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题目链接:
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 458 Accepted Submission(s): 212
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); #define lson o<<1 #define rson o<<1|1 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 LL inf=1e18; const int N=1e6+10; const int maxn=1e5+10; const double eps=1e-12; int n,m,a[maxn],b[maxn],fa[N],fb[N]; int main() { int t; read(t); while(t--) { read(n);read(m); For(i,1,n)read(a[i]); For(i,1,m)read(b[i]); For(i,1,n)fa[a[i]-1]=fa[a[i]]=0,fb[a[i]-1]=fb[a[i]]=0; For(i,1,m)fa[b[i]-1]=fa[b[i]]=0,fb[b[i]-1]=fb[b[i]]=0; For(i,1,n)fa[a[i]]=fa[a[i]-1]+1; For(i,1,m)fb[b[i]]=fb[b[i]-1]+1; int ans=0; for(int i=1;i<=n;i++)ans=max(ans,min(fa[a[i]],fb[a[i]])); printf("%d\n",ans); } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5906395.html