BZOJ 3262

转自:http://blog.csdn.net/lwt36/article/details/50625051

我觉得是整体二分+树状数组,存个模板。

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

struct Point {
    int x, y, z;
    int cnt, sum;
    void read() {
        scanf("%d%d%d", &x, &y, &z);
        cnt = sum = 0;
    }
    bool operator<(const Point& p) const {
        if(x != p.x) return x < p.x;
        if(y != p.y) return y < p.y;
        return z < p.z;
    }
    bool operator==(const Point& p) const {
        return x == p.x && y == p.y && z == p.z;
    }
} p[N];

const int M = 2e5 + 10;
struct BIT {
    int n, b[M];
    void init(int _n) {
        n = _n;
        memset(b, 0, sizeof b);
    }
    void add(int i, int v) {
        for(; i <= n; i += i & -i) b[i] += v;
    }
    int sum(int i) {
        int ret = 0;
        for(; i; i -= i & -i) ret += b[i];
        return ret;
    }
} bit;

bool cmpy(Point a, Point b) {
    return a.y < b.y;
}

void cdq(int l, int r) {
    if(l == r) return;
    int m = l + r >> 1;
    cdq(l, m);
    cdq(m + 1, r);
    sort(p + l, p + m + 1, cmpy);
    sort(p + m + 1, p + r + 1, cmpy);
    int j = l;
    for(int i = m + 1; i <= r; ++i) {
        for(; j <= m && p[j].y <= p[i].y; ++j) bit.add(p[j].z, p[j].cnt);
        p[i].sum += bit.sum(p[i].z);
    }
    for(int i = l; i < j; ++i) bit.add(p[i].z, -p[i].cnt);
}

int n, m, k;
int ans[N];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    while(scanf("%d%d", &n, &k) == 2) {
        bit.init(k);
        for(int i = 1; i <= n; ++i) p[i].read();
        sort(p + 1, p + 1 + n);
        int m = 0;
        for(int i = 1, j; i <= n; ++i) {
            for(j = i; j <= n && p[i] == p[j]; ++j) ++p[i].cnt;
            p[++m] = p[i];
            i = j - 1;
        }
        cdq(1, m);
        memset(ans, 0, sizeof ans);
        for(int i = 1; i <= m; ++i) ans[p[i].sum + p[i].cnt - 1] += p[i].cnt;
        for(int i = 0; i < n; ++i) printf("%d\n", ans[i]);
    }
    return 0;
}

 

posted on 2016-10-12 20:04  very_czy  阅读(283)  评论(0编辑  收藏  举报

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