最长公共子序列

//poj  1458
Common Subsequence
Time Limit: 1000MS        Memory Limit: 10000K
Total Submissions: 69884        Accepted: 29304
Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.
Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.
Sample Input

abcfbc         abfcab
programming    contest 
abcd           mnp
Sample Output

4
2
0

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 1000
char s[N],p[N];
int ls,lp,dp[N][N];
//dp[i][j] :s[0]~s[i-1]与p[0]~p[j-1]的最长公共子序列长度
int main()
{
    while(~scanf("%s%s",&s,&p)){
        ls=strlen(s);lp=strlen(p);
        memset(dp,0,sizeof(dp));
        for(int i =1;i<=ls;i++){
            for(int j=1;j<=lp;j++){
                if(s[i-1]==p[j-1]){
                    dp[i][j] = dp[i-1][j-1]+1;
                }
                else{
                    dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
                }
            }
        }
        printf("%d\n",dp[ls][lp]);
    }
    return 0;
 } 

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 1000
char s[N],p[N];
int ls,lp,dp[N][N];
int flag [N][N];
void print(int i,int j){
    if(i==0&&j==0) return ;
    if(!flag[i][j]) {
        print(i-1,j-1);
        printf("%c",s[i-1]) ;//只有共同才打印 
    }
    else if(flag[i][j]==1){
        print(i-1,j);
    }
    else{
        print(i,j-1);
    }
}
int main()
{
    while(~scanf("%s%s",&s,&p)){
        ls=strlen(s);lp=strlen(p);
        memset(dp,0,sizeof(dp));
        memset(flag,0,sizeof(flag));
        for(int i =1;i<=ls;i++){
            for(int j=1;j<=lp;j++){
                if(s[i-1]==p[j-1]){
                    dp[i][j] = dp[i-1][j-1]+1;
                    flag[i][j] = 0;
                }
                else
                {
                    if(dp[i-1][j]>dp[i][j-1]){
                        flag [i][j] =1;
                        dp[i][j] = dp[i-1][j];
                    }
                    else{
                        flag  [i][j] =-1;
                        dp[i][j] = dp[i] [j-1];
                    }
                
                }
            }
        }
        print(ls,lp);//打印出来 
    }
    return 0;
 } 

 

import java.util.*;
import java.util.concurrent.locks.LockSupport;

public class Main {
    public static int lengthOfLIS(int[] nums) {
        // 希望递增子序列更长，表示该序列增加的更缓慢。也就是希望一定长度下的递增子序列的最后一个值更小
        // dp[l]: 长度为l的递增子序列的末尾最小值
        // 0 2 1 7 :0 2 ,0 1,0 7,2 7 dp[2]=1
        // 因此dp 递增，可以二分
        int n = nums.length;
        int dp[] = new int[n + 1];
        int l = 1;
        dp[l] = nums[0];
        for (int i = 1; i < n; i++) {
            if (nums[i] > dp[l]) {
                dp[++l] = nums[i];
            } else {
                int j = 1, k = l, pos = 0;
                while (j <= k) {
                    int mid = (j + k) >> 1;
                    // 在dp[1-l]中找到最后一个比nums[i]小的数
                    if (dp[mid] < nums[i]) {// 只要小于nums[i]就记录
                        pos = mid;
                        j = mid + 1;
                    } else {
                        k = mid - 1;
                    }
                }

                System.out.println(i + " " + pos);
                // pos+1就是dp[1-l]中找到第一个比nums[i]大的数的ID
                dp[pos + 1] = nums[i];
            }
        }
        return l;
    }

    public static void main(String[] args) {
        int a[] = { 0, 1, 0, 3, 2, 3 };
        System.out.println(lengthOfLIS(a));
    }
}