思路都已经讲得很清楚了,这里提供一个权值线段树优化版本,时间复杂度同样是$O(n\log n)$(不爱用STL的我手动滑稽)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#define ll long long
using namespace std;

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while(isdigit(ch)) {x = 10*x+ch-'0'; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) putchar('-'),x = -x;
    if(x > 9) out(x/10);
    putchar(x%10+'0');
}
//---------------------------------------------------------------
const int N = 2505,M = 1005;
int n,m,ans;

struct cow {
    int l,r;
    bool operator < (const cow &sed) const {
        return l > sed.l;
    }
}c[N];

int sum[M<<2],mx[M<<2];

void A(int u,int l,int r,int x,int k) {
    if(l == r) {
        sum[u] += k; 
        mx[u] = (sum[u] ? l : 0); 
        return;
    }
    int mid = (l+r)>>1;
    if(x <= mid) A(u<<1,l,mid,x,k);
    else A(u<<1|1,mid+1,r,x,k);
    sum[u] = sum[u<<1]+sum[u<<1|1];
    mx[u] = max(mx[u<<1],mx[u<<1|1]);
}

int Q(int u,int l,int r,int x,int y) {
    if(x <= l && y >= r) return mx[u];
    int mid = (l+r)>>1,res = 0;
    if(x <= mid) res = max(res,Q(u<<1,l,mid,x,y));
    if(y > mid) res = max(res,Q(u<<1|1,mid+1,r,x,y));
    return res;
}

int main() {
    int i,a,cnt; in(n); in(m);
    for(i = 1;i <= n; ++i) {
        in(c[i].l); in(c[i].r);
    }
    sort(c+1,c+n+1);
    for(i = 1;i <= m; ++i) {
        in(a); in(cnt);
        A(1,1,1000,a,cnt);
    }
    for(i = 1;i <= n; ++i) {
        int pos = Q(1,1,1000,c[i].l,c[i].r);
        if(!pos) continue;
        ++ans; A(1,1,1000,pos,-1);
    }
    out(ans);
    return 0;
}