題目鏈接:https://vjudge.net/problem/CodeForces-369E
題意:n條線段,m次詢問,給出cnt個點,問這些點在多少線段上
題解:如果每個點對於所在的線段上沒有影響的話,那麼我們就可以建立權值線段樹,直接查詢即可,但是存在多個點在一條直線的情況,所以就沒法這麼搞了,所以我們可以把查詢的點轉化成間隔的線段,計算出有多少線段在這些線段上,這樣我們離線記錄下來,就是一個二維偏序了,按照左右區間關係排序,再用梳狀數組維護下即可。
#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) (x & (-x))
const int N = 3e5 + 10;
const int M = 1e6 + 10;
struct node {
int l, r;
int op, id;
bool operator <(const node &x)const{
if(r != x.r) return r < x.r;
else if(l != x.l) return l > x.l;
else return op < x.op;
}
}a[N * 4];
int n, m;
int len;
int sum[M];
void update(int x) {
while(x) {
sum[x]++;
x -= lowbit(x);
}
}
int query(int x) {
int res = 0;
while(x < M) {
res += sum[x];
x += lowbit(x);
}
return res;
}
int ans[N];
int main() {
scanf("%d %d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d %d", &a[i].l, &a[i].r);
a[i].op = 0;
}
len = n;
int x, y, pre;
for(int i = 1; i <= m; i++) {
scanf("%d", &y);
pre = 1;
while(y--) {
scanf("%d", &x);
if(x != pre) {
len++;
a[len].l = pre;
a[len].r = x - 1;
a[len].op = 1;
a[len].id = i;
}
pre = x + 1;
}
if(pre <= 1000000) {
len++;
a[len].l = pre;
a[len].r = 1000000;
a[len].id = i;
a[len].op = 1;
}
}
sort(a + 1, a + 1 + len);
for(int i = 1; i <= len; i++) {
// cout << a[i].id << " " << a[i].l <<" " << a[i].r<<endl;
if(a[i].op) ans[a[i].id] += query(a[i].l);
else update(a[i].l);
}
for(int i = 1; i <= m; i++)
printf("%d\n", n - ans[i]);
return 0;
}