題目意思

每個案例一個數字n，接下來n行表示張貼海報的起點與終點，求有多少海報沒有被完全遮住，經典區間染色問題。本題資料量比較大，所以需要將起點和終點離散化處理。

Sample Input 1 5 1 4 2 6 8 10 3 4 7 10 Sample Output 4

解題思路

針對每一次update，若當前區間在詢問區間內，直接更新當前區間mark並return，若無交集，直接retrun，反之迭代至下一次。 查詢時，若查到當前區間mark不為0時，標記該mark並return，最後數一數book中標記個數即為所求答案。

AC程式碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
#define lson left,mid,k<<1
#define rson mid+1,right,k<<1|1
using namespace std;
ll ql, qr, val;
int book[200005];
int a[200005];
struct node
{
    int l;
    int r;
    ll mark;
}que[200005
 
];
struct Node
{
    int s;
    int e;
}dot[200005];
void push_down(int k)
{
    if (que[k].mark)
    {
        que[k << 1].mark = que[k].mark;
        que[k << 1 | 1].mark = que[k].mark;
        que[k].mark = 0;
    }
}
void build(int left, int right,int k)
{
    que[k].l = left;
    que[k].r = right;
    que[k].mark = 0
 
;
    if (left == right)
        return;
    int mid = (left + right) >> 1;
    build(lson);
    build(rson);
}
void update(int left, int right, int k)
{
    if (ql <= left && right <= qr)
    {
        que[k].mark = val;
        return;
    }
    if (right < ql || qr < left)
        return;
    push_down(k);
    int mid = (left + right) >> 1;
    update(lson);
    update(rson);
}
void query(int left, int right, int k)
{
    if (que[k].mark)
    {
        book[que[k].mark] = 1;
        return;
    }
    if (left == right)
        return;
    int mid = (left + right) >> 1;
    query(lson);
    query(rson);
}
int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int x = 0, ans = 0;
        memset(book, 0, sizeof book);
        int n;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &dot[i].s, &dot[i].e);
            a[x++] = dot[i].s;
            a[x++] = dot[i].e;
        }
        //離散化
        sort(a, a + x);
        int all = unique(a, a + x) - a;
        build(1, all, 1);
        for (int i = 0; i < n; i++)
        {
            //離散化
            dot[i].s = lower_bound(a, a + all, dot[i].s) - a + 1;
            dot[i].e = lower_bound(a, a + all, dot[i].e) - a + 1;
            ql = dot[i].s;
            qr = dot[i].e;
            val = i + 1;
            update(1, all, 1);
        }
        ql = 1;
        qr = all;
        query(1, all, 1);
        for (int i = 1; i < 200005; i++)
            if (book[i])
                ans++;
        printf("%d\n", ans);
    }
}