HDU1542--Atlantis(扫描线)

Posted 2020-06-13

给N个矩形的端点坐标，求矩形覆盖面积和。

原理很简单，从左到右扫描，线段树记录的是纵向覆盖的长度。区间更新。因为坐标是实数而且很大，所以需要离散化。

WA+RE+CE+MLE+。。。一共错了二十多次。用了最蠢的办法，最后发现错在初始化的时候，构造函数参数我写成了int。。蠢哭。。。

AC代码：

#include <bits/stdc++.h>
#define clr(x,c) memset(x,c,sizeof(x))

using namespace std;
const int N = 20005;

struct ScanLine {
    double x;
    double upY, downY;
    int flag;   // 入边1 出边-1
    bool operator<(const ScanLine a) const {
        return x < a.x;
    }
    ScanLine() {}
    ScanLine(double x, double y1, double y2, int f) : x(x), upY(y1), downY(y2), flag(f) {}
} line[N];

double tr[N];
int cover[N];
double yy[N];

#define lson (o<<1)
#define rson (o<<1|1)
#define mid (l+r>>1)
int yl, yr, v;
void pushup(int o, int l, int r)
{
    if (cover[o]) tr[o] = yy[r] - yy[l];
    else if (l + 1 == r) tr[o] = 0;         // 叶子
    else tr[o] = tr[lson] + tr[rson];
}

void update(int o, int l, int r)
{
    if (yl > r || yr < l) return ;
    if (yl <= l && yr >= r) {
        cover[o] += v;
        pushup(o, l, r);
        return ;
    }
    if (l + 1 == r) return ;                // 不包含的叶子节点要退出，否则死循环T^T
    if (yl <= mid) update(lson, l, mid);
    if (yr > mid) update(rson, mid, r);     // 注意这里不是mid+1 因为mid~mid+1一段的距离也要算
    pushup(o, l ,r);
}

int main()
{
    int n;
    int cas = 0;
    while (~scanf("%d", &n) && n) {
        int cnt = 0;
        double x1, y1, x2, y2;
        for (int i = 0; i < n; ++i) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[++cnt] = ScanLine(x1, y2, y1, 1);
            yy[cnt] = y1;
            line[++cnt] = ScanLine(x2, y2, y1, -1);
            yy[cnt] = y2;
        }
        sort(yy + 1, yy + cnt + 1);
        sort(line + 1, line + cnt + 1);
        int len = unique(yy + 1, yy + cnt + 1) - yy - 1;
        clr(cover, 0);
        clr(tr, 0);
        double ans = 0;
        for (int i = 1; i <= cnt; ++i) {
            ans += tr[1] * (line[i].x - line[i - 1].x);
            yl = lower_bound(yy+1, yy + len + 1, line[i].downY) - yy;
            yr = lower_bound(yy+1, yy + len + 1, line[i].upY) - yy;
            v = line[i].flag;
            update(1, 1, len);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++cas, ans);
    }
	return 0;
}

 

还有一种方法，比较好理解，和平时的线段树比较像，就是对于每个区间求[l, r-1]，算的时候右边加一，这样递归的时候就不用考虑叶子节点了。

#include <bits/stdc++.h>
#define clr(x,c) memset(x,c,sizeof(x))
using namespace std;
const int N = 20005;

struct ScanLine {
    double x;
    double upY, downY;
    int flag;   // 入边1 出边-1
    bool operator<(const ScanLine a) const {
        return x < a.x;
    }
    ScanLine() {}
    ScanLine(double x, double y1, double y2, int f) : x(x), upY(y1), downY(y2), flag(f) {}
} line[N];

double tr[N];
int cover[N];
double yy[N];

#define lson (o<<1)
#define rson (o<<1|1)
#define mid ((l+r)>>1)
int yl, yr, v;

void pushup(int o, int l, int r)
{
    if (cover[o] > 0) tr[o] = yy[r+1] - yy[l];
    else if (l == r) tr[o] = 0;
    else tr[o] = tr[lson] + tr[rson];
}

void update(int o, int l, int r)
{
    if (yl > r || yr < l) return ;
    if (yl <= l && yr >= r) {
        cover[o] += v;
        pushup(o, l, r);
        return ;
    }
    if (yl <= mid) update(lson, l, mid);
    if (yr > mid) update(rson, mid + 1, r);
    pushup(o, l ,r);
}

int main()
{
    int n;
    int cas = 0;
    while (~scanf("%d", &n) && n) {
        int cnt = 0;
        double x1, y1, x2, y2;
        for (int i = 0; i < n; ++i) {
            scanf("%lf%lf%lf%lf", &x1, &y1, &x2, &y2);
            line[++cnt] = ScanLine(x1, y2, y1, 1);
            yy[cnt] = y1;
            line[++cnt] = ScanLine(x2, y2, y1, -1);
            yy[cnt] = y2;
        }
        sort(yy + 1, yy + cnt + 1);
        sort(line + 1, line + cnt + 1);
        int len = unique(yy + 1, yy + cnt + 1) - yy - 1;
        clr(cover, 0);
        clr(tr, 0);
        double ans = 0;
        for (int i = 1; i <= cnt; ++i) {
            ans += tr[1] * (line[i].x - line[i - 1].x);
            yl = lower_bound(yy+1, yy+len+1, line[i].downY) - yy;
            yr = lower_bound(yy+1, yy+len+1, line[i].upY) - yy - 1;
            v = line[i].flag;
            update(1, 1, len-1);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++cas, ans);
    }
	return 0;
}