poj 1080

题意：

给两串DNA序列，按照给定的方法找他们最大的相似度。比如序列AGTGATG和GTTAG，化为AGTGATG和-GTTA-G，相似度最大，为14。

思路：

由低到高的往上递推，动态规划。

 

设dp(i,j)为第一个序列(s1)的前i个数和第二个序列(s2)的前j个数的相似度的最大值。当s1[i-1]==s2[j-1]时，由题目给出的表显然可以得出dp(i,j)=dp(i-1,j-1)+p[s1[i-1]][s2[j-1]];score数组为题目中给出的那个表格。当s1[i-1]!=s2[j-1]时，反证法显然有dp(i,j)=max(d(i-1,j)+score[s1[i-1]]['-'],dp(i,j-1)+score['-'][],d(i-1,j-1)+score[s1[i-1]][s2[j-1]])。于是，两个for就解决问题了。注意初始化数组。

 

代码如下：

 

#include<iostream>
using namespace std;
const int inf=-5;  //无穷小

int score['T'+1]['T'+1];  //积分表

void initial(void)  //打表
{
	score['A']['A']=5;
	score['C']['C']=5;
	score['G']['G']=5;
	score['T']['T']=5;
	score['-']['-']=inf;
	score['A']['C']=score['C']['A']=-1;
	score['A']['G']=score['G']['A']=-2;
	score['A']['T']=score['T']['A']=-1;
	score['A']['-']=score['-']['A']=-3;
	score['C']['G']=score['G']['C']=-3;
	score['C']['T']=score['T']['C']=-2;
	score['C']['-']=score['-']['C']=-4;
	score['G']['T']=score['T']['G']=-2;
	score['G']['-']=score['-']['G']=-2;
	score['T']['-']=score['-']['T']=-1;
	return;
}

int max(int a,int b,int c)
{
	int k=(b>c?b:c);
	return a>k?a:k;   //注意求三个数最大值时，a>b?a:(b>c?b:c)在C++中是错误的
}                     //b的值没有因为(b>c?b:c)而改变，必须把三个数拆开求最大值

int main(int i,int j)
{
	initial();
	
	int test;
	cin>>test;
	while(test--)
	{
		/*Input*/
		
		int len1,len2;
		
		cin>>len1;
		char* s1=new char[len1+1];
		cin>>s1;
		
		cin>>len2;
		char* s2=new char[len2+1];
		cin>>s2;
		
		int **dp=new int*[len1+1];   //申请动态二维数组，第一维
		dp[0]=new int[len2+1];
		
		/*Initial*/
		
		dp[0][0]=0;
		for(i=1;i<=len1;i++)
		{
			dp[i]=new int[len2+1];  //申请动态二维数组，第二维
			dp[i][0]=dp[i-1][0]+score[ s1[i-1] ]['-'];   //注意下标，dp数组是从1开始，s1和s2都是从0开始
		}
		for(j=1;j<=len2;j++)
			dp[0][j]=dp[0][j-1]+score['-'][ s2[j-1] ];
		
		/*Dp*/
		
		for(i=1;i<=len1;i++)
			for(j=1;j<=len2;j++)
			{
				int temp1=dp[i-1][j]+score[ s1[i-1] ]['-'];
				int temp2=dp[i][j-1]+score['-'][ s2[j-1] ];
				int temp3=dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ];
				dp[i][j]=max(temp1,temp2,temp3);
			}
			
			cout<<dp[len1][len2]<<endl;
			
			delete[] dp;
	}
	return 0;
}