Booth reservation stall reservation

A typical greedy

Abstract look, in fact, in the interval, the interval can form a disjoint chain, demand chain of the minimum number of

Solution

Proof is actually very simple, you just find a usable throw into the like, because no matter where a fence lost, the last are still considered to be the last time, so in fact there is no difference

 

#include <. bits / STDC ++ H>
 the using  namespace STD;
 int NUM = . 1 , n-;
 struct Node { int ST, ED, IND, Used;} A [ 50005 ];
 BOOL CMP (Node A, Node B) { return A .st <b.st;} // sorted by start time 
BOOL CMP2 (Node A, Node B) { return a.ind <b.ind;} // a numbered ordering 
The priority_queue <pair < int , int >> Q; // priority queue 
int main () {
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
     scanf("%d%d",&a[i].st,&a[i].ed),a[i].ind=i;//记录序号
    sort(a+1,a+n+1,cmp);
    A [ . 1 ] = .used . 1 ; q.push (the make_pair (-a [ . 1 ] .ed, . 1 )); // initialize a cowshed 
    for ( int I = 2 ; I <= n-; I ++ ) {
         int X = - q.top () First;.
         int tmp = q.top () SECOND;.
         IF (X <A [I] .st) { // can be put into the opinions 
            q.pop ();
            q.push(make_pair(-a[i].ed,tmp));
            a[i].used=tmp;
        }
        the else { // can not be put on to open a 
            q.push (the make_pair (-a [I] .ed, ++ NUM));
            A [I] .used = NUM; // record 
        }
    }
    sort(a+1,a+n+1,cmp2);
    printf("%d\n",num);
    for(int i=1;i<=n;i++)
     printf("%d\n",a[i].used);//输出
}