线段树 hdu1255 覆盖的面积

来源：互联网发布：大数据和股票有哪些编辑：程序博客网时间：2024/04/28 21:48

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,1int 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;}

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