제출 #212330

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
212330		4fecta	건물 4 (JOI20_building4)	C++14	100 / 100	460 ms	33812 KiB

building4

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;

#define ll long long
#define ld long double
#define pii pair<int, int>
#define f first
#define s second
#define readl(_s) getline(cin, (_s));
#define boost() cin.tie(0); cin.sync_with_stdio(0)

const int MN = 1000005;

int n, a[MN][2], mn[MN][2], mx[MN][2], valid[MN][2];
char ans[MN];

int main() {
    boost();

    memset(mn, 0x3f, sizeof(mn));
    cin >> n;
    for (int j = 0; j < 2; j++) for (int i = 1; i <= 2 * n; i++) cin >> a[i][j];
    valid[1][0] = 1;
    valid[1][1] = 1;
    mn[1][0] = mx[1][0] = 0;
    mn[1][1] = mx[1][1] = 1;
    for (int i = 2; i <= 2 * n; i++) {
        for (int j = 0; j < 2; j++) {
            for (int k = 0; k < 2; k++) {
                if (a[i][j] < a[i - 1][k] || !valid[i - 1][k]) continue;
                valid[i][j] = 1;
                mn[i][j] = min(mn[i][j], mn[i - 1][k] + j);
                mx[i][j] = max(mx[i][j], mx[i - 1][k] + j);
            }
        }
    }
    if (valid[2 * n][0] && mn[2 * n][0] <= n && mx[2 * n][0] >= n) {
        int j = 0, rem = n;
        for (int i = 2 * n; i > 0; i--) {
            ans[i] = 'A' + j;
            if (i == 1) break;
            rem -= j;
            if (a[i - 1][0] <= a[i][j] && mn[i - 1][0] <= rem && mx[i - 1][0] >= rem) j = 0;
            else j = 1;
        }
    } else if (valid[2 * n][1] && mn[2 * n][1] <= n && mx[2 * n][1] >= n) {
        int j = 1, rem = n;
        for (int i = 2 * n; i > 0; i--) {
            ans[i] = 'A' + j;
            if (i == 1) break;
            rem -= j;
            if (a[i - 1][0] <= a[i][j] && mn[i - 1][0] <= rem && mx[i - 1][0] >= rem) j = 0;
            else j = 1;
        }
    } else return 0 * printf("-1\n");
    for (int i = 1; i <= 2 * n; i++) printf("%c", ans[i]);

    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 89 / 89