POJ1683_Common Subsequence_DP

题目描述

Common Subsequence

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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, x _ij = 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         abfcabprogramming    contest abcd           mnp

Sample Output

420

解题报告

最长公共子序列

当前配对不成功，取没有当前字符的最长组：

a[i][j]=max(a[i-1][j],a[i][j-1])

成功

a[i][j]=a[i-1][j-1]+1;

01滚动一下就可

代码

#include <iostream>#include <cstring>#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>char s1[1000],s2[1000];int lsc[2][1000],lenth1,lenth2;#define max(a,b) (a>b?a:b);using namespace std;int main(){    while(scanf("%s%s",s1,s2)!=EOF)    {        bool zo=false;        int pre,now;        memset(lsc,0,sizeof(lsc));        lenth1=strlen(s1);        lenth2=strlen(s2);//       for(int i=0;i<lenth1;i++){            {pre=zo;zo=not zo;now=zo;}            for(int j=0;j<lenth2;j++){                if(s1[i]==s2[j])                    lsc[now][j]=(j!=0?lsc[pre][j-1]+1:1);                else                    lsc[now][j]=max(lsc[pre][j],(j!=0?lsc[now][j-1]:0));            }       }        int maximum=-1;         for(int i=0;i<=lenth2;i++)            {maximum=(maximum>lsc[now][i]?maximum:lsc[now][i]);}        printf("%d\n",maximum);    }    return 0;}