題目

5
1 4
2 2
3 3
4 1
5 5

2
3
2
2
1

分析

https://blog.csdn.net/weixin_43993341/article/details/88913374

代碼

 1 #include <bits/stdc++.h>
 2  using namespace std;
 3  int f[100001],maxr[100001];
 4  struct A
 5  {
 6      int a,b,c;
 7  }a[100001];
 8  bool cmp (A a,A b)
 9  {
10      return a.a>b.a;
11  }
12  int main()
13  {
14      ios::sync_with_stdio(false);
15      int n;
16      cin>>n;
17      for (int i=1;i<=n;i++)
18      {
19          cin>>a[i].a>>a[i].b;
20         a[i].c=i;
21      }
22      sort (a+1,a+n+1,cmp);
23      maxr[1]=a[1].b;
24      f[a[1].c]=1;
25      for (int i=2;i<=n;i++)
26      {
27          int mid,t=a[i].b,l=1,r=i;
28          while (l<=r)
29          {
30               mid=(l+r)/2;
31              if (maxr[mid]<=t) r=mid-1;
32              if (maxr[mid]>t) l=mid+1;
33          }
34          maxr[l]=max(maxr[l],t);
35          f[a[i].c]=l;
36      }
37      for (int i=1;i<=n;i++)
38        printf("%d\n",f[i]);
39  }

JZOJ 4214. 【NOIP2015模擬9.12】SERN的野望