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)

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

Problem Description

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

Input

There are multiple test cases. The first line of input contains an integerT, 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f(i)f(i)是以a_ia i结尾的最大值, g(i)g(i)是以b_ibi结尾的最大值. 答案就是\max_{a_i = b_j} {\min(f(i),g(j)}maxai=bj{min(f(i),g(j)}. ff和gg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;
}

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BestCoder Round #87（1003-【思维】【LISLCS】）

1003-LCIS

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