【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());
        }
    }
}