POJ-1080 Human Gene Functions---类似LCS

题目链接：

https://cn.vjudge.net/problem/POJ-1080

题目大意：

给定两组序列，要你求出它们的最大相似度，每个字母与其他字母或自身和空格对应都有一个打分，求在这两个字符串中插入空格，让这两个字符串的匹配分数最大

解题思路：

类似LCS，以dp[i][j]表示s1前i位和s2前j位的最优解。

递推式为：

先不考虑括号

dp[i][j]只由dp[i-1][j-1]递推而来

if(s1[i] == s2[j])dp[i][j] = dp[i - 1][j - 1] + 5

else dp[i][j] = dp[i - 1][j - 1] + Map[id1][id2]此处Map[id1][id2]为s1[i]和s2[j]匹配的分数

考虑括号

dp[i][j]在原来基础上可以由dp[i-1][j]和dp[i][j-1]递推而来

由dp[i-1][j]递推而来的时候是s1[i]和空格匹配

由dp[i][j-1]递推而来的时候是空格和s2[j]匹配

dp[i][j] = max(dp[i][j]，dp[i-1][j] + Map[id1][空格]， dp[i][j - 1] + Map[空格][id2])

注意：

要初始化dp[0][i]和dp[j][0]的值（不能单纯的置为0，置为0是错误的），dp[0][0]=0

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<sstream>
#define Mem(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int INF = 1e9+7;
int dp[105][105];
int Map[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,
};
char s1[105], s2[105];
int main()
{
    map<char, int>id;
    id['A'] = 0;
    id['C'] = 1;
    id['G'] = 2;
    id['T'] = 3;
    id[' '] = 4;
    int T, n, m;
    cin >> T;
    while(T--)
    {
        cin >> n >> (s1 + 1);
        cin >> m >> (s2 + 1);
        Mem(dp, 0);
        for(int i = 1; i <= n; i++)
        {
            dp[i][0] = dp[i - 1][0] + Map[id[s1[i]]][id[' ']];
        }
        for(int i = 1; i <= m; i++)
        {
            dp[0][i] = dp[0][i - 1] + Map[id[' ']][id[s2[i]]];
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                if(s1[i] == s2[j])
                    dp[i][j] = dp[i - 1][j - 1] + 5;
                else
                    dp[i][j] = dp[i - 1][j - 1] + Map[id[s1[i]]][id[s2[j]]];
                //cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
                dp[i][j] = max(dp[i][j], dp[i - 1][j] + Map[id[s1[i]]][id[' ']]);
                dp[i][j] = max(dp[i][j], dp[i][j - 1] + Map[id[' ']][id[s2[j]]]);
                //cout<<i<<" "<<j<<" "<<dp[i][j]<<endl;
            }
            //
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}