hdu5904-LCIS

题目链接

LCIS Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1386    Accepted Submission(s): 608   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 integer 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  2 3 4 5 1 1 2 1   Sample Output 1 5 0   Source BestCoder Round #87

思路:这是一道简单题目，但是我一来就wa了TAT,直接开二维数组dp是不行的，数据范围约束了，所以就有了以下的状态转移方程，

dp[temp]=dp[temp-1]+1;（temp为输入的值，这样能保证是连续上升子序列，还不重复，想想为什么）

代码:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#define in(x) scanf("%d",&x)
#define out(x) printf("%d\n",x)
using namespace std;
const int maxn=200005;
int k[maxn],p[maxn];
int main()
{
	int t,n,m,temp,smax;
	in(t);
	while(t--){
		in(n),in(m);
		smax=-99;
		memset(k,0,sizeof(k));
		memset(p,0,sizeof(p));
		for(int i=0;i<n;i++){
			in(temp);
			k[temp]=k[temp-1]+1;
		}
		for(int i=0;i<m;i++){
			in(temp);
			p[temp]=p[temp-1]+1;
			smax=max(smax,min(k[temp],p[temp]));//取min因为得是最长公共子序列
		}
		out(smax);
	}
	return 0;
}