POJ 题目2127 Greatest Common Increasing Subsequence（LICS，输出路径）

Greatest Common Increasing Subsequence

Time Limit: 10000MS Memory Limit: 65536KTotal Submissions: 9762 Accepted: 2584Case 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 , . . . , S_N of length N is called an increasing subsequence of a sequence A₁ , A₂ , . . . , A_M of length M if there exist 1 <= i₁ < i₂ < . . . < 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 (-2³¹ <= A_i < 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

51 4 2 5 -124-12 1 2 4

Sample Output

21 4

Source

Northeastern Europe 2003, Northern Subregion

ac代码

#include<stdio.h>#include<string.h>int n,m;int dp[1010][1010],path[1010][1010],ans[1010],aj,ai,res,a[1010],b[1010];void LCS(){int i,j,mj;res=0;memset(dp,0,sizeof(dp));memset(path,-1,sizeof(path));for(i=1;i<=n;i++){int ma=0;for(j=1;j<=m;j++){dp[i][j]=dp[i-1][j];if(b[j]<a[i]&&dp[i][j]>ma){ma=dp[i][j];mj=j;}else{if(b[j]==a[i]){dp[i][j]=ma+1;path[i][j]=mj;}}if(res<dp[i][j]){res=dp[i][j];ai=i;aj=j;}}}}int main(){while(scanf("%d",&n)!=EOF){int i,j;for(i=1;i<=n;i++)scanf("%d",&a[i]);scanf("%d",&m);for(j=1;j<=m;j++)scanf("%d",&b[j]);LCS();printf("%d\n",res);int temp=res;while(temp){if(path[ai][aj]>-1){ans[temp--]=b[aj];aj=path[ai][aj];}ai--;}for(i=1;i<=res;i++)printf("%d%c",ans[i],i==res?'\n':' ');}}