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CodeForces Gym 100685J Just Another Disney Problem (STL，排序)

2024-08-11 16:50:26   213人阅读

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)


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