POJ #1080 - Human Gene Functions

A classic 2D DP problem. A disguise of LCS - actually not very hard to decode: it is about 2 sequences‘ matching, though with a weight value of each match.

The point of this problem: how to decode problem statement and how to distill the actuall model behind. Coding is not very hard, but my 80% debug time is spent on a stupid detail: in the 2 level for loop, indices start from 1, but char index of each string should be i - 1, j - 1.

Main reference: http://blog.csdn.net/xiaoxiaoluo/article/details/7366537

The AC code:

//    1080
//    http://blog.csdn.net/xiaoxiaoluo/article/details/7366537
/*
 *    Q1: What is the target value? Similarity Val
 *    Q2: What are the Variables? indices of two strings
 *    So, dp[i][j] = val
 *    in which i is the index of 1st string, j is of the 2nd, and value is similarity
 *
 *    The key is recurrence relations:
 *  Eq1: s0[i] isChar, s1[j] isChar
         dp[i][j] = dp[i-1][j-1] + score[s0[i]][s1[j]]
    Eq2: s0[i] isChar, s1[j] is ‘-‘
         dp[i][j] = dp[i][j-1] + score[‘-‘][s1[j]]
    Eq3: s0[i] is ‘-‘, s1[j] isChar
         dp[i][j] = dp[i-1][j] + score[s0[i]][‘-‘]

    The above eqs are to simulate LCS eqs. ‘-‘ is artificially put to match strings
 */
#include <stdio.h>

#define MAX_LEN 100

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 Inx(char c)
{
    switch (c)
    {
    case ‘A‘: return 0;
    case ‘C‘: return 1;
    case ‘G‘: return 2;
    case ‘T‘: return 3;
    case ‘-‘: return 4;
    }
}

int max2(int a, int b)
{
    return (a > b) ? (a) : (b);
}

int calc(int len0, char in0[MAX_LEN], int len1, char in1[MAX_LEN])
{
    int dp[MAX_LEN + 1][MAX_LEN + 1];

    //    Init
    dp[0][0] = 0;
    for (int i = 1; i <= len0; i ++)
    {
        dp[i][0] = dp[i - 1][0] + score[Inx(in0[i-1])][Inx(‘-‘)]; // eq2
    }
    for (int j = 1; j <= len1; j++)
    {
        dp[0][j] = dp[0][j - 1] + score[Inx(‘-‘)][Inx(in1[j-1])]; // eq1
    }

    //    Go
    for (int i = 1; i <= len0; i ++)
    for (int j = 1; j <= len1; j ++)
    {
        int val0 = dp[i - 1][j - 1] + score[Inx(in0[i-1])][Inx(in1[j-1])];
        int val1 = dp[i][j - 1] + score[Inx(‘-‘)][Inx(in1[j-1])];
        int val2 = dp[i - 1][j] + score[Inx(in0[i-1])][Inx(‘-‘)];
        dp[i][j] = max2(val0, max2(val1, val2));
    }

    return dp[len0][len1];
}
int main()
{
    int n; scanf("%d", &n);
    while (n--)
    {
        int len[2] = { 0 };
        char in0[MAX_LEN] = { 0 };
        char in1[MAX_LEN] = { 0 };

        scanf("%d", len);        scanf("%s", in0);
        scanf("%d", len + 1);    scanf("%s", in1);

        int ret = calc(len[0], in0, len[1], in1);
        printf("%d\n", ret);
    }
    return 0;
}

POJ #1080 - Human Gene Functions