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  2. 【扫描线_求周长】HDU 1828 Picture（两种方法【详】）

【扫描线_求周长】HDU 1828 Picture（两种方法【详】）

由 ↘锁芯ラ 提交于 2020-01-20 08:37:30

 HDU 1828 Picture

  • 题意：n个矩形块(有叠加部分)，求所有矩形块形成的大的二维图形的周长。
  • 思路：首先说求横向边长度 = 当前扫描线覆盖的横向长度 - 上一次扫描线覆盖的横向长度。再就是垂直长度。

新开三个数组

  1. lc[ rt ]: rt 区间的左端点有没有被扫面线覆盖
  2. rc[ rt ]: rt 区间的右端点有没有被扫描线覆盖
  3. num[ rt ]: rt 区间有几条线段

num数组的更新涉及到区间合并：如果左儿子区间的右端点和右儿子区间的左端点都被覆盖了，那么说明两个区间可以合并成一个。所以此时num需要减一

那么垂直长度就是两条扫描线间的高度 * 2 * num[ 1 ]//每个线段有两条边

  • 注意：按照y坐标排序时，如果y坐标相同，那么一定要先扫入边，后扫出边。G++过不了就是因为忽略了这个原因，会将实际没用的出边多计算两次贡献。【下面代码最后给的第三组样例就可以把先扫出边、后扫入边的hack掉，可以自己调试蛤儿qwq】
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) x & (-x)

#define MID (l + r ) >> 1
#define lsn rt << 1
#define rsn rt << 1 | 1
#define Lson lsn, l, mid
#define Rson rsn, mid + 1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define eps  1e-6

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxN = 10000 + 5;
int n;
int ABS(int x)
{
    return x >= 0 ? x : -x;
}
struct node{
    int lx, rx, high, add;
    node() {}
    node(int a, int b, int c, int d):lx(a), rx(b), high(c), add(d){}
    friend bool operator < (node n1, node n2) { return n1.high == n2.high ? n1.add > n2.add : n1.high < n2.high; }
}info[maxN];
int discx[maxN];
int tree[maxN << 2], cover[maxN << 2];
bool lc[maxN << 2], rc[maxN << 2];
int num[maxN << 2];
void build_tree(int rt, int l, int r)
{
    tree[rt] = 0;
    cover[rt] = 0;
    lc[rt] = rc[rt] = false;
    num[rt] = 0;
    if(l == r) return ;
    int mid = MID;
    build_tree(Lson);
    build_tree(Rson);
}
void pushup(int rt, int l, int r)
{
    if(cover[rt])
    {
        tree[rt] = discx[r + 1] - discx[l];
        lc[rt] = rc[rt] = true;
        num[rt] = 1;
    }
    else if(l == r)
    {
        tree[rt] = 0;
        lc[rt] = rc[rt] = false;
        num[rt] = 0;
    }
    else
    {
        tree[rt] = tree[lsn] + tree[rsn];
        num[rt] = num[lsn] + num[rsn];
        lc[rt] = lc[lsn]; rc[rt] = rc[rsn];
        if(lc[rsn] && rc[lsn]) num[rt] -- ;
    }
}
void update(int rt, int l, int r, int ql, int qr, int val)
{
    if(ql <= l && qr >= r)
    {
        cover[rt] += val;
        pushup(rt, l, r);
        return ;
    }
    int mid = MID;
    if(qr <= mid) update(QL, val);
    else if(ql > mid) update(QR, val);
    else { update(QL, val); update(QR, val); }
    pushup(rt, l, r);
}
int main()
{
    while(~scanf("%d", &n))
    {
        int tot = 0;
        for(int i = 1; i <= n; i ++ )
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            info[++ tot] = node(x1, x2, y1, 1);
            discx[tot] = x1;
            info[ ++ tot] = node(x1, x2, y2, -1);
            discx[tot] = x2;
        }
        sort(info + 1, info + tot + 1);
        sort(discx + 1, discx + tot + 1);
        int UP = unique(discx + 1, discx + tot + 1) - discx - 1;
        build_tree(1, 1, UP - 1);
        int ans = 0, last = 0;
        for(int i = 1; i <= tot; i ++ )
        {
            int lx = lower_bound(discx + 1, discx + UP + 1, info[i].lx) - discx;
            int rx = lower_bound(discx + 1, discx + UP + 1, info[i].rx) - discx - 1;
            update(1, 1, UP - 1, lx, rx, info[i].add);
            ans += ABS(tree[1] - last);
            last = tree[1];
            if(i < tot)
                ans += num[1] * 2 * (info[i + 1].high - info[i].high);
        }
        ans += last;
        printf("%d\n", ans);
    }
    return 0;
}
/*
4
0 0 2 2
1 1 3 3
2 2 4 4
3 3 5 5
 ans 20
3
0 2 2 5
1 3 5 7
3 0 6 4
 ans 28
2
1 0 2 1
0 1 2 2
 ans 8
 */

 下面的做法是两边扫描线，一次横向一次纵向，两次加起来就是答案。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define lowbit(x) x & (-x)

#define MID (l + r ) >> 1
#define lsn rt << 1
#define rsn rt << 1 | 1
#define Lson lsn, l, mid
#define Rson rsn, mid + 1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define eps  1e-6

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxN = 10000 + 5;
int n;
struct node{
    int lx, rx, high, add;
    node() {}
    node(int a, int b, int c, int d):lx(a), rx(b), high(c), add(d){}
    friend bool operator < (node n1, node n2) { return n1.high == n2.high ? n1.add > n2.add : n1.high < n2.high; }
}info_x[maxN];
struct NODE{
    int ly, hy, wide, add;
    NODE() {}
    NODE(int a, int b, int c, int d):ly(a), hy(b), wide(c), add(d) {}
    friend bool operator < (NODE n1, NODE n2) { return n1.wide == n2.wide ? n1.add > n2.add : n1.wide < n2.wide; }
}info_y[maxN];
int discx[maxN];
int discy[maxN];
int tree[maxN << 2], cover[maxN << 2];
void build_tree(int rt, int l, int r)
{
    tree[rt] = 0;
    cover[rt] = 0;
    if(l == r) return ;
    int mid = MID;
    build_tree(Lson);
    build_tree(Rson);
}
void pushup(int rt, int l, int r)
{
    if(cover[rt]) tree[rt] = discx[r + 1] - discx[l];
    else if(l == r) tree[rt] = 0;
    else tree[rt] = tree[lsn] + tree[rsn];
}
void update(int rt, int l, int r, int ql, int qr, int val)
{
    if(ql <= l && qr >= r)
    {
        cover[rt] += val;
        pushup(rt, l, r);
        return;
    }
    int mid = MID;
    if(qr <= mid) update(QL, val);
    else if(ql > mid) update(QR, val);
    else { update(QL, val); update(QR, val); }
    pushup(rt, l, r);
}

void pushup_y(int rt, int l, int r)
{
    if(cover[rt]) tree[rt] = discy[r + 1] - discy[l];
    else if(l == r) tree[rt] = 0;
    else tree[rt] = tree[lsn] + tree[rsn];
}
void update_y(int rt, int l, int r, int ql, int qr, int val)
{
    if(ql <= l && qr >= r)
    {
        cover[rt] += val;
        pushup_y(rt, l, r);
        return ;
    }
    int mid = MID;
    if(qr <= mid) update_y(QL, val);
    else if(ql > mid) update_y(QR, val);
    else { update_y(QL, val); update_y(QR, val); }
    pushup_y(rt, l, r);
}
int ansx[maxN], ansy[maxN];
int ABS(int x)
{
    return x >= 0 ? x : -x;
}
int main()
{
    while(~scanf("%d", &n))
    {
        int tot = 0;
        for(int i = 1; i <= n; i ++ )
        {
            int x1, y1, x2, y2;
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            info_x[++ tot] = node(x1, x2, y1, 1);
            discx[tot] = x1;
            info_y[tot] = NODE(y1, y2, x1, 1);
            discy[tot] = y1;
            info_x[++ tot] = node(x1, x2, y2, -1);
            discx[tot] = x2;
            info_y[tot] = NODE(y1, y2, x2, -1);
            discy[tot] = y2;
        }
        sort(info_x + 1, info_x + tot + 1);
        sort(discx + 1, discx + tot + 1);
        int UP_x = unique(discx + 1, discx + tot + 1) - discx - 1;
        build_tree(1, 1, UP_x - 1);
        int ans_x = 0;
        for(int i = 1; i <= tot; i ++ )
        {
            int lx = lower_bound(discx + 1, discx + UP_x + 1, info_x[i].lx) - discx;
            int rx = lower_bound(discx + 1, discx + UP_x + 1, info_x[i].rx) - discx - 1;
            update(1, 1, UP_x - 1, lx, rx, info_x[i].add);
            ansx[i] = tree[1];
            if(i == 1)
                ans_x += ansx[i];
            else
                ans_x += ABS(ansx[i] - ansx[i - 1]);
        }
        ans_x += ansx[tot];

        sort(info_y + 1, info_y + tot + 1);
        sort(discy + 1, discy + tot + 1);
        int UP_y = unique(discy + 1, discy + tot + 1) - discy - 1;
        build_tree(1, 1, UP_y - 1);
        int ans_y = 0;
        for(int i = 1; i <= tot; i ++ )
        {
            int ly = lower_bound(discy + 1, discy + UP_y + 1, info_y[i].ly) - discy;
            int hy = lower_bound(discy + 1, discy + UP_y + 1, info_y[i].hy) - discy - 1;
            update_y(1, 1, UP_y - 1, ly, hy, info_y[i].add);
            ansy[i] = tree[1];
            if(i == 1)
                ans_y += ansy[i];
            else
                ans_y += ABS(ansy[i] - ansy[i - 1]);
        }
        ans_y += ansy[tot];
        printf("%d\n", ans_x + ans_y);
    }
    return 0;
}
/*
4
0 0 2 2
1 1 3 3
2 2 4 4
3 3 5 5
 ans 20
3
0 2 2 5
1 3 5 7
3 0 6 4
 ans 28
 */

 

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周长
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