NYOJ 760 See LCS again(基础dp+哈希表)(复习)

See LCS again
时间限制：1000 ms  |  内存限制：65535 KB 难度：3描述 There are A, B two sequences, the number of elements in the sequence is n、m;
Each element in the sequence are different and less than 100000.
Calculate the length of the longest common subsequence of A and B.
输入The input has multicases.Each test case consists of three lines;
The first line consist two integers n, m (1 < = n, m < = 100000);
The second line with n integers, expressed sequence A;
The third line with m integers, expressed sequence B;输出For each set of test cases, output the length of the longest common subsequence of A and B, in a single line.样例输入

5 4
1 2 6 5 4
1 3 5 4

样例输出

3

//注意理解题意啊，给我看红色区，本来打算利用一下数组下标的，但又觉得数字重复怎么办，哼，看了思路才恍然大悟
//其实就是利用数组下标，建立哈希函数对应关系，进而转化为最长上升子序列问题#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;

int a[100000+10];
int dp[100000+10];
int len;
int binarysearch(int key)
{
int left=1,right=len,mid;
while(left<=right)
{
mid=(left+right)>>1;
if(dp[mid]>key)
right=mid-1;
else
left=mid+1;
}
return left;
}
int main()
{
int n,m,x;
while(~scanf("%d%d",&n,&m))
{
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
{
scanf("%d",&x);
a[x]=i;
}
scanf("%d",&x);
dp[1]=a[x];
len=1;
for(int i=2; i<=m; i++)
{
scanf("%d",&x);
int t=binarysearch(a[x]);
dp[t]=a[x];
len=max(len,t);
}
printf("%d\n",len);
}
return 0;
} //ac//STL的算法
lower_bound（数组首地址，数组长度，查找元素val）函数是一个下确界，
[ first，last )前开后闭区间，从首元素开始查找，找到第一个大于等于val，
upper_bound（数组首地址，数组长度，查找元素val）函数是一个下确界，
[ first，last )前开后闭区间，从首元素开始查找，找到第一个大于val，#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>

using namespace std;

int g[100000+10];
int dp[100000+10];

int main()
{
int n,m,x;
while(~scanf("%d%d",&n,&m))
{
memset(dp,0,sizeof(dp));
for(int i=1; i<=n; i++)
{
scanf("%d",&x);
dp[x]=i;
}
int k=0;
for(int i=1; i<=m; i++)
{
scanf("%d",&x);
if(dp[x])
g[k++]=dp[x];
}
int p=0;
dp[p++]=g[0];
for(int i=1; i<k; i++)
{
if(dp[p-1]<g[i])
dp[p++]=g[i];
else
{
x=lower_bound(dp,dp+p,g[i])-dp;
dp[x]=g[i];
}
}
printf("%d\n",p);
}
return 0;
} //ac