【dp】POJ - 1080 Human Gene Functions
2017-12-04 22:36
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【dp】POJ - 1080 Human Gene Functions
【poj链接】【vj链接】
题目大意
给两个字符串包括A,C,G,T,一一对应的组合有权重,加入’-‘使两字符串长度相等,求最大权重和,注意:两个’-‘不可对应。Input
The input consists of T test cases. The number of test cases ) (T is given in the first line of the input file. Each test case consists of two lines: each line contains an integer, the length of a gene, followed by a gene sequence. The length of each gene sequence is at least one and does not exceed 100.
Output
The output should print the similarity of each test case, one per line.
Sample Input
2 7 AGTGATG 5 GTTAG 7 AGCTATT 9 AGCTTTAAA
Sample Output
14 21
样例一解释
AGTGATG -GTTA-G This alignment gives a score of (-3)+5+5+(-2)+5+(-1) +5=14
解题思路
把字符串转化成int数组a,b方便表示和计算,dp[i][j][k]表示a里前i个和b里前j个都分别延长成长度k的最大权重,状态转移方程为if(i>0&&k>=j+1) dp[i][j][k]=max(dp[i][j][k],dp[i-1][j][k-1]+v[a[i]][4]); if(j>0&&k>=i+1)dp[i][j][k]=max(dp[i][j][k],dp[i][j-1][k-1]+v[4][b[j]]); if(i>0&&j>0&&k<=i+j-1)dp[i][j][k]=max(dp[i][j][k],dp[i-1][j-1][k-1]+v[a[i]][b[j]]);
是以最后结尾是一个a里面的数和’-‘结尾还是b里面的还是a,b里面的来分类的。
一个坑点是k的范围,可能dp[i][j][k-1]根本不存在。
后来发现并不需要k,只要两维来表示a里前i个和b里前j个的最大权重就可以了。
【别人的博客】
AC代码
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int dp[107][107][207],a[107],b[107];
char s[107];
int change(char c)
{
if(c=='A') return 0;
if(c=='C') return 1;
if(c=='G') return 2;
if(c=='T') return 3;
}
int v[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,-1,0
};//我觉得这样比较方便
int main()
{
int t,i,j,k;
scanf("%d",&t);
while(t--)
{
int n1,n2;
scanf("%d%s",&n1,s);
for(i=0;i<n1;i++)
a[i+1]=change(s[i]);
scanf("%d%s",&n2,s);
for(i=0;i<n2;i++)
b[i+1]=change(s[i]);
for(i=0;i<=n1;i++)
{
for(j=0;j<=n2;j++)
{
for(k=max(i,j);k<=i+j;k++)//注意k的范围
{
if(i==0&&j==0) {dp[i][j][k]=0;continue;}//注意初始化
dp[i][j][k]=-1000000;//注意清0
if(i>0&&k>=j+1) dp[i][j][k]=max(dp[i][j][k],dp[i-1][j][k-1]+v[a[i]][4]); if(j>0&&k>=i+1)dp[i][j][k]=max(dp[i][j][k],dp[i][j-1][k-1]+v[4][b[j]]); if(i>0&&j>0&&k<=i+j-1)dp[i][j][k]=max(dp[i][j][k],dp[i-1][j-1][k-1]+v[a[i]][b[j]]);}
}
}
int maxx=-1000000;
for(k=max(n1,n2);k<=n1+n2;k++)
if(dp[n1][n2][k]>maxx) maxx=dp[n1][n2][k];
printf("%d\n",maxx);
}
return 0;
}
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