HDU1159 && POJ1458 Common Subsequence (LCS模版题)

标签：poj   hdu   1159   1458   c++   

题目大意：求出两个串的公共子序列的长度

LCS的入门题，读懂题了直接模板就可以

#include <iostream>
#include <cstring>

using namespace std;

const int N=1000;
int a[N][N];

int LCS(const char *s1, const char *s2)
{// s1:0...m, s2:0...n
    int m = strlen(s1), n = strlen(s2);
    int i, j;
    a[0][0] = 0;
    for( i=1; i <= m; ++i ) a[i][0] = 0;
    for( i=1; i <= n; ++i ) a[0][i] = 0;
    for( i=1; i <= m; ++i )
        for( j=1; j <= n; ++j ){
            if(s1[i-1]==s2[j-1]) a[i][j] = a[i-1][j-1]+1;
            else if(a[i-1][j]>a[i][j-1])a[i][j]= a[i-1][j];
            else a[i][j] = a[i][j-1];
        }
    return a[m][n];
}

int main()
{
    char s1[N], s2[N];
    while(cin>>s1>>s2)
        cout<<LCS(s1, s2)<<endl;
    return 0;
}

HDU1159 && POJ1458 Common Subsequence (LCS模版题)

标签：poj   hdu   1159   1458   c++   

原文地址：http://blog.csdn.net/wangxinxin_/article/details/45101885