BestCoder Round #87（1003-【思维】【LIS&&LCS】）

题目链接：​​点击打开链接​​

1003-LCIS

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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Total Submission(s): 243    Accepted Submission(s): 98

Problem Description

a1,a2,...,an and b1,b2,...,bm. He wants find a longest common subsequence that consists of consecutive values in increasing order.

Input

T, indicating the number of test cases. For each test case:

The first line contains two integers n and m (1≤n,m≤100000) -- the length of two sequences. The second line contains n integers: a1,a2,...,an (1≤ai≤106). The third line contains n integers: b1,b2,...,bm (1≤bi≤106).

There are at most 1000 test cases and the sum of n and m does not exceed 2×106.

Output

For each test case, output the length of longest common subsequence that consists of consecutive values in increasing order.

Sample Input

3 3 3 1 2 3 3 2 1 10 5 1 23 2 32 4 3 4 5 6 1 1 2 3 4 5 1 1 2 1

Sample Output

1 5 0

Source

​​BestCoder Round #87 ​​

题解：

令是以结尾的最大值, 是以结尾的最大值. 答案就是. 和随便dp一下就出来了.

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN=1e5+10;
int n,m;
int a[MAXN],b[MAXN];
int dp1[MAXN],dp2[MAXN];
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d %d",&n,&m);
    for(int i=0;i<MAXN;i++) 
    {
      a[i]=b[i]=dp1[i]=dp2[i]=0;
    }
//    这里用 for来初始化数组也可以，用 memset一般初始化一些较大的结构体和数组比较省时
//    像空间比较小的如 100，用 for就比较快
//    memset(a,0,sizeof(a));
//    memset(b,0,sizeof(b));
//    memset(dp1,0,sizeof(dp1));
//    memset(dp2,0,sizeof(dp2));
    for(int i=0;i<n;i++)
    {
      scanf("%d",&a[i]);
      dp1[a[i]]=max(dp1[a[i]],dp1[a[i]-1]+1);
    }
    for(int i=0;i<m;i++)
    {
      scanf("%d",&b[i]);
      dp2[b[i]]=max(dp2[b[i]],dp2[b[i]-1]+1);
    }
    int ans=0;
    for(int i=0;i<n;i++)
    {
      ans=max(ans,min(dp1[a[i]],dp2[a[i]]));
    }
//    for(int i=0;i<m;i++) 任意找一个(a或 b) dp都可以 
//    {
//      ans=max(ans,min(dp1[b[i]],dp2[b[i]]));
//    } 
    printf("%d\n",ans);
  }
  return 0;
}