HDU 1080 Human Gene Functions

最长公共子序列的变形

题目大意：给出两个基因序列，求这两个序列的最大相似度。

题目中的表格给出了两两脱氧核苷酸的相似度。

状态转移方程为：

dp[i][j] = max(dp[i-1][j]+Similarity(s1[i], '-'),
                     dp[i][j-1]+Similarity(s2[j], '-'),
                     dp[i-1][j-1]+Similarity(s1[i], s2[j]));

注意边界的初始化。

//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

char s1[105], s2[105];
int dp[105][105];

int table[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 max(int a, int b, int c)
{
    return max(max(a, b), c);
}

int f(char c)
{
    if(c == 'A')    return 0;
    if(c == 'C')    return 1;
    if(c == 'G')    return 2;
    if(c == 'T')    return 3;
    if(c == '-')    return 4;
}

int Similarity(char c1, char c2)
{    return table[f(c1)][f(c2)];    }

int main(void)
{
    #ifdef LOCAL
        freopen("1080in.txt", "r", stdin);
    #endif

    int T;
    scanf("%d", &T);
    while(T--)
    {
        int len1, len2, i, j;
        dp[0][0] = 0;
        scanf("%d %s", &len1, s1+1);
        scanf("%d %s", &len2, s2+1);
        for(i = 1; i <= len1; ++i)
            dp[i][0] = dp[i-1][0] + Similarity(s1[i], '-');
        for(i = 1; i <= len2; ++i)
            dp[0][i] = dp[0][i-1] + Similarity(s2[i], '-');
        for(i = 1; i <= len1; ++i)
            for(j = 1; j <= len2; ++j)
                dp[i][j] = max(dp[i-1][j]+Similarity(s1[i], '-'),
                                dp[i][j-1]+Similarity(s2[j], '-'),
                                dp[i-1][j-1]+Similarity(s1[i], s2[j]));

        printf("%d
", dp[len1][len2]);
    }
    return 0;
}