답안 #476976

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
476976 2021-09-29T15:34:56 Z urosk Regions (IOI09_regions) C++14
100 / 100
2721 ms 48152 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 3 ms 200 KB Output is correct
2 Correct 1 ms 200 KB Output is correct
3 Correct 3 ms 200 KB Output is correct
4 Correct 7 ms 200 KB Output is correct
5 Correct 8 ms 360 KB Output is correct
6 Correct 18 ms 484 KB Output is correct
7 Correct 31 ms 524 KB Output is correct
8 Correct 44 ms 656 KB Output is correct
9 Correct 72 ms 1360 KB Output is correct
10 Correct 106 ms 1728 KB Output is correct
11 Correct 105 ms 2240 KB Output is correct
12 Correct 110 ms 3396 KB Output is correct
13 Correct 99 ms 3104 KB Output is correct
14 Correct 215 ms 4160 KB Output is correct
15 Correct 221 ms 8360 KB Output is correct
# 결과 실행 시간 메모리 Grader output
1 Correct 1070 ms 10232 KB Output is correct
2 Correct 1057 ms 8948 KB Output is correct
3 Correct 1707 ms 15472 KB Output is correct
4 Correct 305 ms 5128 KB Output is correct
5 Correct 384 ms 8304 KB Output is correct
6 Correct 539 ms 8044 KB Output is correct
7 Correct 858 ms 9272 KB Output is correct
8 Correct 934 ms 20072 KB Output is correct
9 Correct 1632 ms 26280 KB Output is correct
10 Correct 2168 ms 35744 KB Output is correct
11 Correct 2721 ms 31376 KB Output is correct
12 Correct 1154 ms 25116 KB Output is correct
13 Correct 1592 ms 27756 KB Output is correct
14 Correct 1903 ms 29188 KB Output is correct
15 Correct 2452 ms 38572 KB Output is correct
16 Correct 2391 ms 48152 KB Output is correct
17 Correct 2286 ms 45580 KB Output is correct