poj 1080 zoj 1027(最长公共子序列变种)

来源：互联网发布：淘宝的心怎么快速升级编辑：程序博客网时间：2024/06/05 03:34

http://poj.org/problem?id=1080

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=27

/*zoj 1027   poj 1080 思路： 三种状态,取最大值：s1[i]和s2[j]配 ：dp[i-1][j-1]+cost[my[s1[i]]][my[s2[j]]];s1[i]和'-' 配： dp[i-1][j]+cost[my[s1[i]]][my['-']];s2[j]和'-' 配： dp[i][j-1]+cost[my['-']][my[s2[j]]];注意边界：d[i][0]= cost[my[s1[i]]][my['-']]+dp[i-1][0]; 只能全部和'-'配，同理： dp[0][j]=cost[my['-']][my[s2[j]]]+dp[0][j-1];*/#include <iostream>#include <map>#include <cstring>#include <string>using namespace std;int cost[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,-100000005}};int dp[1005][1005];string s1,s2;map<char,int> my;int main(int argc, char *argv[]){int t,i,n,j,m;my['A']=0; my['C']=1; my['G']=2; my['T']=3; my['-']=4;cin>>t;while(t--){cin>>n>>s1>>m>>s2;s1='0'+s1; s2='0'+s2;dp[0][0]=0;for(i=1;i<=n;i++)dp[i][0]=cost[my[s1[i]]][my['-']]+dp[i-1][0];for(j=1;j<=m;j++)dp[0][j]=cost[my['-']][my[s2[j]]]+dp[0][j-1];for(i=1;i<=n;i++)for(j=1;j<=m;j++){int t1,t2,t3;t1=dp[i-1][j-1]+cost[my[s1[i]]][my[s2[j]]];t2=dp[i-1][j]+cost[my[s1[i]]][my['-']];t3=dp[i][j-1]+cost[my['-']][my[s2[j]]];t1=max(t1,t2);  t1=max(t1,t3);dp[i][j]=t1;}cout<<dp[n][m]<<endl;}return 0;}