题意：给定你大小未知的n个数，你允许有不超过一万次的询问，每次询问两个数，第i个数是否比第j个数小？然后后台会返回给你一个结果YES或者NO（即一行输入），
然后经过多次询问后，你需要给出一个正确的原未知序列的升序排列。
析：当时是真没看懂题意是啥意思，然后就放过了，如果看懂了，并不是很难么，这不就是一个排序么，你可以问后台要数据，然后你决定怎么排序，
我们只要处理这个排序就好，但是不能用sort进行排序，因为这个的复杂度并总不是nlogn，如果是极端数据就卡不过了，也就是说可能会超过询问10000次，
要么我们自己写个快排要么用stabble_sort，这个排序函数来解决。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
string s;
bool cmp(int a, int b){
cout << "1 " << a << " " << b << endl;
cin >> s;
return s[0] == ‘Y‘;
}
int main(){
cout.flush();
cin >> n;
for(int i = 0; i < n; ++i) a[i] = i+1;
stable_sort(a, a+n, cmp);
cout << "0";
for(int i = 0; i < n; ++i) cout << " " << a[i];
cout << endl;
return 0;
}
CodeForces Gym 100685J Just Another Disney Problem (STL，排序)
原文：http://www.cnblogs.com/dwtfukgv/p/5831417.html