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  • 【Codeforces 922D】Robot Vacuum Cleaner

    【链接】 我是链接,点我呀:)
    【题意】

    让你把n个字符串重新排序,然后按顺序连接在一起 使得这个组成的字符串的"sh"子序列最多

    【题解】

    ```java /* * 假设A的情况好于B * 也就对应了A排在B的前面 * 那么 * As*Bh>Ah*Bs ① * 现在假设B又比C来得好 * 那么有 * Bs*Ch>Bh*Cs ② * 由①可得 * As/Ah>Bs/Bh * 由②可得 * Bs/Bh>Cs/Ch * 那么有 * As/Ah>As/Ch * 即As*Ch>Ah*As * 也就是说A排在C的前面比较优秀 * 也就是说这个大小关系有传递性 * 那么就直接排个序就好啦,按照两两之间排序的规则来写个比较函数就好了。 */ ``` StringBuilder比直接用字符串的"+"来得快

    【代码】

    import java.io.*;
    import java.util.*;
    
    public class Main {
        
        
        static InputReader in;
        static PrintWriter out;
            
        public static void main(String[] args) throws IOException{
            //InputStream ins = new FileInputStream("E:\rush.txt");
            InputStream ins = System.in;
            in = new InputReader(ins);
            out = new PrintWriter(System.out);
            //code start from here
            new Task().solve(in, out);
            out.close();
        }
        
        static int N = (int)1e5;
        static class Task{
            
            class Pair{
            	int s,h,i;
            	public Pair(int s,int h,int i) {
            		this.s = s;
            		this.h = h;
            		this.i = i;
            	}
            }
            
            public void solve(InputReader in,PrintWriter out) {
                Pair a[] = new Pair[N+10];
                String s[] = new String[N+10];
                int n;
            	n = in.nextInt();
            	for (int i = 1;i <= n;i++) {
            		s[i] = in.next();
            		int ss = 0,hh = 0;
            		for (int j = 0;j < (int)s[i].length();j++) {
            			char key = s[i].charAt(j);
            			if (key=='s') ss++;
            			if (key=='h') hh++;
            		}
            		a[i] = new Pair(ss,hh,i);
            	}
            	Arrays.sort(a, 1,n+1,new Comparator<Pair>() {
    
    				@Override
    				public int compare(Pair A, Pair B) {
    					// TODO Auto-generated method stub
    					long As,Ah,Bs,Bh;
    					As = A.s;Ah = A.h;
    					Bs = B.s;Bh = B.h;
    					long temp1 =As*Bh;long temp2 =Ah*Bs;
    					if (temp1>temp2) {
    						return -1;
    					}else if (temp1==temp2) {
    						return 0;
    					}else {
    						return 1;
    					}
    				}
            		
    			});
            	StringBuilder sb = new StringBuilder();
            	for (int i = 1;i <= n;i++) {
            		sb.append(s[a[i].i]);
            	}
            	String v = sb.toString();
            	n = v.length();
            	long ans = 0;
            	long cnts = 0;
            	for (int i = 0;i < n;i++) {
            		if (v.charAt(i)=='s') {
            			cnts++;
            		}else if (v.charAt(i)=='h'){
            			ans = ans + cnts;
            		}
            	}
            	out.println(ans);
            }
        }
    
        
    
        static class InputReader{
            public BufferedReader br;
            public StringTokenizer tokenizer;
            
            
    
            
    
            
            public InputReader(InputStream ins) {
                br = new BufferedReader(new InputStreamReader(ins));
                tokenizer = null;
            }
            
            public String next(){
                while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                    try {
                    tokenizer = new StringTokenizer(br.readLine());
                    }catch(IOException e) {
                        throw new RuntimeException(e);
                    }
                }
                return tokenizer.nextToken();
            }
            
            public int nextInt() {
                return Integer.parseInt(next());
            }
        }
    }
    
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    [LeetCode] Insert Interval
    [LeetCode] Permutation Sequence
    [LeetCode] Rotate List
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    [LeetCode] Simplify Path(可以不用看)
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/10582043.html
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