POJ 1080 && HDU 1080 Human Gene Functions（dp）

Description
给定两组DNA序列，要你求出它们的最大相似度
每个字母与其他字母或自身和空格对应都有一个打分，求在这两个字符串中插入空格，让这两个字符串的匹配分数最大

Input
T组用例，每组用例两个DNA长度（不超过100）及其序列
Output
对于每组用例，输出两个DNA序列的最大匹配分数
Sample Input
2
7 AGTGATG
5 GTTAG
7 AGCTATT
9 AGCTTTAAA
Sample Output
14
21
Solution
设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[][]为s1[i]和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,必须全面考虑到所有情况，
当i=j=0时，dp[i][j]=0
当i=0时，dp[0,j] = dp[0][j-1] + score[‘-‘][ s2[j-1] ]
当j=0时，dp[i,0] = dp[i-1][0] + score[ s1[i-1] ][‘-‘]
Code

#include<cstdio>#include<iostream>using namespace std;#define INF (1<<29)#define maxn -5#define Max(a,b) (a>b?a:b)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 main(){    initial();      int T;    cin>>T;    while(T--)    {        int i,j,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];//申请二维动态数组，第一维         dp[0]=new int[len2+1];          dp[0][0]=0;//初始化          for(i=1;i<=len1;i++)//初始化          {              dp[i]=new int[len2+1];//申请二维动态数组，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(i=1;i<=len2;i++)//初始化             dp[0][i]=dp[0][i-1]+score['-'][s2[i-1]];//注意下标，dp数组是从1开始，s1和s2都是从0开始        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,(Max(temp2,temp3)));            }        cout<<dp[len1][len2]<<endl;        delete[] dp;//释放动态数组     }    return 0;}