POJ 1080 - Human Gene Functions


问题描述

无。

解题思路

LCS的变形而已

注意LCS的子串可以是离散的,不必连续,用动态规划

dp[i][j] 为取s1第i个字符,s2第j个字符时的最大分值

则决定dp为最优的情况有三种( score[][]s1[i]s2[j] 两符号的分数):

  • 1、 s1取第i个字母,s2取 '-'dp[i-1][j]+score[ s1[i-1] ]['-'];
  • 2、 s1取 '-',s2取第j个字母: dp[i][j-1]+score['-'][ s2[j-1] ];
  • 3、 s1取第i个字母,s2取第j个字母: dp[i-1][j-1]+score[ s1[i-1] ][ s2[j-1] ];

即 :

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

注意初始化。

不仅仅只有

dp[0][0] = 0

也不仅仅是

dp[0][0] = 0
dp[1][0] = score[ s1[i-1] ]['-']
dp[0][1] = score['-'][ s2[j-1] ]

必须全面考虑到所有情况:

  • i=j=0 时,dp[i][j]=0
  • i=0 时,dp[0,j] = dp[0][j-1] + score['-'][ s2[j-1] ]
  • j=0 时,dp[i,0] = dp[i-1][0] + score[ s1[i-1] ]['-']

AC 源码

//Memory Time 
//300K   0MS  

#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;
}

相关资料


文章作者: EXP
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