hdu1080 Human Gene Functions（带权值最长公共子序列）

最新推荐文章于 2021-05-22 22:29:28 发布

原创最新推荐文章于 2021-05-22 22:29:28 发布 · 586 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#hdu

hdu 专栏收录该内容

231 篇文章

订阅专栏

本文探讨了如何解决带权值的最长公共子序列问题，详细介绍了在原有算法基础上结合权值的更新方式，包括s1[i]等于或不等于s2[j]的情况，并强调了初始化阶段权重的处理以及避免将字母转换为数字的优化技巧。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

带权值的最长公共子序列。

原本s1[i] == s2[j]时dp[i][j] = dp[i - 1][j - 1] + 1，现在变为dp[i][j] = dp[i - 1][j - 1] + Map[s1[i]][s2[j]]。

原本s1[i] != s2[j]时dp[i][j] = max(dp[i][j - 1]，dp[i - 1][j] )，现在变为dp[i][j] = max(dp[i][j - 1] + Map['-'][s2[j]]，dp[i - 1][j] + Map[s1[i]]['-'])。

而且既然都加上权值，就可以放在一起比较。同样初始化也要加权值。

还有打表注意的技巧，何必将字母变为数字？多麻烦。直接用字母表示下标就好了~

#include <stdio.h>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 105;
const int INF = 1<<27;

int Map[6][6];

void init()
{
    Map['A']['A'] = Map['T']['T'] = Map['G']['G'] = Map['C']['C'] = 5;
    Map['A']['C'] = Map['C']['A'] = Map['A']['T'] = Map['T']['A'] = Map['T']['-'] = Map['-']['T'] = -1;
    Map['G']['A'] = Map['A']['G'] = Map['T']['C'] = Map['C']['T'] = Map['T']['G'] = Map['G']['T'] = Map['G']['-'] = Map['-']['G'] = -2;
    Map['C']['G'] = Map['G']['C'] = Map['A']['-'] = Map['-']['A'] = -3;
    Map['C']['-'] = Map['-']['C'] = -4;
}
int main()
{
  //  freopen("in.txt", "r", stdin);
    init();
    int dp[N][N];
    int T, j, i, n, l1, l2;
    char s1[N], s2[N];
    scanf("%d", &T);
    while(T --)
    {
        scanf("%d %s", &l1, s1 + 1);
        scanf("%d %s", &l2, s2 + 1);
        for(i = 0; i <= l1; i ++)
            for(j = 0; j <= l2; j ++)
                dp[i][j] = -INF;
        dp[0][0] = 0;
        for(i = 1; i <= l1; i ++) dp[i][0] = dp[i - 1][0] + Map[s1[i]]['-'];
        for(i = 1; i <= l2; i ++) dp[0][i] = dp[0][i - 1] + Map[s2[i]]['-'];
        for(i = 1; i <= l1; i ++)
            for(j = 1; j <= l2; j ++)
                dp[i][j] = max(dp[i - 1][j - 1] + Map[s1[i]][s2[j]], max(dp[i - 1][j] + Map[s1[i]]['-'], dp[i][j - 1] + Map['-'][s2[j]]));
        printf("%d\n", dp[l1][l2]);
    }
    return 0;
}