답안 #815399

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
815399	2023-08-08T14:46:34 Z	jophyyjh	고대 책들 (IOI17_books)	C++14	50 / 100	100 ms	17228 KB

books

/**
 * Cute problem~ I started by experimenting different examples. Nope. I noticed nothing,
 * and was stuck for a while. I initially guessed that maybe Aryan never moves a book to
 * an intermediate table, rather than directly to its destination, but I quickly dismissed
 * the thought because of (2,3,0,1). In fact, the min num of steps to solve (2,3,0,1) is 8,
 * not 10. Then, I wonder, can I prove that 8 is the minimum? Yep! That was how I found the
 * way. I chose to consider some monovariant/invariant. A common one for sorting is
 * sum{|i - p_i|}, i.e. the sum of distances between a book and its destination. This sum
 * is exactly 8 for (2,3,0,1), so 8 is indeed the min.
 * 
 * sum{|i-p_i|} is a minimum because every step of Aryan can only bring a book at most 1
 * table closer to its destination. The min is achieved iff Aryan takes a book towards its
 * right place at every step, regardless of whether a book has been put down in an
 * intermediate table. What's a convenient way to achieve sum{|i-p_i|}? Note that a
 * permutation can be seen as the collection of some cycles. For each cycle, we can just
 * pick the books in it, in order. Let's call such traversal the "min-cost traversal" of
 * the cycle. For example, if the perm is just one cycle of size n, then the sum is indeed
 * sum{|i-p_i|}, achieved with just one complete "min-cost traversal".
 * 
 * For convenience, let's call sum{|i-p_i|} the "principal cost". Note that the principal
 * cost isn't always the min num of steps. Let c_1, c_2, ..., c_k be all the non-trivial
 * cycles (len>1, containing elements that have to be moved) representing our perm. Let the
 * "repr range" of c_i be [smallest_j_in_c_i, greatest_j_in_c_i]. We've already
 * established that a single cycle can be handled with one min-cost traversal. If the repr
 * ranges of c_i and c_j intersect, it can be proved that their min-cost traversal can be
 * "combined" (with no additional steps needed), which means the k cycles' repr ranges can
 * now be seen as a few disjoint repr ranges. The gaps between these disjoint ranges are
 * definitely not covered in the principal cost, but Aryan must walk through each gap at
 * least twice (since he has to return to s too).
 * 
 * Finally, if p[i] == i, i.e. i isn't in any non-trivial cycle, we have to start from s
 * and walk to a j with p[j] != j. We can prove that this j can be/is the closest j to s
 * with p[j] != j. These 3 parts of cost make up the final cost.
 * 
 * Time Complexity: O(n * log(n))           (maths, permutation)
 * Implementation 1                 (Full solution)
*/

#include <bits/stdc++.h>
#include "books.h"

typedef long long   ll;
typedef std::vector<int>    vec;


ll minimum_walk(vec perm, int s) {
    int n = perm.size();

    std::vector<bool> visited(n, false);
    vec ls, rs;
    int cs = 0, closest = n;
    ll cost = 0;
    for (int src = 0; src < n; src++) {
        if (visited[src] || src == perm[src])
            continue;
        int l = n, r = -1;
        for (int pt = src;; pt = perm[pt]) {
            visited[pt] = true;
            closest = std::min(closest, std::abs(s - pt));
            cost += std::abs(pt - perm[pt]), l = std::min(l, pt), r = std::max(r, pt);
            if (perm[pt] == src)
                break;
        }
        ls.push_back(l);
        rs.push_back(r);
        cs++;
    }
    if (cs == 0)
        return 0;
    cost += 2 * closest;
    
    std::sort(ls.begin(), ls.end());
    std::sort(rs.begin(), rs.end());
    for (int t = ls.front(), i = 0, j = 0, level = 0; t < rs.back(); t++) {
        while (i < cs && ls[i] <= t)
            level++, i++;
        while (j < cs && rs[j] <= t)
            level--, j++;
        if (level == 0)
            cost += 2;
    }
    return cost;
}

Subtask #1
12.0 / 12.0

#	결과	실행 시간	메모리	Grader output
1	Correct	0 ms	212 KB	Output is correct
2	Correct	0 ms	212 KB	Output is correct
3	Correct	0 ms	212 KB	Output is correct
4	Correct	0 ms	212 KB	Output is correct
5	Correct	0 ms	212 KB	Output is correct
6	Correct	0 ms	212 KB	Output is correct
7	Correct	0 ms	212 KB	Output is correct
8	Correct	0 ms	212 KB	Output is correct
9	Correct	0 ms	212 KB	Output is correct
10	Correct	1 ms	300 KB	Output is correct
11	Correct	1 ms	212 KB	Output is correct
12	Correct	0 ms	212 KB	Output is correct
13	Correct	0 ms	300 KB	Output is correct
14	Correct	0 ms	212 KB	Output is correct
15	Correct	0 ms	212 KB	Output is correct
16	Correct	0 ms	340 KB	Output is correct
17	Correct	0 ms	212 KB	Output is correct
18	Correct	0 ms	212 KB	Output is correct

Subtask #2
10.0 / 10.0

#	결과	실행 시간	메모리	Grader output
1	Correct	0 ms	212 KB	Output is correct
2	Correct	0 ms	212 KB	Output is correct
3	Correct	0 ms	212 KB	Output is correct
4	Correct	0 ms	212 KB	Output is correct
5	Correct	0 ms	212 KB	Output is correct
6	Correct	0 ms	212 KB	Output is correct
7	Correct	0 ms	212 KB	Output is correct
8	Correct	0 ms	212 KB	Output is correct
9	Correct	0 ms	212 KB	Output is correct
10	Correct	1 ms	300 KB	Output is correct
11	Correct	1 ms	212 KB	Output is correct
12	Correct	0 ms	212 KB	Output is correct
13	Correct	0 ms	300 KB	Output is correct
14	Correct	0 ms	212 KB	Output is correct
15	Correct	0 ms	212 KB	Output is correct
16	Correct	0 ms	340 KB	Output is correct
17	Correct	0 ms	212 KB	Output is correct
18	Correct	0 ms	212 KB	Output is correct
19	Correct	0 ms	212 KB	Output is correct
20	Correct	0 ms	212 KB	Output is correct
21	Correct	1 ms	212 KB	Output is correct
22	Correct	0 ms	300 KB	Output is correct
23	Correct	1 ms	296 KB	Output is correct
24	Correct	1 ms	212 KB	Output is correct
25	Correct	1 ms	304 KB	Output is correct
26	Correct	1 ms	212 KB	Output is correct
27	Correct	0 ms	300 KB	Output is correct
28	Correct	0 ms	212 KB	Output is correct
29	Correct	0 ms	212 KB	Output is correct

Subtask #3
28.0 / 28.0

#	결과	실행 시간	메모리	Grader output
1	Correct	0 ms	212 KB	Output is correct
2	Correct	0 ms	212 KB	Output is correct
3	Correct	0 ms	212 KB	Output is correct
4	Correct	0 ms	212 KB	Output is correct
5	Correct	0 ms	212 KB	Output is correct
6	Correct	0 ms	212 KB	Output is correct
7	Correct	0 ms	212 KB	Output is correct
8	Correct	0 ms	212 KB	Output is correct
9	Correct	0 ms	212 KB	Output is correct
10	Correct	1 ms	300 KB	Output is correct
11	Correct	1 ms	212 KB	Output is correct
12	Correct	0 ms	212 KB	Output is correct
13	Correct	0 ms	300 KB	Output is correct
14	Correct	0 ms	212 KB	Output is correct
15	Correct	0 ms	212 KB	Output is correct
16	Correct	0 ms	340 KB	Output is correct
17	Correct	0 ms	212 KB	Output is correct
18	Correct	0 ms	212 KB	Output is correct
19	Correct	0 ms	212 KB	Output is correct
20	Correct	0 ms	212 KB	Output is correct
21	Correct	1 ms	212 KB	Output is correct
22	Correct	0 ms	300 KB	Output is correct
23	Correct	1 ms	296 KB	Output is correct
24	Correct	1 ms	212 KB	Output is correct
25	Correct	1 ms	304 KB	Output is correct
26	Correct	1 ms	212 KB	Output is correct
27	Correct	0 ms	300 KB	Output is correct
28	Correct	0 ms	212 KB	Output is correct
29	Correct	0 ms	212 KB	Output is correct
30	Correct	91 ms	14924 KB	Output is correct
31	Correct	92 ms	14972 KB	Output is correct
32	Correct	74 ms	14952 KB	Output is correct
33	Correct	91 ms	16704 KB	Output is correct
34	Correct	90 ms	16708 KB	Output is correct
35	Correct	91 ms	16840 KB	Output is correct
36	Correct	91 ms	16716 KB	Output is correct
37	Correct	81 ms	15208 KB	Output is correct
38	Correct	86 ms	15028 KB	Output is correct
39	Correct	78 ms	14944 KB	Output is correct
40	Correct	91 ms	14940 KB	Output is correct
41	Correct	87 ms	14936 KB	Output is correct
42	Correct	86 ms	14924 KB	Output is correct
43	Correct	100 ms	17228 KB	Output is correct

Subtask #4
0 / 20.0

#	결과	실행 시간	메모리	Grader output
1	Incorrect	0 ms	212 KB	3rd lines differ - on the 1st token, expected: '3304', found: '2744'
2	Halted	0 ms	0 KB	-

Subtask #5
0 / 30.0

#	결과	실행 시간	메모리	Grader output
1	Correct	0 ms	212 KB	Output is correct
2	Correct	0 ms	212 KB	Output is correct
3	Correct	0 ms	212 KB	Output is correct
4	Correct	0 ms	212 KB	Output is correct
5	Correct	0 ms	212 KB	Output is correct
6	Correct	0 ms	212 KB	Output is correct
7	Correct	0 ms	212 KB	Output is correct
8	Correct	0 ms	212 KB	Output is correct
9	Correct	0 ms	212 KB	Output is correct
10	Correct	1 ms	300 KB	Output is correct
11	Correct	1 ms	212 KB	Output is correct
12	Correct	0 ms	212 KB	Output is correct
13	Correct	0 ms	300 KB	Output is correct
14	Correct	0 ms	212 KB	Output is correct
15	Correct	0 ms	212 KB	Output is correct
16	Correct	0 ms	340 KB	Output is correct
17	Correct	0 ms	212 KB	Output is correct
18	Correct	0 ms	212 KB	Output is correct
19	Correct	0 ms	212 KB	Output is correct
20	Correct	0 ms	212 KB	Output is correct
21	Correct	1 ms	212 KB	Output is correct
22	Correct	0 ms	300 KB	Output is correct
23	Correct	1 ms	296 KB	Output is correct
24	Correct	1 ms	212 KB	Output is correct
25	Correct	1 ms	304 KB	Output is correct
26	Correct	1 ms	212 KB	Output is correct
27	Correct	0 ms	300 KB	Output is correct
28	Correct	0 ms	212 KB	Output is correct
29	Correct	0 ms	212 KB	Output is correct
30	Correct	91 ms	14924 KB	Output is correct
31	Correct	92 ms	14972 KB	Output is correct
32	Correct	74 ms	14952 KB	Output is correct
33	Correct	91 ms	16704 KB	Output is correct
34	Correct	90 ms	16708 KB	Output is correct
35	Correct	91 ms	16840 KB	Output is correct
36	Correct	91 ms	16716 KB	Output is correct
37	Correct	81 ms	15208 KB	Output is correct
38	Correct	86 ms	15028 KB	Output is correct
39	Correct	78 ms	14944 KB	Output is correct
40	Correct	91 ms	14940 KB	Output is correct
41	Correct	87 ms	14936 KB	Output is correct
42	Correct	86 ms	14924 KB	Output is correct
43	Correct	100 ms	17228 KB	Output is correct
44	Incorrect	0 ms	212 KB	3rd lines differ - on the 1st token, expected: '3304', found: '2744'
45	Halted	0 ms	0 KB	-

답안 #815399

Compilation message

Subtask #1 12.0 / 12.0

Subtask #2 10.0 / 10.0

Subtask #3 28.0 / 28.0

Subtask #4 0 / 20.0

Subtask #5 0 / 30.0

Subtask #1
12.0 / 12.0

Subtask #2
10.0 / 10.0

Subtask #3
28.0 / 28.0

Subtask #4
0 / 20.0

Subtask #5
0 / 30.0