【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】

        /** 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 herenew 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>() {@Overridepublic int compare(Pair A, Pair B) {// TODO Auto-generated method stublong 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());}}
}