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:
    Excel2010表格里设置每页打印时都有表头
    新手常见Python运行时错误
    如何查看某个端口被谁占用
    ubuntu更换阿里源
    c# 值类型与引用类型(转)
    vs2015 企业版 专业版 密钥
    csdn中使用Git的一些注意问题
    在notepad++中快速插入当前时间方法
    EF6 code first 新建项目注意问题
    vs2015新建web应用程序空模板和添加webapi的模板生成文件的比较
  • 原文地址:https://www.cnblogs.com/zero-begin/p/4488474.html
Copyright © 2011-2022 走看看