HDU1159-Common Subsequence

标签：

描述：

　　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. 

　　The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by anynumber of white spaces. The input data are correct. 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. 

代码：

　　设序列X={x1,x2,…,xm}和Y={y1,y2,…,yn}的最长公共子序列为Z={z1,z2,…,zk} ，则

　　　　(1)若xm=yn，则zk=xm=yn，且zk-1是xm-1和yn-1的最长公共子序列。

　　　　(2)若xm≠yn且zk≠xm，则Z是xm-1和Y的最长公共子序列。

　　　　(3)若xm≠yn且zk≠yn，则Z是X和yn-1的最长公共子序列。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stdlib.h>
#include <math.h>
using namespace std;
#define N 1000

int main(){
    char x[N],y[N];
    int dp[N][N];
    while ( scanf("%s%s",&x,&y)!=EOF ){
        //TLE:memset(dp,0,sizeof(dp));
        for( int i=0;i<N;i++ ){
            dp[0][i]=0;dp[i][0]=0;
        }

        for( int i=1;i<=strlen(x);i++ ){
            for( int j=1;j<=strlen(y);j++ ){
                if( x[i-1]==y[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[strlen(x)][strlen(y)]);
    }
    system("pause");
    return 0;
}

HDU1159-Common Subsequence

标签：

原文地址：http://www.cnblogs.com/lucio-yr/p/4749036.html