hdu1080 Human Gene Functions

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

题目大意：基因工程，给你两段基因序列，给定任意两个字符间的相似度，可以根据需要插入减号（-）与字符匹配，求这两段基因序列的最大相似度。

算法分析：一道dp题，类似于求两个字符串的编辑距离，主要找到状态转移方程：f[i][j] = Max(f[i][j - 1] + Match('-', s2[j]), f[i - 1][j] + Match('-', s1[i]), f[i - 1][j - 1] + Match(s1[i], s2[j]));

 

/*dp题
状态转移方程：f[i][j] = Max(f[i][j - 1] + Match('-', s2[j]), f[i - 1][j] + Match('-', s1[i]), f[i - 1][j - 1] + Match(s1[i], s2[j]));
f[i][j]表示第一个字符串的前i个字符和第二个字符串的前j个字符的最大相似度
Match(a, b) 表示字符a与字符b的相似度，程序中预处理，存入数组mat[][]中
*/

#include<stdio.h>
#define NN 102

int Max(int a, int b)
{
    return a > b ? a : b;
}
int main()
{
    int T, i, j, len1, len2;
    char s1[NN], s2[NN];
    int f[NN][NN], mat[NN][NN];
    mat['A']['A'] = mat['C']['C'] = mat['G']['G'] = mat['T']['T'] = 5;
    mat['A']['C'] = mat['C']['A'] = mat['A']['T'] = mat['T']['A'] = mat['-']['T'] = -1;
    mat['A']['G'] = mat['G']['A'] = mat['C']['T'] = mat['T']['C'] = mat['G']['T'] = mat['T']['G'] = mat['-']['G'] = -2;
    mat['-']['A'] = mat['C']['G'] = mat['G']['C'] = -3;
    mat['-']['C'] = -4;

    scanf("%d", &T);
    while (T--){
        scanf("%d%s%d%s", &len1, s1, &len2, s2);
        f[0][0] = 0;
        for (i = 1; i <= len1; i++)
            f[i][0] = f[i - 1][0] + mat['-'][s1[i - 1]];
        for (j = 1; j <= len2; j++)
            f[0][j] = f[0][j - 1] + mat['-'][s2[j - 1]];
        for (i = 1; i <= len1; i++){
            for (j = 1; j <= len2; j++){
                f[i][j] = Max(f[i][j - 1] + mat['-'][s2[j - 1]], Max(f[i - 1][j] + mat['-'][s1[i - 1]], f[i - 1][j - 1] + mat[s1[i - 1]][s2[j - 1]]));
            }
        }

        printf("%d\n", f[len1][len2]);
    }
    return 0;
}