這道題最關鍵的點就在離散化吧。

假如有三張海報[1, 10] [10, 13][15, 20] 僅僅三個區間就得占用到20了。

但是離散化後就可以是[1, 2] [2, 3] [4, 5] n到1e4 不重疊的話最大也只到2e4

那麽就可以做了

離散化技巧需要好好消化

代碼如下

#include <cstring>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define lp p<<1
#define rp p<<1|1
using
 
 namespace std;
const int N = 20000+5;
int tree[N<<2];
int a[N], ans;
bool vis[N];
pair<int,int> p[N];

inline void pushdown(int p) {
    if (tree[p]) {
        tree[lp] = tree[rp] = tree[p];
        tree[p] = 0;
    }
}
void build(int p, int l, int r) {
    if (tree[p]) {
        
 
if (!vis[tree[p]]) {
            vis[tree[p]] = 1;
            ans++;
        }
        return;
    }
    int mid = l + r >> 1;
    build(lp, l, mid);
    build(rp, mid + 1, r);
}
void change(int p, int l, int r, int x, int y, int z) {
    if (x <= l && y >= r) {
        tree[p] 
 
= z;
        return;
    }
    pushdown(p);
    int mid = l + r >> 1;
    if (x <= mid) change(lp, l, mid, x, y, z);
    if (y > mid) change(rp, mid + 1, r, x, y, z);
}
int main()
{
    int T; scanf("%d", &T);
    while (T--) {
        int n;
        scanf("%d", &n);
        for (int i = 1; i <= 2 * n; i++) {
            scanf("%d", &p[i].first);
            p[i].second = i;
        }
        sort(p + 1, p + 2 * n + 1);
        int last = 0, cnt = 0;
        for (int i = 1; i <= 2 * n; i++) {
            if (p[i].first == last) {
                a[p[i].second] = cnt;
            } else {
                a[p[i].second] = ++cnt, last = p[i].first;
            }
        }
        memset(tree, 0, sizeof(tree));
        for (int i = 1; i <= 2 * n; i += 2) {
            change(1, 1, cnt, a[i], a[i+1], (i + 1) / 2);
        }
        memset(vis, 0, sizeof(vis));
        ans = 0;
        build(1, 1, cnt);
        printf("%d\n", ans);
    }
    return 0;
}

Mayor's posters（線段樹+離散化）