답안 #402353

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
402353 2021-05-11T15:55:35 Z fvogel499 Plahte (COCI17_plahte) C++14
0 / 160
75 ms 104396 KB
#include <iostream>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <algorithm>

using namespace std;

#define mp pair<pii, pii>
#define pii pair<int, int>
#define f first
#define s second

#define pow2 (1 << 20) // just over 1 million
#define siz 80000

set<pii> bt [pow2*2];

vector<int> graph [siz+1]; // add extra space for virtual root
set<int> colorsAtNode [siz];
int nbColours [siz]; // different colors at node rect

struct Event {
    Event(bool lIsRectSide, bool lIsStartSide, int lx, int lidx) {
        isRectSide = lIsRectSide;
        isStartSide = lIsStartSide;
        x = lx;
        idx = lidx;
    }
    bool isRectSide, isStartSide; // iSStartSide is only for rectangles
    int x, idx; // idx is rectangle or ball index
};

// used for sorting
bool sf(Event a, Event b) {
    return (a.x < b.x or (a.x == b.x and (a.isRectSide and a.isStartSide) and !b.isRectSide) or (a.x == b.x and !a.isRectSide and (b.isRectSide and !b.isStartSide)));
    // second conditions are to avoid tricky edge cases where point is located on segment. We must make sure the point is between the start side and the end side
}

// reinit segtree
void btClear() {
    for (int i = 0; i < pow2*2; i++) bt[i].clear();
}

// insert pair in bt interval
void btAdd(int rBegin, int rEnd, pii rPair, int qNode, int qBegin, int qEnd) {
    if (rBegin > qEnd or rEnd < qBegin) return;
    else if (rBegin <= qBegin and qEnd <= rEnd) bt[qNode].insert(rPair);
    else {
        int qMid = (qBegin+qEnd)/2;
        btAdd(rBegin, rEnd, rPair, qNode*2, qBegin, qMid);
        btAdd(rBegin, rEnd, rPair, qNode*2+1, qMid+1, qEnd);
    }
}

// remove pair from bt interval
void btRemove(int rBegin, int rEnd, pii rPair, int qNode, int qBegin, int qEnd) {
    if (rBegin > qEnd or rEnd < qBegin) return;
    else if (rBegin <= qBegin and qEnd <= rEnd) bt[qNode].erase(rPair);
    else {
        int qMid = (qBegin+qEnd)/2;
        btRemove(rBegin, rEnd, rPair, qNode*2, qBegin, qMid);
        btRemove(rBegin, rEnd, rPair, qNode*2+1, qMid+1, qEnd);
    }
}

// point request for maximum pair
pii btGetMax(int idx) {
    idx += pow2;
    pii mP = pii(-1, -1); // mP = max pair, init with lowest possible value, NOTE: time starts at 0
    while (idx != 0) {
        if (!bt[idx].empty()) mP = max(mP, *(prev(bt[idx].end())));
        idx /= 2; // loop for all parents
    }
    return mP;
}

// small-to-large merging technique to count number of different colors
set<int>& dfs(int cn) { // cn = current node
    for (int nn : graph[cn]) { // nn = new node
        set<int>& ls = dfs(nn); // pass by reference
        if (colorsAtNode[cn].size() < ls.size()) swap(colorsAtNode[cn], ls);
        for (int j : ls) colorsAtNode[cn].insert(j); // small-to-large merging
    }
    nbColours[cn] = colorsAtNode[cn].size();
    return colorsAtNode[cn];
}

int main() {
    // int n, nbBalls;
    // cin >> n >> nbBalls;

    // mp rect [n];
    // vector<Event> sweep;
    // for (int i = 0; i < n; i++) {
    //     cin >> rect[i].f.f >> rect[i].f.s >> rect[i].s.f >> rect[i].s.s;
    //     sweep.push_back(Event(true, true, rect[i].f.f, i));
    //     sweep.push_back(Event(true, false, rect[i].s.f, i));
    // }

    // int ballColor [nbBalls];
    // int ballY [nbBalls];
    // for (int i = 0; i < nbBalls; i++) {
    //     int lx;
    //     cin >> lx >> ballY[i] >> ballColor[i];
    //     sweep.push_back(Event(false, -1, lx, i)); // second parameter has no influence
    // }

    // sort(sweep.begin(), sweep.end(), sf); // sort by x-coordinate

    // int par [n]; // the immediate rectangle a rectangle is contained inside
    // int ballImpactRect [nbBalls]; // the smallest rectangle the ball hits
    // int t = 0; // time variable
    // int insertedT [n]; // remember time when deleting elements
    btClear();
    // for (Event i : sweep) {
    //     if (i.isRectSide and i.x == rect[i.idx].f.f) {
    //         pii j = btGetMax(rect[i.idx].f.s); // it doesn't matter on which point of interval [ rect[i.idx].f.s ; rect[i.idx].s.s ] we look at since there are no sides crossing
    //         if (j.second == -1) j.second = n; // n is the virtual root
    //         par[i.idx] = j.second;
    //         insertedT[i.idx] = t;
    //         btAdd(rect[i.idx].f.s, rect[i.idx].s.s, pii(insertedT[i.idx], i.idx), 1, 0, pow2-1); // add this side as latest rectangle in this interval
    //     }
    //     else if (i.isRectSide and i.x == rect[i.idx].s.f) {
    //         btRemove(rect[i.idx].f.s, rect[i.idx].s.s, pii(insertedT[i.idx], i.idx), 1, 0, pow2-1);
    //     }
    //     else {
    //         pii j = btGetMax(ballY[i.idx]);
    //         if (j.second == -1) j.second = n; // n is the virtual root
    //         ballImpactRect[i.idx] = j.second;
    //     }
    //     t++;
    // }

    // for (int i = 0; i < n; i++) {
    //     graph[par[i]].push_back(i);
    // }

    // for (int i = 0; i < nbBalls; i++) {
    //     colorsAtNode[ballImpactRect[i]].insert(ballColor[i]);
    // }

    // dfs(n); // start dfs from virtual root

    // for (int i = 0; i < n; i++) {
    //     cout << nbColours[i] << endl;
    // }

    int d = 0;
    d++;
}
# 결과 실행 시간 메모리 Grader output
1 Incorrect 66 ms 104388 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 68 ms 104388 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 64 ms 104392 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 67 ms 104292 KB Output isn't correct
2 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Incorrect 75 ms 104396 KB Output isn't correct
2 Halted 0 ms 0 KB -