POJ 2528 Mayor's posters 线段树成段更新+离散化

题目来源：POJ 2528 Mayor‘s posters

题意：很多张海报贴在墙上 求可以看到几张海报 看那两张图就行了 第一张俯视图

思路：最多2W个不同的数 离散化一下 然后成段更新 a[rt] = i代表这个区间是第i张报纸 更新玩之后一次query cover[i]=1代表可以看到第i张报纸

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 20010;
int a[maxn<<2];
int cover[maxn<<2];
struct Point
{
	int x, id, v;
}b[maxn];
int n;
bool cmp1(Point a, Point b)
{
	return a.x < b.x;
}
bool cmp2(Point a, Point b)
{
	return a.id < b.id;
}
void build(int l, int r, int rt)
{
	a[rt] = 0;
	//printf("%d %d\n", l, r);
	if(l+1 == r)
		return;
	int m = (l + r) >> 1;
	build(l, m, rt<<1);
	build(m, r, rt<<1|1);
}
void update(int x, int y, int l, int r, int rt, int num)
{
	;
	if(l == x && r == y)
	{
		a[rt] = num;
		return;
	}
	int m = (l + r) >> 1;
	if(a[rt])
	{
		a[rt<<1] = a[rt];
		a[rt<<1|1] = a[rt];
		a[rt] = 0;
	}
	if(y <= m)
		update(x, y, l, m, rt<<1, num);
	else if(x >= m)
		update(x, y, m, r, rt<<1|1, num);
	else
	{
		update(x, m, l, m, rt<<1, num);
		update(m, y, m, r, rt<<1|1, num);
	}
}

void query(int l, int r, int rt)
{
	if(l+1 == r || a[rt])
	{
			cover[a[rt]] = 1;
		//printf("%d\n", a[rt]);
		return;
	}
	int m = (l + r) >> 1;
	if(a[rt])
	{
		a[rt<<1] = a[rt];
		a[rt<<1|1] = a[rt];
		a[rt] = 0;
	}
	query(l, m, rt<<1);
	query(m, r, rt<<1|1);
}
int main()
{
	int T;

	scanf("%d", &T);
	while(T--)
	{
		scanf("%d", &n);
		for(int i = 0; i < n; i++)
		{
			scanf("%d %d", &b[i<<1].x, &b[i<<1|1].x);
			b[i<<1|1].x++;
			b[i<<1].id = i<<1;
			b[i<<1|1].id = i<<1|1;
		}
		sort(b, b+2*n, cmp1);
		int now = b[0].x;
		int sum = 1;
		for(int i = 0; i < 2*n; i++)
		{
			if(b[i].x != now)
			{
				now = b[i].x;
				sum++;
			}
			b[i].v = sum;
		}
		sort(b, b+2*n, cmp2);
		//printf("%d\n", sum);
		//memset(a, 0, sizeof(a));
		build(1, sum, 1);
		//puts("sdsf");
		for(int i = 0; i < n; i++)
		{
			//printf("**%d %d\n", b[i<<1].v, b[i<<1|1].v);
			update(b[i<<1].v, b[i<<1|1].v, 1, sum, 1, i+1);
			//puts("ddsf");
		}
		memset(cover, 0, sizeof(cover));

		query(1, sum, 1);
		int ans = 0;
		for(int i = 1; i <= n; i++)
			if(cover[i])
				ans++;
		printf("%d\n", ans);
	}
	return 0;
}

POJ 2528 Mayor's posters 线段树成段更新+离散化,布布扣,bubuko.com

POJ 2528 Mayor's posters 线段树成段更新+离散化