最长单调递增公共子序列（路径记录+poj2127+zoj2432）Greatest Common Increasing Subsequence
Greatest Common Increasing Subsequence
| Time Limit: 10000MS | Memory Limit: 65536K | |
| Total Submissions: 10340 | Accepted: 2727 | |
| 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 S 1 , S 2 , . . . , S N of length N is called an increasing subsequence of a sequence A 1 , A 2 , . . . , A M of length M if there exist 1 <= i 1 < i 2 < . . . < i N <= M such that S j = A ij for all 1 <= j <= N , and S j < S j+1 for all 1 <= j < N .
Sequence S 1 , S 2 , . . . , S N of length N is called an increasing subsequence of a sequence A 1 , A 2 , . . . , A M of length M if there exist 1 <= i 1 < i 2 < . . . < i N <= M such that S j = A ij for all 1 <= j <= N , and S j < S j+1 for all 1 <= j < N .
Input
Each sequence is described with M --- its length (1 <= M <= 500) and M integer numbers A
i (-2
31 <= A
i < 2
31 ) --- 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
链接：http://poj.org/problem?id=2127
思路：也是一道模板题，求最长递增公共子序列，并记录路径返回输出；
状态：dp[i][j]表示以s1串的前i个字符s2串的前j个字符且以s2[j]为结尾构成的LCIS的长度。
状态转移：当 s1[i-1]!=s2[j-1]时,按照lcs可知由2个状态转移过来,dp[i-1][j],dp[i][j-1],因为dp[i][j]是以s2[j]为结尾构成的LCIS的长度。所以s2[j-1]一定会包含在里面，所以舍去dp[i][j-1],只由dp[i-1][j] 转移过来。
当s1[i-1]==s2[j-1]，这时肯定要找前面s1[ii-1]==s2[jj-1]的最长且比s2[j-1]小的状态转移过来.
若s1[i-1]!=s2[j-1] 那么dp[i][j]=dp[i][j-1]
若s1[i-1]==s2[j-1] 那么dp[i][j]=MAX{dp[i-1][k]+1(0<k<j),s2[k-1]<s2[j-1]};
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n,m,a[505],b[505];
int dp[505][505];
int rd[505][505];
void LICS()
{
int ans=0;
// int dp[505]={0};
int mac=0;//用来记录路径的中间变量
int I,J;//最后的坐标
int lu[505];//用来最后输出路径
for(int i=1;i<=n;i++)
{
int len=0; //长度
for(int j=1;j<=m;j++)
{
int k=dp[i][j]=dp[i-1][j];
if(len<k && a[i]>b[j])
{
len=k;//记录相等前小于a[i]的最大长度
mac=j;
}
if(a[i]==b[j])
{
dp[i][j]=len+1;
rd[i][j]=mac;//记录最大长度下，上一个j的坐标
}
if(ans<dp[i][j])//更新最大长度，并记录结尾的坐标
{
ans=dp[i][j];
I=i;
J=j;
}
}
}
/* printf("(%d,%d)\n",I,J);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
printf("%d ",rd[i][j]);
}
printf("\n");
}*/
printf("%d\n",ans);
int Len=ans;
if(Len>0)
{
lu[Len--]=J;
}
while(Len && I && J)
{
// printf("(%d,%d),rd=%d\n",I,J,rd[I][J]);
if(rd[I][J]>0)
{
lu[Len--]=rd[I][J];
J=rd[I][J];
}
I--;
}
for(int i=1;i<=ans;i++)
{
printf("%d ",b[lu[i]]);
}
printf("\n");
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%d",&b[i]);
}
LICS();
}
return 0;
}
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int n,m,a[505],b[505];
//int dp[505][505];
int rd[505][505];
/*****
题意：求最长上升公共子序列（LCIS），并记录路径.
状态：dp[i][j]表示以s1串的前i个字符s2串的
前j个字符且以s2[j]为结尾构成的LCIS的长度。
状态转移：当 s1[i-1]!=s2[j-1]时,按照lcs可知由2个状态转移过来,
dp[i-1][j],dp[i][j-1],因为dp[i][j]是以s2[j]为结尾构
成的LCIS的长度。所以s2[j-1]一定会包含在里面，所以
舍去dp[i][j-1],只由dp[i-1][j] 转移过来。
当s1[i-1]==s2[j-1]，这时肯定要找前面s1[ii-1]==s2[jj-1]的最长且比s2[j-1]小的状态转移过来.
若s1[i-1]!=s2[j-1] 那么dp[i][j]=dp[i][j-1]
若s1[i-1]==s2[j-1] 那么dp[i][j]=MAX{dp[i-1][k]+1(0<k<j),s2[k-1]<s2[j-1]};
*/
void LICS()
{
int ans=0;
int dp[505]={0};
int mac=0;//用来记录路径的中间变量
int I,J;//最后的坐标
int lu[505];//用来最后输出路径
// memset(dp,0,sizeof(dp));//全局变量默认初始化，可以不写这句
for(int i=1;i<=n;i++)
{
int len=0; //长度
for(int j=1;j<=m;j++)
{
/**按照lcs可知由2个状态转移过来,dp[i-1][j],dp[i][j-1],
因为dp[i][j]是以s2[j]为结尾构成的LCIS的长度。
所以s2[j-1]一定会包含在里面，所以舍去dp[i][j-1],
只由dp[i-1][j] 转移过来。*/
/***由于只和dp[i-1][j]有关，所以优化空间降维***/
// int k=dp[i][j]=dp[i-1][j];
int k=dp[j];
if(len<k && a[i]>b[j])
{
len=k;//记录相等前小于a[i]的最大长度
mac=j;
}
if(a[i]==b[j])
{
dp[j]=len+1;
rd[i][j]=mac;//记录最大长度下，上一个j的坐标
}
if(ans<dp[j])//更新最大长度，并记录结尾的坐标
{
ans=dp[j];
I=i;
J=j;
}
}
}
/* printf("(%d,%d)\n",I,J);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
printf("%d ",rd[i][j]);
}
printf("\n");
}*/
printf("%d\n",ans);
int Len=ans;
if(Len>0)
{
lu[Len--]=J;
}
while(Len && I && J)
{
// printf("(%d,%d),rd=%d\n",I,J,rd[I][J]);
if(rd[I][J]>0)
{
lu[Len--]=rd[I][J];
J=rd[I][J];
}
I--;
}
for(int i=1;i<=ans;i++)
{
printf("%d ",b[lu[i]]);
}
printf("\n");
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%d",&b[i]);
}
LICS();
}
return 0;
}