Chapter06-Phylogenetic Trees Inherited（POJ 2414）（减少国家DP）
Phylogenetic Trees Inherited
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Time Limit: 3000MS |
Memory Limit: 65536K |
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Total Submissions: 480 |
Accepted: 297 |
Special Judge |
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Description
Among other things,Computational Molecular Biology deals with processing genetic sequences.Considering the evolutionary relationship of two sequences, we can say thatthey are closely related if they do not differ very much. We might representthe
relationship by a tree, putting sequences from ancestors above sequencesfrom their descendants. Such trees are called phylogenetic trees.
Whereas one task of phylogenetics is to infer a tree from given sequences,we'll simplify things a bit and provide a tree structure - this will be acomplete binary tree. You'll be given the n leaves of the tree. Sure you know,n is always a power of 2. Each leaf
is a sequence of amino acids (designated bythe one-character-codes you can see in the figure). All sequences will be ofequal length l. Your task is to derive the sequence of a common ancestor withminimal costs.
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The costs are determined asfollows: every inner node of the tree is marked with a sequence of length l,the cost of an edge of the tree is the number of positions at which the twosequences at the ends of the edge differ, the total cost is the sum of the costsat all edges. The sequence of a common ancestor of all sequences is then foundat the root of the tree. An optimal common ancestor is a common ancestor withminimal total costs.
Input
The input file containsseveral test cases. Each test case starts with two integers n and l, denotingthe number of sequences at the leaves and their length, respectively. Input isterminated by n=l=0. Otherwise, 1<=n<=1024 and 1<=l<=1000. Thenfollow n words of length l over the amino acid alphabet. They represent the leavesof a complete binary tree, from left to right.
Output
For each test case, output aline containing some optimal common ancestor and the minimal total costs.
Sample Input
4 3
AAG
AAA
GGA
AGA
4 3
AAG
AGA
AAA
GGA
4 3
AAG
GGA
AAA
AGA
4 1
A
R
A
R
2 1
W
W
2 1
W
Y
1 1
Q
0 0
Sample Output
AGA 3
AGA 4
AGA 4
R 2
W 0
Y 1
Q 0
Source
【题目大意】给出全然二叉树的全部叶子节点，每一个叶子节点上的字符串的长度是固定长度的，分别比較两个孩子的字符串，选取合适的字符串作为这个孩子的父节点。假如两个叶子节点各自是AAG，GAG。那么我们的父节点能够选择AAG或者GAG，由于他们的权值为1(权值是由父节点依据字符串字母的顺序依次与两个孩子比較的区别之和)，由叶子节点依次往根节点求值，求得最小值，并打印出字符串，假设存在多个字符串具有同样的值。仅仅需打印出当中一个就可以；
【分析：】由于每一个字符串的长度都是一样的，那么我们能够每次取一个字母进行求值，然后遍历字符串的长度。依次获得字符串每一位的和；针对一个字母。每次求父节点的最大值为1，就是两个都不同样，并且仅仅要两个都同样。我们必须取同样的字母就可以获得最小的值。
看到全然二叉树。显然是要用数组来实现，与之前的一层一层稍有不同的是，每一层分别要算出全部父节点的状态值；
由于不须要遍历状态值，所以全部的元素仅仅须要依照其自然顺序就可以。即Z的下标为Z-‘A’；
Java代码例如以下：
import java.util.Arrays;
import java.util.Scanner;
public class Main {
// 节点的最大个数是2*1024-1,
private int dp[][]= new int[1000][2*1024];
private int arrayLen;
private int stringLen;
public void initial(int arrayLen, int stringLen) {
this.arrayLen = arrayLen;
this.stringLen = stringLen;
for (int i = 0; i < stringLen; i++) {
Arrays.fill(dp[i], 0,arrayLen,0);
}
// Arrays.fill(value, 0,arrayLen,0);
}
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner cin = new Scanner(System.in);
int m, n;// 节点数和字符串长度
m = cin.nextInt();
n = cin.nextInt();
Main ma = new Main();
String str = null;
while (!(m == 0 && n == 0)) {
ma.initial(2 * m, n);
for (int i = 0; i < m; i++) {
str = cin.next();
for (int j = 0; j < n; j++) {
ma.dp[j][m + i] = 1 << ((int) (str.charAt(j) - 'A'));
}
}
if (m == 1) {
System.out.println(str + " 0");
} else {
ma.getMinValueAndPrint();
}
m = cin.nextInt();
n = cin.nextInt();
}
}
private void getMinValueAndPrint() {
// 从第0个字母開始
int temp;
int countSum=0;
for (int i = 0; i < stringLen; i++) {
temp = arrayLen / 4;
while (temp >= 1) {
for (int j = temp; j < temp * 2; j++) {
if ((dp[i][2 * j] & dp[i][2 * j + 1]) == 0) {
dp[i][j] = dp[i][2 * j] | dp[i][2 * j + 1];
countSum++;
} else {
dp[i][j] = dp[i][2 * j] & dp[i][2 * j + 1];
}
}
temp = temp/2;
}
}
temp = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < stringLen; i++) {
// 26表示26个字母；
temp = dp[i][1];
for (int count = 0; count < 26; count++) {
if ((temp & 1) != 0) {
sb.append((char)(count + 'A'));
break;
}
temp = temp>>1;
}
}
System.out.println(sb.toString() + " " + countSum);
}
}
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