POJ 1080 Human Gene Functions

来源：互联网发布：c 编程思想编辑：程序博客网时间：2024/04/30 06:51

动态规划问题

最长公共子串的变形主要注意初始化

#include<iostream>using namespace std;#define N 105int v[N][N],dp[N][N];char s1[N],s2[N];int max(int a,int b){return a>b?a:b;}void init(){v['A']['A']=5;v['C']['C']=5;v['G']['G']=5;v['T']['T']=5;v['A']['C']=-1;v['A']['G']=-2;v['A']['T']=-1;v['A']['-']=-3;v['C']['A']=-1;v['C']['G']=-3;v['C']['T']=-2;v['C']['-']=-4;v['G']['A']=-2;v['G']['C']=-3;v['G']['T']=-2;v['G']['-']=-2;v['T']['A']=-1;v['T']['C']=-2;v['T']['G']=-2;v['T']['-']=-1;v['-']['A']=-3;v['-']['C']=-4;v['-']['G']=-2;v['-']['T']=-1;}int main(){int la,lb,t,i,j;cin>>t;init();while(t--){cin>>la>>s1;cin>>lb>>s2;for(i=1;i<=la;i++)dp[0][i]=dp[0][i-1]+v['-'][s2[i-1]];  //dp[i][j]表示s1前i个字符 s2前j个字符时 最大的相似度  dp[i][0]表示s1前i个与s2都不符合 for(i=1;i<=lb;i++)dp[i][0]=dp[i-1][0]+v[s1[i-1]]['-'];

for(i=1;i<=la;i++)for(j=1;j<=lb;j++){dp[i][j]=max(max(dp[i-1][j]+v[s1[i-1]]['-'],dp[i][j-1]+v['-'][s2[j-1]]),dp[i-1][j-1]+v[s1[i-1]][s2[j-1]]);}cout<<dp[la][lb]<<endl;}return 0;}