ACM/ICPC 之 DP-基因相似度(POJ1080-ZOJ1027)

题意：两端基因片段，各有明确的碱基序列，现有一个碱基匹配的相似度数组，设计程序使得该相似度最大。

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//POJ1080-ZOJ1027
//题解：将s1碱基和s2碱基看做等长，添加一个碱基为'-'，即每次都将上下碱基相互匹配后再和'-'匹配，找到最优解即可
//即分为三种状态：
//    1.s1[i]和s2[j]匹配
//    2.s1[i]和 '-' 匹配
//    3. '-' 和s2[j]匹配
#include <iostream>
#include <cstdio>
#include <map>
using namespace std;

#define MAX 105
#define max(x,y) ((x)>(y)?(x):(y))

char s1[MAX], s2[MAX];
int len1, len2;
int dp[MAX][MAX];    //dp[i][j]：s1基因的前i个碱基和s2基因的前j个碱基中的最优匹配解

map<char, int> gene;    //碱基和adjust下标一一映射
int adjust[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 main()
{
    int T;
    scanf("%d", &T);

    gene['A'] = 0; gene['C'] = 1;
    gene['G'] = 2; gene['T'] = 3;
    gene['-'] = 4;

    while (T--)
    {
        scanf("%d %s", &len1, s1);
        scanf("%d %s", &len2, s2);

        memset(dp, 0, sizeof(dp));

        for (int i = 1; i <= len1; i++)
            dp[i][0] = dp[i-1][0] + adjust[gene[s1[i - 1]]][gene['-']];    //上碱基各碱基仅匹配'-'时
        for (int j = 1; j <= len2; j++)
            dp[0][j] = dp[0][j-1] + adjust[gene['-']][gene[s2[j - 1]]];    //下碱基各碱基仅匹配'-'时

        for (int i = 1; i <= len1; i++)
            for (int j = 1; j <= len2; j++)
            {
                dp[i][j] = max(dp[i][j - 1] + adjust[gene['-']][gene[s2[j - 1]]],
                    dp[i - 1][j] + adjust[gene[s1[i - 1]]][gene['-']]);        //上下碱基各匹配'-'时
                dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + adjust[gene[s1[i - 1]]][gene[s2[j - 1]]]);    //上下碱基相互配对时
            }
        printf("%d
", dp[len1][len2]);
    }


    return 0;
}