poj1080 - Human Gene Functions (dp)

标签：cst   hat   添加   std   return   mini   str   sed   最大的   

题面
It is well known that a human gene can be considered as a sequence, consisting of four nucleotides, which are simply denoted by four letters, A, C, G, and T. Biologists have been interested in identifying human genes and determining their functions, because these can be used to diagnose human diseases and to design new drugs for them.
题意
在两个字符串中,添加‘-‘,使得长度相同,并且对应的权值和最大.
注意两个‘-‘ 不能对应
思路
转换为lcs问题,类似的地方在于,求出对应最大的权值,但是同时考虑不再lcs里面的,就是和‘-‘相对应

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

#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define lson l,mid,ls
#define rson mid+1,r,rs
#define ls (rt<<1)
#define rs ((rt<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int loveisblue = 486;
const int maxn = 205;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

const int mat[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,-1,5}
};


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

char s1[maxn],s2[maxn];
int dp[maxn][maxn];
int main() {
    ios::sync_with_stdio(true);
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

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

    return 0;
}

poj1080 - Human Gene Functions (dp)

标签：cst   hat   添加   std   return   mini   str   sed   最大的   

原文地址：https://www.cnblogs.com/ZGQblogs/p/11569735.html