线段树 hdu1255 覆盖的面积

和之前做过的fzu的一道线段树维护的内容恰好相反

这题求的是覆盖次数大于等于2的面积

思路：维护两个值，一个是覆盖次数大于等于1的面积，一个是覆盖次数大于等于2的面积

然后在push_up的时候仔细分析一下，想清楚更新顺序就做完了..

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,rear,1

int const MX = 1e4 + 5;

int rear, cnt[MX << 2];
double A[MX], S1[MX << 2], S2[MX << 2];

struct Que {
    int d;
    double top, L, R;
    Que() {}
    Que(double _top, double _L, double _R, int _d) {
        top = _top; L = _L; R = _R; d = _d;
    }
    bool operator<(const Que &b)const {
        return top < b.top;
    }
} Q[MX];

int BS(double x) {
    int L = 1, R = rear, m;
    while(L <= R) {
        m = (L + R) >> 1;
        if(A[m] == x) return m;
        if(A[m] > x) R = m - 1;
        else L = m + 1;
    }
    return -1;
}

void push_up(int l, int r, int rt) {
    if(cnt[rt]) {
        S1[rt] = A[r + 1] - A[l];
        if(cnt[rt] == 1) {
            S2[rt] = S1[rt << 1] + S1[rt << 1 | 1];
        } else S2[rt] = S1[rt];
    } else if(l == r) S1[rt] = S2[rt] = 0;
    else {
        S1[rt] = S1[rt << 1] + S1[rt << 1 | 1];
        S2[rt] = S2[rt << 1] + S2[rt << 1 | 1];
    }
}

void update(int L, int R, int d, int l, int r, int rt) {
    if(L <= l && r <= R) {
        cnt[rt] += d;
        push_up(l, r, rt);
        return;
    }

    int m = (l + r) >> 1;
    if(L <= m) update(L, R, d, lson);
    if(R > m) update(L, R, d, rson);
    push_up(l, r, rt);
}

int main() {
    //freopen("input.txt", "r", stdin);
    int n, T;
    scanf("%d", &T);
    while(T--) {
        rear = 0;
        memset(cnt, 0, sizeof(cnt));
        memset(S1, 0, sizeof(S1));
        memset(S2, 0, sizeof(S2));

        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            double x1, y1, x2, y2;
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            Q[i] = Que(y1, x1, x2, 1);
            Q[i + n] = Que(y2, x1, x2, -1);

            A[++rear] = x1; A[++rear] = x2;
        }
        sort(Q + 1, Q + 1 + 2 * n);
        sort(A + 1, A + 1 + rear);
        rear = unique(A + 1, A + 1 + rear) - A - 1;

        double ans = 0, last = 0;
        for(int i = 1; i <= 2 * n; i++) {
            ans += (Q[i].top - last) * S2[1];
            update(BS(Q[i].L), BS(Q[i].R) - 1, Q[i].d, root);
            last = Q[i].top;
        }
        printf("%.2lf\n", ans);
    }
    return 0;
}