洛谷P1140 相似基因【线性dp】

题目：https://www.luogu.org/problemnew/show/P1140

题意：

给定两串基因串（只包含ATCG），在其中插入任意个‘-’使得他们匹配。（所以一共是5种字符）

这5种字符两两之间有一个匹配数值，要求使这两个字符串的匹配值之和最大。

思路：

dp[i][j]表示匹配了s1中前i个字符和s2中前j个字符的最大匹配值。

完成s1[i]与s2[j]的匹配只有三种可能。

1.s1[i]之前已经匹配好，s2[j]配‘-’。即dp[i][j] = dp[i][j-1]+sco[s2[j]]['-']

2.s2[j]之前已经匹配好，s1[i]配‘-’。即d[i][j] = dp[i-1][j]+sco[s1[i]]['-']

3.s1[i]配s2[j]。即dp[i][j] = dp[i - 1][j - 1] + sco[s1[i]][s2[j]]

#include<cstdio>
#include<cstdlib>
#include<map>
#include<set>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath> 
#include<stack>
#include<queue>
#include<iostream>

#define inf 0x7fffffff
using namespace std;
typedef long long LL;
typedef pair<string, string> pr;

const int sco[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},
};
const int maxn = 105;
int len1, len2;
char s1[maxn], s2[maxn];
int dp[maxn][maxn];

int main()
{
    scanf("%d %s", &len1, s1 + 1);
    scanf("%d %s", &len2, s2 + 1);
    map<char, int>mp;
    mp['A'] = 0;mp['C'] = 1;mp['G'] = 2; mp['T'] = 3;
    for(int i = 0; i <= len1; i++){
        for(int j = 0; j <= len2; j++){
            dp[i][j] = -2e8;
        }
    }
    
    dp[0][0] = 0;
    for(int i = 1; i <= len1; i++){
        dp[i][0] = dp[i - 1][0] + sco[mp[s1[i]]][4];
    }
    for(int i = 1; i <= len2; i++){
        dp[0][i] = dp[0][i - 1] + sco[mp[s2[i]]][4];
    }
    for(int i = 1; i <= len1; i++){
        for(int j = 1; j <= len2; j++){
            dp[i][j] = max(dp[i][j], dp[i][j - 1] + sco[mp[s2[j]]][4]);
            dp[i][j] = max(dp[i][j], dp[i - 1][j] + sco[mp[s1[i]]][4]);
            dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + sco[mp[s1[i]]][mp[s2[j]]]);
        }
    }
    printf("%d
", dp[len1][len2]);
    return 0; 
}