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Common Subsequence

abcfbc abfcab
programming contest 
abcd mnp

4
2
0

#include<stdio.h>
#include<string>
#include<vector>
#include<iostream>
using namespace std;
int main(int argc, char *argv[])
{
    string a,b;
    while(cin>>a>>b)
    {
        vector<vector<int> > c;
        int x=a.size();
        int y=b.size();
        int SIZE=x>y?x:y;
        SIZE+=1;
        c.resize(SIZE);
        for(int i=0;i<SIZE;++i)
            c[i].resize(SIZE,0);
        for(int i=0;i<=x;++i)
            c[i][0]=0;
        for(int i=0;i<=y;++i)
            c[0][i]=0;
        for(int i=1;i<=x;++i)
            for(int j=1;j<=y;++j)
            {
                if(a[i-1]==b[j-1])
                {
                    c[i][j]=c[i-1][j-1]+1;
                }
                else
                {
                    c[i][j]=c[i][j-1]>c[i-1][j]?c[i][j-1]:c[i-1][j];
                }
            }
        cout<<c[x][y]<<endl;

    }
    return 0;
}

hdu 1159 Common Subsequence （dp求LCS）

标签：hdu   1159   

原文地址：http://blog.csdn.net/wdkirchhoff/article/details/41988325