洛谷 P1140 相似基因 题解
每日一题 day23 打卡
Analysis
dp[i][j]表示序列A中前i个与序列B中前j个匹配的相似度最大值
所以,dp方程很容易想到:
1.让a[i]与b[j]匹配
2.让a[i]与B序列中一个空位匹配
3.让b[j]与A序列中一个空位匹配
dp[i][j]=max(dp[i][j],dp[i-1][j-1]+form[a[i]][b[j]],dp[i-1][j]+form[a[i]][5],dp[i][j-1]+form[5][b[j]]);
对于初始化,就是分别让每个a[i]和每个b[j]与一个空串匹配
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #define int long long 6 #define maxn 100+10 7 #define INF 2147483647 8 using namespace std; 9 inline int read() 10 { 11 int x=0; 12 bool f=1; 13 char c=getchar(); 14 for(; !isdigit(c); c=getchar()) if(c=='-') f=0; 15 for(; isdigit(c); c=getchar()) x=(x<<3)+(x<<1)+c-'0'; 16 if(f) return x; 17 return 0-x; 18 } 19 inline void write(int x) 20 { 21 if(x<0){putchar('-');x=-x;} 22 if(x>9)write(x/10); 23 putchar(x%10+'0'); 24 } 25 int n,m; 26 char s1[maxn],s2[maxn]; 27 int str1[maxn],str2[maxn],dp[maxn][maxn]; 28 int form[maxn][maxn]= 29 { 30 {0,0,0,0,0,0,0}, 31 {0,5,-1,-2,-1,-3}, 32 {0,-1,5,-3,-2,-4}, 33 {0,-2,-3,5,-2,-2}, 34 {0,-1,-2,-2,5,-1}, 35 {0,-3,-4,-2,-1,0} 36 }; 37 inline int max_four(int x,int y,int z,int o) 38 { 39 return max(max(x,y),max(z,o)); 40 } 41 signed main() 42 { 43 n=read(); 44 for(int i=1;i<=n;i++) cin>>s1[i]; 45 m=read(); 46 for(int i=1;i<=m;i++) cin>>s2[i]; 47 for(int i=1;i<=n;i++) 48 for(int j=1;j<=m;j++) 49 dp[i][j]=-INF; 50 for(int i=1;i<=n;i++) 51 { 52 if(s1[i]=='A') str1[i]=1; 53 else if(s1[i]=='C') str1[i]=2; 54 else if(s1[i]=='G') str1[i]=3; 55 else if(s1[i]=='T') str1[i]=4; 56 } 57 for(int i=1;i<=m;i++) 58 { 59 if(s2[i]=='A') str2[i]=1; 60 else if(s2[i]=='C') str2[i]=2; 61 else if(s2[i]=='G') str2[i]=3; 62 else if(s2[i]=='T') str2[i]=4; 63 } 64 for(int i=1;i<=n;i++) dp[i][0]=dp[i-1][0]+form[str1[i]][5]; 65 for(int i=1;i<=m;i++) dp[0][i]=dp[0][i-1]+form[5][str2[i]]; 66 for(int i=1;i<=n;i++) 67 for(int j=1;j<=m;j++) 68 { 69 dp[i][j]=max_four(dp[i][j],dp[i-1][j-1]+form[str1[i]][str2[j]],dp[i-1][j]+form[str1[i]][5],dp[i][j-1]+form[5][str2[j]]); 70 } 71 write(dp[n][m]); 72 return 0; 73 }
请各位大佬斧正(反正我不认识斧正是什么意思)