poj——1080（dp）

来源：互联网发布：java培训骗局编辑：程序博客网时间：2024/04/20 17:24

题目地址：http://poj.org/problem?id=1080

解析：dp[i][j]来由：

1.dp[i][j-1]转移，则‘-’和s2[j]匹配。

2.dp[i-1][j]转移，则s1[i]和‘-’匹配。

3.dp[i-1][j-1],则s1[i]和s2[j]匹配。三者取最大值即可。

#include <iostream>#include <cmath>#include <string>#include <cstring>#include <cstdlib>#include <ctime>#include <algorithm>#include <cstdio>#include <map>#include <vector>#include <set>#include <queue>#include <stack>using namespace std;typedef long long ll;#define INF 0x7fffffff#define MAX(a,b) a>b?a:b#define MIN(a,b) a>b?b:a#define N 103int len2,len1;int  dp[N][N];char s1[N],s2[N];int  match[N][N];void init(){match['A']['T']=-1;match['T']['A']=-1;match['A']['A']=5;match['T']['T']=5;match['A']['C']=-1;match['C']['A']=-1;match['A']['G']=-2;match['G']['A']=-2;match['C']['G']=-3;match['G']['C']=-3;match['G']['G']=5;match['C']['C']=5;match['C']['T']=-2;match['T']['C']=-2;match['T']['G']=-2;match['G']['T']=-2;match['A']['-']=-3;match['-']['A']=-3;match['-']['C']=-4;match['C']['-']=-4;match['G']['-']=-2;match['-']['G']=-2;match['-']['T']=-1;match['T']['-']=-1;}void solve(){int i,j;dp[0][0]=0;for(i=1;i<=len1;i++)  dp[i][0]=dp[i-1][0]+match[s1[i]]['-'];    for(i=1;i<=len2;i++)      dp[0][i]=dp[0][i-1]+match['-'][s2[i]];    for(i=1;i<=len1;i++)     for(j=1;j<=len2;j++)      {     dp[i][j]=max(dp[i-1][j-1]+match[s1[i]][s2[j]],max(dp[i-1][j]+match[s1[i]]['-'],dp[i][j-1]+match['-'][s2[j]]));     }     cout<<dp[len1][len2]<<endl;}int main(){  int t;  cin>>t;  init();      while(t--){      cin>>len1>>s1+1;      cin>>len2>>s2+1;      solve();      }return 0;}