poj_2945 Find the Clones （Trie树 内存分配）

【题目描述】

Doubleville, a small town in Texas, was attacked by the aliens. They have abducted some of the residents and taken them to the a spaceship orbiting around earth. After some (quite unpleasant) human experiments, the aliens cloned the victims, and released multiple copies of them back in Doubleville. So now it might happen that there are 6 identical person named Hugh F. Bumblebee: the original person and its 5 copies. The Federal Bureau of Unauthorized Cloning (FBUC) charged you with the task of determining how many copies were made from each person. To help you in your task, FBUC have collected a DNA sample from each person. All copies of the same person have the same DNA sequence, and different people have different sequences (we know that there are no identical twins in the town, this is not an issue).

【题目要求】

The input contains several blocks of test cases. Each case begins with a line containing two integers: the number 1 ≤ n ≤ 20000 people, and the length 1 ≤ m ≤ 20 of the DNA sequences. The next n lines contain the DNA sequences: each line contains a sequence of m characters, where each character is either `A', `C', `G' or `T'. 
The input is terminated by a block with n = m = 0 .

For each test case, you have to output n lines, each line containing a single integer. The first line contains the number of different people that were not copied. The second line contains the number of people that were copied only once (i.e., there are two identical copies for each such person.) The third line contains the number of people that are present in three identical copies, and so on: the i -th line contains the number of persons that are present in i identical copies. For example, if there are 11 samples, one of them is from John Smith, and all the others are from copies of Joe Foobar, then you have to print `1' in the first andthe tenth lines, and `0' in all the other lines.

【样例输入】

9 6
AAAAAA
ACACAC
GTTTTG
ACACAC
GTTTTG
ACACAC
ACACAC
TCCCCC
TCCCCC
0 0

【样例输出】

1
2
0
1
0
0
0
0
0

【我的解法】

这依然是一道典型的Trie树应用题，我们这里想要讨论的是：由于题目中有大量的输入block，对于每个block都需要重新建一棵新的Trie树，如果采用动态内存分配来实现，最后如果不加以释放，是否会占用大量内存空间而导致题目无法通过？而如果我们每一次都释放用过的空间，无疑需要遍历整棵树，如果block的数目很大，又有可能在时间上无法满足要求。

因此，我们决定采用静态内存分配，开定一个可以存储最多20000*20=400000个结点的数组，用来重复存储每一次新建的Trie树。

程序如下：

#include <stdio.h>

typedef struct
{
    char c;
    int num;
    long int fChild,rCousin;
} node;

void newNode(node tree[], long int p, char ch)
{
    tree[p].c=ch;
    tree[p].num=0;
    tree[p].fChild=-1;
    tree[p].rCousin=-1;
}

void print(node tree[], long int p, int a[])
{
    a[tree[p].num]++;
    if (tree[p].fChild>0) print(tree,tree[p].fChild,a);
    if (tree[p].rCousin>0) print(tree,tree[p].rCousin,a);
}

int main()
{
    node tree[400001];
    int m,n;
    scanf("%d%d",&n,&m);
    while (n>0)
    {
        long int total=0,i,j,k;
        char s[25];
        newNode(tree,0,'0');
        for (i=0;i<n;i++)
        {
            scanf("%s",s);
            long int x=0,y;
            for (j=0;j<m;j++)
            {
                y=tree[x].fChild;
                if (y==-1)
                {
                    for (k=j;k<m;k++)
                    {
                        total++;
                        newNode(tree,total,s[k]);
                        tree[x].fChild=total;
                        x=total;
                    }
                    break;
                }
                while (y!=-1)
                {
                    if (tree[y].c==s[j]){x=y;break;}
                    x=y;
                    y=tree[y].rCousin;
                }
                if (y==-1)
                {
                    total++;
                    newNode(tree,total,s[j]);
                    tree[x].rCousin=total;
                    x=total;
                    for (k=j+1;k<m;k++)
                    {
                        total++;
                        newNode(tree,total,s[k]);
                        tree[x].fChild=total;
                        x=total;
                    }
                    break;
                }
            }
            tree[x].num++;
        }
        int a[20001]={0};
        print(tree,0,a);
        for (i=1;i<=n;i++) printf("%d\n",a[i]);
        scanf("%d%d",&n,&m);
    }
    return 0;
}

继而，我又写了一个动态内存分配的程序，且每次都不释放空间，用以换取时间，来测试一下是否能通过评测。

程序如下：

#include <stdio.h>
#include <stdlib.h>

typedef struct node
{
    char c;
    int num;
    struct node* fChild;
    struct node* rCousin;
}* triTree;

triTree newTree(char ch)
{
    triTree p=(triTree) malloc(sizeof(struct node));
    p->c=ch;
    p->num=0;
    p->fChild=NULL;
    p->rCousin=NULL;
    return p;
}

void print(triTree t, int a[])
{
    a[t->num]++;
    if (t->fChild!=NULL) print(t->fChild,a);
    if (t->rCousin!=NULL) print(t->rCousin,a);
}

int main()
{
    int m,n,i,j,k;
    char s[25];
    scanf("%d%d",&n,&m);
    while (n>0)
    {
        triTree t=newTree('0');
        for (i=0;i<n;i++)
        {
            scanf("%s",s);
            triTree x=t,y;
            for (j=0;j<m;j++)
            {
                y=x->fChild;
                if (y==NULL)
                {
                    for (k=j;k<m;k++){x->fChild=newTree(s[k]); x=x->fChild;}
                    break;
                }
                while (y!=NULL)
                {
                    if (y->c==s[j]){x=y;break;}
                    x=y;
                    y=y->rCousin;
                }
                if (y==NULL)
                {
                    x->rCousin=newTree(s[j]);
                    x=x->rCousin;
                    for (k=j+1;k<m;k++){x->fChild=newTree(s[k]); x=x->fChild;}
                    break;
                }
            }
            x->num++;
        }
        int a[20001]={0};
        print(t,a);
        for (i=1;i<=n;i++) printf("%d\n",a[i]);
        scanf("%d%d",&n,&m);
    }
    return 0;
}

两个程序都通过了评测，结果如下（第一个为静态内存分配，第二个为动态内存分配）：

可以看到，在面对需要多次建立大Trie树的情况时，静态分配不仅大大减少了内存的开销，而且在时间上也很有优势。

也许多次开辟新的内存单元，同时大量使用指针调用，远没有静态数组索引来得快。这是值得我们思考和探寻的问题。