poj 2127 Greatest Common Increasing Subsequence （最长公共递增子序列）

Greatest Common Increasing Subsequence Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 8888   Accepted: 2345 Case Time Limit: 2000MS   Special Judge Description You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.  Sequence S1 , S2 , . . . , SN of length N is called an increasing subsequence of a sequence A1 , A2 , . . . , AM of length M if there exist 1 <= i1 < i2 < . . . < iN <= M such that Sj = Aij for all 1 <= j <= N , and Sj < Sj+1 for all 1 <= j < N . Input Each sequence is described with M --- its length (1 <= M <= 500) and M integer numbers Ai (-231 <= Ai < 231 ) --- the sequence itself. Output On the first line of the output file print L --- the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them. Sample Input 5 1 4 2 5 -12 4 -12 1 2 4 Sample Output 2 1 4 Source Northeastern Europe 2003, Northern Subregion

题意：

给出两个序列a、b，求最长公共递增子序列。

思路：

dp[j]表示以b[j]结尾的LCIS的长度。

每次用a[i]去更新dp[j]，则

if(a[i]==b[j]) dp[j]=max(dp[j],dp[k]);  (1<=k<j&&b[k]<b[j])

维护好k的值，那么就可以将复杂度降到O(n*m)了。

由于要记录路径，用一维的dp、pre不太好处理，我用的是二维的，递归输出路径。

思路来源：推荐

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int n,m,k,ans,cnt,tot,flag;
int ai,aj;
int a[505],b[505];
int dp[505][505],px[505][505],py[505][505];

void output(int x,int y,int last)
{
if(x==0||y==0) return ;
output(px[x][y],py[x][y],y);
if(y!=last)
{
cnt++;
if(cnt==1) printf("%d",b[y]);
else printf(" %d",b[y]);
}
}
void solve()
{
int i,j,k,t;
memset(dp,0,sizeof(dp));
memset(px,0,sizeof(px));
memset(py,0,sizeof(py));
ans=0;
for(i=1;i<=n;i++)
{
k=0;
for(j=1;j<=m;j++)
{
dp[i][j]=dp[i-1][j];
px[i][j]=i-1,py[i][j]=j;
if(b[j]==a[i])
{
if(dp[i][j]<dp[i][k]+1)
{
dp[i][j]=dp[i][k]+1;
px[i][j]=i,py[i][j]=k;
}
}
else if(b[j]<a[i])
{
if(dp[i][k]<dp[i][j]) k=j;
}
if(dp[i][j]>ans)
{
ans=dp[i][j];
ai=i,aj=j;
}
}
}
printf("%d\n",ans);
if(ans==0) return ;
cnt=0;
output(ai,aj,-1);
printf("\n");
}
int main()
{
int i,j,t;
while(~scanf("%d",&n))
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(i=1;i<=m;i++)
{
scanf("%d",&b[i]);
}
solve();
}
return 0;
}
/*
7
1 2 4 0 1 4 5
6
0 3 1 2 4 5
*/