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 shoshould 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 100int 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;}

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