POJ 1080 Human Gene Functions

动态规划问题

最长公共子串的变形  主要注意初始化

 

#include<iostream>
using namespace std;
#define N 105
int v[N][N],dp[N][N];
char s1[N],s2[N];
int max(int a,int b)
{
	return a>b?a:b;
}
void init()
{
	v['A']['A']=5;
	v['C']['C']=5;
	v['G']['G']=5;
	v['T']['T']=5;
	v['A']['C']=-1;
	v['A']['G']=-2;
	v['A']['T']=-1;
	v['A']['-']=-3;
	v['C']['A']=-1;
	v['C']['G']=-3;
	v['C']['T']=-2;
	v['C']['-']=-4;
	v['G']['A']=-2;
	v['G']['C']=-3;
	v['G']['T']=-2;
	v['G']['-']=-2;
	v['T']['A']=-1;
	v['T']['C']=-2;
	v['T']['G']=-2;
	v['T']['-']=-1;
	v['-']['A']=-3;
	v['-']['C']=-4;
	v['-']['G']=-2;
	v['-']['T']=-1;
}
int main()
{
	int la,lb,t,i,j;
	cin>>t;
	init();
	while(t--)
	{
		cin>>la>>s1;
		cin>>lb>>s2;
		for(i=1;i<=la;i++)dp[0][i]=dp[0][i-1]+v['-'][s2[i-1]];  //dp[i][j]表示s1前i个字符 s2前j个字符时 最大的相似度  dp[i][0]表示s1前i个与s2都不符合 
		for(i=1;i<=lb;i++)dp[i][0]=dp[i-1][0]+v[s1[i-1]]['-'];

		for(i=1;i<=la;i++)
			for(j=1;j<=lb;j++)
			{
				dp[i][j]=max(max(dp[i-1][j]+v[s1[i-1]]['-'],dp[i][j-1]+v['-'][s2[j-1]]),dp[i-1][j-1]+v[s1[i-1]][s2[j-1]]);
			}
		cout<<dp[la][lb]<<endl;

	}
	return 0;
	
}