답안 #533062

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
533062 2022-03-04T15:20:47 Z MinhAnhnd Regions (IOI09_regions) C++17
100 / 100
2671 ms 48032 KB
/* IOI 2009 "region" problem.
 *
 * Solution by Bruce Merry.
 *
 * This is the second solution described in the writeup. Briefly:
 * - queries are cached so that duplicate queries can be answered again quickly
 * - each new quality is answered in either O(A log B), O(B log A) or O(A + B),
 *   whichever is deemed fastest.
 */

#include <cstdio>
#include <algorithm>
#include <vector>
#include <climits>
#include <map>

using namespace std;

typedef long long ll;

/* An employee in the tree */
struct node
{
    int id;     /* New employee ID from a pre-order walk (0-based) */
    int region; /* Employee's home region */
    int low;    /* Lowest ID of managees */
    int high;   /* Highest ID of managees */
    vector<int> children;   /* Supervisees */
};

struct region
{
    vector<int> ids;                /* Sorted list of (new) employee IDs */
    /* Sorted of intervals with the same nesting level. Each pair is
     * (ID, depth) where ID is the left end-point of the interval (inclusive).
     * The right end-point is implicit from the following interval.
     */
    vector<pair<int, int> > ranges;
    int depth; /* Working depth during the DFS. */

    region() : ids(), ranges(), depth(0) {}
};

static int N, R, Q;
static vector<node> nodes;
static vector<region> regions;

/* Does a pre-order walk over the subtree rooted at root. id_pool contains
 * the next unused employee ID, and on return it will be updated to again
 * be the next available ID.
 *
 * This procedure builds the regions arrays, after which the tree is no
 * longer needed.
 */
static void process_tree(int root, int &id_pool)
{
    int id = id_pool++;
    int r = nodes[root].region;
    regions[r].ids.push_back(id);
    regions[r].depth++;
    /* Depth changed, so after this point we need a new range */
    regions[r].ranges.push_back(make_pair(id_pool, regions[r].depth));

    /* Recursively process children */
    for (size_t i = 0; i < nodes[root].children.size(); i++)
        process_tree(nodes[root].children[i], id_pool);
    regions[r].depth--;
    /* Undo the depth change, and start another interval after the last
     * managee.
     */
    regions[r].ranges.push_back(make_pair(id_pool, regions[r].depth));
}

/* Find smallest n such that 2^n <= x */
static int log2(int x)
{
    int ans = -1;
    while (x)
    {
        x >>= 1;
        ans++;
    }
    return ans;
}

/* Query in O(R2 log R1) time, by counting for each employee in r2. */
static ll query_by_id(const region &r1, const region &r2)
{
    ll ans = 0;
    for (size_t i = 0; i < r2.ids.size(); i++)
    {
        int pos = r2.ids[i];
        vector<pair<int, int> >::const_iterator site;
        /* Find the first range that starts at pos or later. This will
         * actually be the range after the one we want.
         */
        site = lower_bound(r1.ranges.begin(), r1.ranges.end(), make_pair(pos, INT_MAX));
        if (site == r1.ranges.begin())
        {
            /* pos is less than the start of the first range, so has no
             * manager in r2.
             */
            continue;
        }
        --site; /* Now we have the range we want. */
        ans += site->second;
    }
    return ans;
}

/* Query in O(R1 log R2) time, by counting for each employee in r1 */
static ll query_by_range(const region &r1, const region &r2)
{
    ll ans = 0;
    for (size_t i = 0; i + 1 < r1.ranges.size(); i++)
    {
        int pos1 = r1.ranges[i].first;
        int pos2 = r1.ranges[i + 1].first;
        ll depth = r1.ranges[i].second;

        /* Each employee from r2 in [pos1, pos2) has depth managers
         * from r1. Find the intersections of [pos1, pos2) with the
         * employee list for r2.
         */
        vector<int>::const_iterator first, last;
        first = lower_bound(r2.ids.begin(), r2.ids.end(), pos1);
        last = lower_bound(r2.ids.begin(), r2.ids.end(), pos2);
        ans += depth * (last - first);
    }
    return ans;
}

/* Query in O(R1 + R2) time, by counting for each employee in r1
 * but with a linear sweep instead of a binary search.
 */
static ll query_stitch(const region &r1, const region &r2)
{
    ll ans = 0;

    /* Find the first employee id that is in the first range */
    vector<int>::const_iterator id = r2.ids.begin();
    if (r1.ranges.empty())
        return 0;
    while (id != r2.ids.end() && *id < r1.ranges[0].first)
        id++;

    /* Iterate over the ranges as above */
    for (size_t i = 0; i + 1 < r1.ranges.size() && id != r2.ids.end(); i++)
    {
        int pos2 = r1.ranges[i + 1].first;
        ll depth = r1.ranges[i].second;

        /* Find the end of the section of employees from this range */
        vector<int>::const_iterator old_id = id;
        while (id != r2.ids.end() && *id < pos2)
            id++;
        ans += depth * (id - old_id);
    }
    return ans;
}

int main()
{
    scanf("%d %d %d", &N, &R, &Q);

    /* Load input and build tree */
    nodes.resize(N);
    regions.resize(R);
    scanf("%d", &nodes[0].region);
    nodes[0].region--;
    for (int i = 1; i < N; i++)
    {
        int parent;
        scanf("%d %d", &parent, &nodes[i].region);
        parent--;
        nodes[i].region--;
        nodes[parent].children.push_back(i);
    }

    /* Turn the tree into regions */
    int id_pool = 0;
    process_tree(0, id_pool);

    /* Process queries */
    map<pair<int, int>, ll> cache;
    for (int q = 0; q < Q; q++)
    {
        int r1, r2;
        scanf("%d %d", &r1, &r2);
        r1--;
        r2--;
        pair<int, int> key(r1, r2);
        if (cache.count(key))
        {
            /* Answer query from the cache */
            printf("%lld\n", cache[key]);
            fflush(stdout);
            continue;
        }

        /* Fudge factor to estimate the relative cost of binary search
         * versus linear walk. This will depend to some extent on the
         * memory system.
         */
        static const int LOG_FACTOR = 5;

        /* Pick the best query method */
        const region &region1 = regions[r1];
        const region &region2 = regions[r2];
        int size1 = region1.ids.size();
        int size2 = region2.ids.size();
        int costs[3] = {
            size1 * (log2(size2) + 2) * LOG_FACTOR,
            size2 * (log2(size1) + 2) * LOG_FACTOR,
            size1 + size2
        };
        ll ans = 0;
        switch (min_element(costs, costs + 3) - costs)
        {
        case 0:
            ans = query_by_range(region1, region2);
            break;
        case 1:
            ans = query_by_id(region1, region2);
            break;
        case 2:
            ans = query_stitch(region1, region2);
            break;
        }
        printf("%lld\n", ans);
        fflush(stdout);
        cache[key] = ans;
    }
    return 0;
}

Compilation message

regions.cpp: In function 'int main()':
regions.cpp:164:10: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
  164 |     scanf("%d %d %d", &N, &R, &Q);
      |     ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
regions.cpp:169:10: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
  169 |     scanf("%d", &nodes[0].region);
      |     ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
regions.cpp:174:14: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
  174 |         scanf("%d %d", &parent, &nodes[i].region);
      |         ~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
regions.cpp:189:14: warning: ignoring return value of 'int scanf(const char*, ...)' declared with attribute 'warn_unused_result' [-Wunused-result]
  189 |         scanf("%d %d", &r1, &r2);
      |         ~~~~~^~~~~~~~~~~~~~~~~~~
# 결과 실행 시간 메모리 Grader output
1 Correct 0 ms 200 KB Output is correct
2 Correct 1 ms 200 KB Output is correct
3 Correct 2 ms 200 KB Output is correct
4 Correct 7 ms 200 KB Output is correct
5 Correct 5 ms 348 KB Output is correct
6 Correct 14 ms 564 KB Output is correct
7 Correct 22 ms 512 KB Output is correct
8 Correct 41 ms 652 KB Output is correct
9 Correct 53 ms 1452 KB Output is correct
10 Correct 92 ms 1744 KB Output is correct
11 Correct 126 ms 2268 KB Output is correct
12 Correct 124 ms 3292 KB Output is correct
13 Correct 161 ms 3228 KB Output is correct
14 Correct 180 ms 4024 KB Output is correct
15 Correct 274 ms 8404 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 871 ms 10248 KB Output is correct
2 Correct 923 ms 8920 KB Output is correct
3 Correct 1685 ms 15472 KB Output is correct
4 Correct 292 ms 5192 KB Output is correct
5 Correct 429 ms 8472 KB Output is correct
6 Correct 572 ms 7852 KB Output is correct
7 Correct 830 ms 9420 KB Output is correct
8 Correct 961 ms 20120 KB Output is correct
9 Correct 1577 ms 26140 KB Output is correct
10 Correct 2232 ms 35680 KB Output is correct
11 Correct 2671 ms 31460 KB Output is correct
12 Correct 1069 ms 24864 KB Output is correct
13 Correct 1475 ms 27644 KB Output is correct
14 Correct 1862 ms 29020 KB Output is correct
15 Correct 2411 ms 38372 KB Output is correct
16 Correct 2448 ms 48032 KB Output is correct
17 Correct 2531 ms 45476 KB Output is correct