poj 1080:Human Gene Functions

来源：互联网发布：大金的淘宝店编辑：程序博客网时间：2024/06/05 20:29

解题思路：

（1）打表

（2）动态规划边界状态

（3）动态规划状态转移

#include<iostream>#include<cstring>using namespace std;int main(){int v['T'+1]['T'+1];v['A']['A'] = 5;v['C']['C'] = 5;v['G']['G'] = 5;v['T']['T'] = 5;v['A']['C'] = v['C']['A'] = -1;v['A']['G'] = v['G']['A'] = -2;v['A']['T'] = v['T']['A'] = -1;v['A']['-'] = v['-']['A'] = -3;v['C']['G'] = v['G']['C'] = -3;v['C']['T'] = v['T']['C'] = -2;v['C']['-'] = v['-']['C'] = -4;v['G']['T'] = v['T']['G'] = -2;v['G']['-'] = v['-']['G'] = -2;v['T']['-'] = v['-']['T'] = -1;v['-']['-'] = -9999;int m;cin>>m;while(m--){char a[110],b[110];int lena=0,lenb=0;cin>>lena;cin>>a;cin>>lenb;cin>>b;int dp[110][110];dp[0][0]=0;for(int i=1;i<=lena;i++)dp[0][i] = v['-'][a[i-1]] + dp[0][i-1];for(int i=1;i<=lenb;i++)dp[i][0] = v['-'][b[i-1]] + dp[i-1][0];for(int i=1;i<=lenb;i++){for(int j=1;j<=lena;j++){int p = dp[i-1][j] + v['-'][b[i-1]];int q = dp[i][j-1] + v['-'][a[j-1]];int r = dp[i-1][j-1] + v[b[i-1]][a[j-1]];int t = max(p,q);t = max(t,r);dp[i][j] = t;} } cout<<dp[lenb][lena]<<endl; }return 0;}

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