2018-10-17多校连测2
T1
题意：
与[Usaco2017 Feb]Why Did the Cow Cross the Road相同，只是数据范围加大了，
将原来的n=2e4加到了2e5，很显然n^2的暴力会TLE，用set在longn时间内做完即可。
典型的贪心，以r为第一关键字，l为第二关键字从小到大排序。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<set> 6 using namespace std; 7 inline int read() 8 { 9 int x=0,w=1; char ch=getchar(); 10 while (!isdigit(ch)){if(ch=='-')w=-1;ch=getchar();} 11 while (isdigit(ch)) x=x*10+ch-'0',ch=getchar(); 12 return x*w; 13 } 14 const int N=200005; 15 int n,m; 16 multiset<int> s; 17 multiset<int>::iterator it; 18 struct note 19 { 20 int l,r; 21 }cow[N]; 22 bool cmp (note a,note b){return a.r==b.r?a.l<b.l:a.r<b.r;} 23 int main() 24 { 25 freopen("dream.in","r",stdin); 26 freopen("dream.out","w",stdout); 27 n=read(),m=read(); 28 for (int u=1;u<=m;u++) cow[u].l=read(),cow[u].r=read(); 29 for (int u=1;u<=n;u++) 30 { 31 int x=read(); 32 s.insert(x); 33 } 34 sort(cow+1,cow+1+m,cmp); 35 int ans=0; 36 for (int u=1;u<=m;u++) 37 { 38 it=s.lower_bound(cow[u].l); 39 if (it!=s.end()&&*it<=cow[u].r) 40 { 41 ans++; 42 s.erase(it); 43 } 44 } 45 printf("%d\n",ans); 46 return 0; 47 }