Submission #499919

#	Time	Username	Problem	Language	Result	Execution time	Memory
499919		gozonite	Pairs (IOI07_pairs)	C++14	60 / 100	88 ms	5080 KiB

pairs

#include <iostream>
#include <vector>
#include <string>
#include <fstream>
#include <algorithm>
#include <climits>
#include <cstdlib>
#include <cstdio>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <bitset>
#include <deque>
#include <queue>
#include <tuple>
#include <cmath>
#include <cctype>
#include <stack>
#include <cassert>
using namespace std;
using ll = long long;

struct vert {
    int x, y, t;
};
bool cmp(const vert& a, const vert& b) {
    if (a.x != b.x) return a.x < b.x;
    return a.t < b.t;
}

int B, N, D, M;
int inp[3][100001];

vert verts[200001];
int bit1[150002]={};
void addv(int k, int v) {
    for(; k <= 150000; k += k&-k) bit1[k] += v;
}
int sumr(int k) {
    int s = 0;
    for (; k >= 1; k -= k&-k) s += bit1[k];
    return s;
}

int main() {
    cin >> B >> N >> D >> M;
    if (B==1) {
        for (int i = 1; i <= N; i++) cin >> inp[0][i];
        sort(inp[0]+1, inp[0]+1+N);
        int idx = 1;
        ll cnt = 0;
        for (int i = 1; i <= N; i++) {
            while (idx < N && inp[0][i]+D >= inp[0][idx+1]) idx++;
            cnt += (idx-i);
        }
        cout << cnt << endl;
    } else if (B==2) {
        for (int i = 1; i <= N; i++) cin >> inp[0][i] >> inp[1][i];
        for (int i = 1; i <= N; i++) {
            verts[2*i-1] = {inp[0][i]-inp[1][i], inp[0][i]+inp[1][i], 0};
            verts[2*i] = {inp[0][i]-inp[1][i]+D, inp[0][i]+inp[1][i], 1};
        }
        sort(verts+1, verts+1+2*N, cmp);
        ll ans = 0;
        for (int i = 1; i <= 2*N; i++) {
            int y = verts[i].y;
            if (verts[i].t == 0) {
                int fy = max(2, y-D), sy = min(2*M, y+D);
                ans += sumr(sy)-sumr(fy-1);
                addv(y, 1);
            } else {
                addv(y, -1);
            }
        }
        cout << ans << endl;    
    } else if (B==3) {
        for (int i = 1; i <= N; i++) cin >> inp[0][i] >> inp[1][i] >> inp[2][i];
    }

    return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 6 / 6
• Subtask 3: 6 / 6
• Subtask 4: 6 / 6
• Subtask 5: 6 / 6
• Subtask 6: 7 / 7
• Subtask 7: 7 / 7
• Subtask 8: 8 / 8
• Subtask 9: 8 / 8
• Subtask 10: 0 / 10
• Subtask 11: 0 / 10
• Subtask 12: 0 / 10
• Subtask 13: 0 / 10