1;
                        temp++;
                        if(temp == k) break;
                    }
                    i++;
                }
            }
        }
        }
       
                for(int i = 1; i <= n ;i++){
                    for(int j = 1; j <= n ;j++){
                        if(a[i][j] == 0)
                        printf("S");
                        else if(a[i][j] == 1)
                        printf("L");
                    }
                    printf("
    ");
                }
                return 0;
            }
    

      

  • 相关阅读:
    [crontab]修改默认编辑器
    [mysql]忘记用户密码或者误删用户账号
    [vim]多行注释和多行删除
    [mysql]my.cnf在哪里
    [python]有中文字符程序异常的解决方案
    [Linux]虚拟机无法安装deepin15.9的解决方案
    Elasticsearch5.X IN Windows 10 系列文章(2)
    Elasticsearch5.X IN Windows 10 系列文章(1)
    HTTP Error 502.5
    centos7 yum install redis
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4488474.html
Copyright © 2011-2022 走看看