hdu 5904 LCIS (dp)

Posted by 111qqz on Sunday, September 25, 2016

TOC

题目链接

题意:
给定两个序列,求它们的最长公共递增子序列的长度, 并且这个子序列的值是连续的
思路:以值为连续做入手点。

很显然个鬼咯
dp[a[i]]表示以a[i]结尾的最大长度。
dp[a[i]] = dp[a[i-1]] + 1
对于b序列一样。

答案为 MAX(min(dp[i],dp2[i])) ( 1=<i <= 1E6)

/* ***********************************************
Author :111qqz
Created Time :Mon 26 Sep 2016 04:02:24 AM CST
File Name :code/hdu/5904.cpp
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair
using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=1E6+7;
int dp[N],dp2[N];
int n,m;
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif
    int T;
    scanf("%d",&T);
    while (T--)
    {
        ms(dp,0);
        ms(dp2,0);
        scanf("%d%d",&n,&m);
        int mx = 0;
        for ( int i = 1 ; i <= n ; i++)
        {
        int x;
        scanf("%d",&x);
        dp[x] = dp[x-1] + 1;
        mx = max(x,mx);
        }
        for ( int i = 1 ; i <= m ; i++)
        {
        int x;
        scanf("%d",&x);
        dp2[x] = dp2[x-1] + 1;
        mx = max(x,mx);
        }
        int ans = 0 ;
        for ( int i = 1 ; i <= mx  ; i++) ans = max(ans,min(dp[i],dp2[i]));
        printf("%d\n",ans);
    }
  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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