제출 #673718

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
673718		stevancv	정렬하기 (IOI15_sorting)	C++14	100 / 100	280 ms	23532 KiB

sorting

#include <bits/stdc++.h>
#include "sorting.h"
#define ll long long
#define ld long double
#define sp ' '
#define en '\n'
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
using namespace std;
const int N = 2e5 + 2;
const int mod = 1e9 + 7;
int findSwapPairs(int n, int s[], int m, int x[], int y[], int p[], int q[]) {
    int zero = 1;
    for (int i = 0; i < n; i++) zero &= s[i] == i;
    if (zero == 1) return 0;
    vector<int> inv(n);
    for (int i = 0; i < n; i++) inv[s[i]] = i;
    auto Can = [&] (int mid) {
      	for (int i = 0; i < m; i++) p[i] = q[i] = 0;
        vector<int> a(n);
        iota(a.begin(), a.end(), 0);
        for (int i = mid; i >= 0; i--) {
            swap(a[x[i]], a[y[i]]);
        }
        vector<int> b(n);
        for (int i = 0; i < n; i++) b[a[i]] = i;
        int j = mid;
        for (int i = 0; i < n; i++) {
            if (b[i] == inv[i]) continue;
            if (j < 0) return 0;
            p[j] = b[i]; q[j--] = inv[i];
            swap(a[b[i]], a[inv[i]]);
            swap(b[i], b[a[b[i]]]);
        }
        vector<int> f(n);
        iota(f.begin(), f.end(), 0);
        vector<int> invf = f;
        for (int i = 0; i <= mid; i++) {
            swap(f[x[i]], f[y[i]]);
            swap(invf[f[x[i]]], invf[f[y[i]]]);
            p[i] = invf[p[i]];
            q[i] = invf[q[i]];
        }
        return 1;
    };
    int l = 0, r = m - 1, ans = m - 1;
    while (l <= r) {
        int mid = l + r >> 1;
        if (Can(mid)) {
            ans = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }
    Can(ans);
    return ans + 1;
}

컴파일 시 표준 에러 (stderr) 메시지

sorting.cpp: In function 'int findSwapPairs(int, int*, int, int*, int*, int*, int*)':
sorting.cpp:48:21: warning: suggest parentheses around '+' inside '>>' [-Wparentheses]
   48 |         int mid = l + r >> 1;
      |                   ~~^~~

• Subtask 1: 8 / 8
• Subtask 2: 12 / 12
• Subtask 3: 16 / 16
• Subtask 4: 18 / 18
• Subtask 5: 20 / 20
• Subtask 6: 26 / 26