最长公共子序列 poj1458
Common Subsequence
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 30423 | Accepted: 11856 |
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
问题分析: 经典的DP问题。
输入串a与串b;
Dp[i][j]表示a串中1~i与b串中1~j的子串的最长公共子序列。
Max{dp[i-1][j], dp[i][j-1]} (a[i]!=b[j])
Dp[i][j]=
Dp[i-1][j-1]+1 (a[i]==b[j])
最后,a,b的最长公共子序列为dp[strlen(a)][strlen(b)]
代码实现:
#include<iostream> #include<string.h> #include<stdio.h> using namespace std; char a[1000]; char b[1000]; int dp[1000][1000]; //dp[i][j] is a[1~i] and b[1~j]'s common string's longest lenth int main(){ while(scanf("%s",a+1)!=EOF){ //cin one of the string scanf("%s",b+1); //cin the other int lena=strlen(a+1); int lenb=strlen(b+1); for(int i=0;i<=lena;i++){ dp[i][0]=0; } for(int j=0;j<=lenb;j++){ dp[0][j]=0; } for(int i=1;i<=lena;i++){ for(int j=1;j<=lenb;j++){ if(a[i]==b[j]){ dp[i][j]=dp[i-1][j-1]+1; } else{ dp[i][j]=((dp[i-1][j])>(dp[i][j-1]))?(dp[i-1][j]):(dp[i][j-1]); } } } printf("%d\n",dp[lena][lenb]); } return 0; }