To avoid unsightly burns while tanning, each of the C (1 ≤ C ≤ 2500) cows must cover her hide with sunscreen when they‘re at the beach. Cow i has a minimum and maximum SPF rating (1 ≤ minSPF_i ≤ 1,000; minSPF_i ≤ maxSPF_i ≤ 1,000) that will work. If the SPF rating is too low, the cow suffers sunburn; if the SPF rating is too high, the cow doesn‘t tan at all........

The cows have a picnic basket with L (1 ≤ L ≤ 2500) bottles of sunscreen lotion, each bottle i with an SPF rating SPF_i (1 ≤ SPF_i ≤ 1,000). Lotion bottle i can cover cover_i cows with lotion. A cow may lotion from only one bottle.

What is the maximum number of cows that can protect themselves while tanning given the available lotions?

* Line 1: Two space-separated integers: C and L
* Lines 2..C+1: Line i describes cow i‘s lotion requires with two integers: minSPF_i and maxSPF_i 
* Lines C+2..C+L+1: Line i+C+1 describes a sunscreen lotion bottle i with space-separated integers: SPF_i and cover_i

A single line with an integer that is the maximum number of cows that can be protected while tanning

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <queue>#include <functional>#include <utility>using namespace std;/*3 2       // C只牛  L个防晒霜 C 行  3 10      minSPF, maxSPF    2 51 5L行 6 2      每种数量, 固定的阳光强度 4 1*/ const int maxn = 100000;int C, L;                 // C只奶牛 , Ltypedef pair<int, int> P; // minSPF, maxSPF //小值先出 priority_queue<int, vector<int>, greater<int> > q;P cow[maxn], bot[maxn];  //牛(min,max) 和 防晒霜(固定的阳光数, 数量) void solve();void input();void input(){    scanf("%d%d", &C, &L);    for (int i = 0; i < C; i++) {        scanf("%d%d", &cow[i].first, &cow[i].second);    }        for (int i = 0; i < L; i++) {        scanf("%d%d", &bot[i].first, &bot[i].second);    }}void solve(){    input();    sort(cow, cow + C);    //按照最小值阳光强度升序     sort(bot, bot + L);    //(first)按照能固定的阳光强度升序     int j = 0, ans = 0;    //当  L 个 防晒霜     for (int i = 0; i < L; i++)    {        //将最小值阳光 和 固定的阳光强度 比较         while (j < C && cow[j].first <= bot[i].first) {            q.push(cow[j].second);    //添加到优先队列  (最小值阳光)             j++;        }                        //如果 最小值阳光 序列不空，当前 防晒霜未用完         while (!q.empty() && bot[i].second) {            int x = q.top(); q.pop();            // 最小值 小于  能固定的阳光, 方案不存在             if (x < bot[i].first) continue;            //否则,              ans++;            //当前防晒霜--             bot[i].second--;         }    }        printf("%d\n", ans);}int main(){    solve();        return 0;}