【POJ1080】Human Gene Functions (Dynamic Programming)

Title: 【POJ1080】Human Gene Functions

This question can be thought of using DP to do it at a glance.

First, put the table in the question in a function:

int s(char x,char y)
{
	if(x=='A')
	{
		if(y=='A') return 5;
		if(y=='C') return -1;
		if(y=='G') return -2;
		if(y=='T') return -1;
		if(y=='-') return -3;
	}
	if(x=='C')
	{
		if(y=='A') return -1;
		if(y=='C') return 5;
		if(y=='G') return -3;
		if(y=='T') return -2;
		if(y=='-') return -4;
	}
	if(x=='G')
	{
		if(y=='A') return -2;
		if(y=='C') return -3;
		if(y=='G') return 5;
		if(y=='T') return -2;
		if(y=='-') return -2;
	}
	if(x=='T')
	{
		if(y=='A') return -1;
		if(y=='C') return -2;
		if(y=='G') return -2;
		if(y=='T') return 5;
		if(y=='-') return -1;
	}
	if(x=='-')
	{
		if(y=='A') return -3;
		if(y=='C') return -4;
		if(y=='G') return -2;
		if(y=='T') return -1;
		if(y=='-') return 0;
	}
	return 0;
}

Then, we have to figure out a way to derive a state transition equation, which is the core part of the entire code.

We denote the maximum similarity that can be obtained when the first string (denoted as s1) matches the i-th position and the second string (denoted as s2) matches the j-th position as f[i][j] , then for f[i][j], it may be obtained from 3 ways:

①f[i][j]=f[i-1][j]+s(s1[i],'-')

②f[i][j]=f[i][j-1]+s('-',s2[j])

③f[i][j]=f[i-1][j-1]+s(s1[i],s2[j])

This should be relatively obvious, so it will not be proved or explained.

From the above, the state transition equation of this question is f[i][j]=max{ f[i-1][j]+s(s1[i],'-') , f[i][j -1]+s('-',s2[j]) , f[i-1][j-1]+s(s1[i],s2[j]) }.

Now that the state transition equation has been introduced, the rest is easy to do, just go to the code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#define N 100
using namespace std;
int n,m,f[N+5][N+5];
string s1,s2;
int s(char x,char y)
{
	if(x=='A')
	{
		if(y=='A') return 5;
		if(y=='C') return -1;
		if(y=='G') return -2;
		if(y=='T') return -1;
		if(y=='-') return -3;
	}
	if(x=='C')
	{
		if(y=='A') return -1;
		if(y=='C') return 5;
		if(y=='G') return -3;
		if(y=='T') return -2;
		if(y=='-') return -4;
	}
	if(x=='G')
	{
		if(y=='A') return -2;
		if(y=='C') return -3;
		if(y=='G') return 5;
		if(y=='T') return -2;
		if(y=='-') return -2;
	}
	if(x=='T')
	{
		if(y=='A') return -1;
		if(y=='C') return -2;
		if(y=='G') return -2;
		if(y=='T') return 5;
		if(y=='-') return -1;
	}
	if(x=='-')
	{
		if(y=='A') return -3;
		if(y=='C') return -4;
		if(y=='G') return -2;
		if(y=='T') return -1;
		if(y=='-') return 0;
	}
	return 0;
}
intmain()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n),cin>>s1,scanf("%d",&m),cin>>s2;
		s1="-"+s1,s2="-"+s2;//Add a '-' before s1 and s2
		for(int i=1;i<=n;i++) f[i][0]=f[i-1][0]+s(s1[i],'-');//Preprocessing, next same
		for(int i=1;i<=m;i++) f[0][i]=f[0][i-1]+s('-',s2[i]);
		for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) f[i][j]=max(max(f[i-1][j]+s(s1[i],'-'),f[i][j-1]+s('-',s2[j])),f[i-1][j-1]+s(s1[i],s2[j]));//DP过程
		printf("%d\n",f[n][m]);//Output the answer
	}
	return 0;
}

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