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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#458142#8831. Chemistry Classucup-team3678#WA 53ms10948kbC++142.0kb2024-06-29 16:00:142024-06-29 16:00:16

Judging History

你现在查看的是最新测评结果

  • [2024-06-29 16:00:16]
  • 评测
  • 测评结果:WA
  • 用时:53ms
  • 内存:10948kb
  • [2024-06-29 16:00:14]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

const int N = 4e5 + 5;

long long a[N], f[N];
long long mx[1 << 20];

void build(int k, int l, int r) {
    mx[k] = 1e18;
    if (l == r) {
        return;
    }
    int mid = (l + r) >> 1;
    build(k << 1, l, mid);
    build(k << 1 | 1, mid + 1, r);
}

void modify(int k, int l, int r, int x, long long y) {
    mx[k] = min(mx[k], y);
    if (l == r) return;
    int mid = (l + r) >> 1;
    if (x <= mid) modify(k << 1, l, mid, x, y);
    else modify(k << 1 | 1, mid + 1, r, x, y);
}

long long query(int k, int l, int r, int x, int y) {
    if (l > y || r < x) return 1e18;
    if (l >= x && r <= y) return mx[k];
    int mid = (l + r) >> 1;
    return min(query(k << 1, l, mid, x, y), query(k << 1 | 1, mid + 1, r, x, y));
}

signed main() {
    int T; scanf("%d", &T);
    while (T--) {
        int n; scanf("%d", &n);
        long long A, B; scanf("%lld%lld", &A, &B);
        for (int i = 1; i <= 2 * n; ++i) {
            scanf("%lld", &a[i]), f[i] = 1e18;
        }
        sort(a + 1, a + n + 1);
        a[0] = 0;
        for (int i = 2; i <= n + n; i += 2) {
            if (a[i] - a[i - 1] > A) {
                a[0] = 1;
            }
        }
        if (a[0]) {
            puts("-1");
            continue;
        }
        build(1, 0, n + n), modify(1, 0, n + n, 0, 0ll);
        for (int i = 2, j = 1, k = 1; i <= 2 * n; i += 2) {
            while (a[i] - a[j] > A) j += 2;
            while (k <= i && a[i] - a[k] > B) k += 2;
            f[i] = f[i - 2] + (a[i] - a[i - 1] > B);
            if (j + 1 != i) {
                if (a[i - 1] - a[i - 2] > B) {
                    j = k = i - 1;
                } else {
                    f[i] = min(f[i], query(1, 0, n + n, j - 1, n + n) + 1);
                    f[i] = min(f[i], query(1, 0, n + n, k - 1, n + n));
                }
            }
            modify(1, 0, n + n, i, f[i]);
        }
        printf("%lld\n", n - f[n + n]);
    }
    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 1ms
memory: 7836kb

input:

4
1 2 1
42 69
2 3 1
1 2 3 4
2 5 1
6 1 3 4
5 19 1
1 7 8 9 10 11 12 13 14 20

output:

-1
2
1
4

result:

ok 4 number(s): "-1 2 1 4"

Test #2:

score: -100
Wrong Answer
time: 53ms
memory: 10948kb

input:

1
199996 67013419502794 1
403716252634677166 895717933735068492 410002430455111886 844431179242134559 322988383133810700 133475121268220299 481706326769800263 606871141911985391 195911124687409946 959578180866483093 930547702157856949 877914383714875160 994158366044742636 890855755285236186 69498488...

output:

-1

result:

wrong answer 1st numbers differ - expected: '0', found: '-1'