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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