poj 2806: 公共子序列

解题思路：

动态规划

#include<iostream>#include<cstring>using namespace std;int main(){char a[205],b[205];int v[205][205];while(cin>>a>>b){memset(v,0,sizeof(v));for(int i=1;i<=strlen(a);i++)for(int j=1;j<=strlen(b);j++){if(a[i-1]==b[j-1]){v[i][j] = max(v[i-1][j-1]+1,v[i-1][j]);v[i][j] = max(v[i][j],v[i][j-1]);}else{v[i][j] = max(v[i-1][j-1],v[i-1][j]);v[i][j] = max(v[i][j],v[i][j-1]);}}cout<<v[strlen(a)][strlen(b)]<<endl;}return 0;}