Human Gene Functions

http://poj.org/problem?id=1080

题意：已知一个人类基因的积分表，给出两个字符串，计算这两个串的最大相似程度；

思路：最长公共子序列的变形，用score[][] 表示积分值，dp[i][j]表示字符串s1[1,2,,,i]和s2[1,2,,,j]的分值，对一个dp[i][j],有三种可能：

> s1取第i个字符，s2取‘-’时，dp[i][j] = dp[i-1][j] + score[s1[i]][0];

>s1取‘-’,s2取第j个字符时，dp[i][j] = dp[i][j-1] + score[0][s2[j]];

>s1取第i个字符，s2取第j个字符时，dp[i][j] = dp[i-1][j-1] + score[s1[i]][s2[j]];

取上述三个数的最大值即是dp[i][j];

注意初始化也与LCS不同，

dp[0][0] = 0;

当i= 0时，dp[0][j] = dp[0][j-1] + score[0][s2[j]];

当j = 0时，dp[i][0] = dp[i-1][0] + score[s1[i]][0];

#include<stdio.h>
#include<string.h>

int score[5][5] ={0,-3,-4,-2,-1,
                 -3,5,-1,-2,-1,
                 -4,-1,5,-3,-2,
                 -2,-2,-3,5,-2,
                 -1,-1,-2,-2,5,};//积分表

int change(char ch)
{
    if(ch == 'A')
        return 1;
    else if(ch == 'C')
        return 2;
    else if(ch == 'G')
        return 3;
    else if(ch == 'T')
        return 4;
}

int max(int a,int b,int c)
{
    if(a >= b && a >= c)
        return a;
    if(b >= a && b >= c)
        return b;
    if(c >= a && c >= b)
        return c;
}

int a[110],b[110],len1,len2,dp[110][110];
char s1[110],s2[110];

int main()
{
    int test,i,j;
    scanf("%d",&test);
    while(test--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));

        scanf("%d %s",&len1,s1);
        scanf("%d %s",&len2,s2);

        for(i = 0; i < len1; i++)
            a[i+1] = change(s1[i]);
        for(i = 0; i < len2; i++)
            b[i+1] = change(s2[i]);
        
        //初始化    
        memset(dp,0,sizeof(dp));
        dp[0][0] = 0;
        for(int i = 1; i <= len2; i++)
            dp[0][i] = dp[0][i-1]+score[0][b[i]];
        for(int i = 1; i <= len1; i++)
            dp[i][0] = dp[i-1][0]+score[a[i]][0];

        for(i = 1; i <= len1; i++)
        {
            for(j = 1; j <= len2; j++)
            {
                int tmp1 = dp[i-1][j]+score[a[i]][0];//s1取第i个字母，s2取‘-’；
                int tmp2 = dp[i][j-1]+score[0][b[j]];//s1取‘-’，s2取第j个字母；
                int tmp3 = dp[i-1][j-1]+score[a[i]][b[j]];//s1取第i个字母，s2取第j个字母；
                int tmp = max(tmp1,tmp2,tmp3);
                dp[i][j] += tmp;
            }
        }
        printf("%d
",dp[len1][len2]);
    }
    return 0;
}