hdoj1080 Human Gene Functions(DP解決LCS問題!)

來源:http://acm.hdu.edu.cn/showproblem.php?pid=1080

類似於求最長公共字串，在dp的初始化和狀態轉移方程的書寫上有少許差別。

dp[i][j] = dp[i-1,j] + dist[s1[i],'-'];s1[i]對上'-'

dp[i][j] = dp[i,j-1] + dist['-',s2[j]];'-'對上s2[j]

dp[i][j] = dp[i-1,j-1] + dist[s1[i],s2[j]];s1[i]對上s2[j]
dp[i,j]=max( dp[i-1,j] + dist[s1[i],'-'], dp[i,j-1]+dist['-',s2[j]], dp[i-1,j-1]+dist[s1[i],s2[j]]);

有關LCS問題的講解鏈接:http://blog.csdn.net/rrrfff/article/details/7523437

本題的另外講解鏈接:http://blog.sina.com.cn/s/blog_93ced90c01016rjv.html

代碼如下:

#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
int max3(int a,int b,int c)
{
	int x;
	x=(a>b)?a:b;
	x=(c>x)?c:x;
	return x;
}
int score[5][5]={
5, -1, -2, -1, -3,
-1, 5, -3, -2, -4,
-2, -3, 5, -2, -2,
-1, -2, -2, 5, -1,
-3, -4, -2, -1, 0};
int dp[105][105];
int a1[105],a2[105];
int main()
{
	char s1[105],s2[105];
	int t,i,j,l1,l2;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%s%d%s",&l1,s1+1,&l2,s2+1);
		for(i=1;i<=l1;i++)
		{
			if(s1[i]=='A')a1[i]=0;
			if(s1[i]=='C')a1[i]=1;
			if(s1[i]=='G')a1[i]=2;
			if(s1[i]=='T')a1[i]=3;
		}
			for(i=1;i<=l2;i++)
		{
			if(s2[i]=='A')a2[i]=0;
			if(s2[i]=='C')a2[i]=1;
			if(s2[i]=='G')a2[i]=2;
			if(s2[i]=='T')a2[i]=3;
		}
		memset(dp,0,sizeof(dp));
		for(i=0;i<=l1;i++)
		for(j=0;j<=l2;j++)
		{
			if (i==0&&j==0) dp[i][j]=0;
            else if (j==0&&i>0) dp[i][j]=dp[i-1][j]+score[a1[i]][4];
            else if (i==0&&j>0) dp[i][j]=dp[i][j-1]+score[4][a2[j]];
            else
            dp[i][j] = max3(dp[i-1][j-1]+score[a1[i]][a2[j]],dp[i][j-1]+score[4][a2[j]],dp[i-1][j]+score[a1[i]][4]);
		}
		printf("%d\n",dp[l1][l2]);
	}
	return 0;
}