题目链接:
Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 131072/131072 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> //#include <bits/stdc++.h> #include <stack> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=1e9; const int N=1e5+10; const int maxn=500+10; const double eps=1e-8; int a[N],d[N],g[N]; int main() { int t; read(t); while(t--) { int n; read(n); For(i,1,n)read(a[i]); for(int i=1;i<=n;i++)g[i]=inf; for(int i=1;i<=n;i++) { int k=lower_bound(g+1,g+n+1,a[i])-g; d[i]=k; g[k]=a[i]; } for(int i=1;i<n;i++)printf("%d ",d[i]);printf("%d\n",d[n]); } return 0; }
原文:http://www.cnblogs.com/zhangchengc919/p/5699893.html