八数码问题使用HashTable优化查找后的版本
在《双向广度优先搜索算法框架及八数码问题例程》一文中，给出了使用双向广度优先搜索算法解决八数码问题的一个
简单版本，查找都是线性的，没有优化。这里给出使用HashTable优化后的版本！
/*
* Author: puresky
* Date: 2010.01.12
* Purpose: solve EIGTH NUMBER problem!
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAXN 10000
#define SWAP(a, b) {char t = a; a = b; b = t;}
typedef struct _Node Node;
struct _Node
{
char tile[10]; // represent the tile as a string ending with '\0'
char pos; // the position of 'x'
char dir; //the moving direction of 'x'
int parent; //index of parent node
};
int head[2], tail[2];
Node queue[2][MAXN];// two queues for double directoin BFS
//shift of moving up, down, left ,right
int shift[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
//for output direction!
char dir[4][2] = {{'u', 'd'}, {'d', 'u'}, {'l', 'r'}, {'r', 'l'}};
//test case
char start[10] = "23415x768";
char end[10] = "12345678x";
/*=================hash table start=========================================*/
#define HASH_TABLE_MAX_SIZE 10000
typedef struct HashNode_Struct HashNode;
struct HashNode_Struct
{
char* sKey;
int nValue;
HashNode* pNext;
};
HashNode* ht[2][HASH_TABLE_MAX_SIZE];//hash table data strcutrue
//initialize hash table
void hash_table_init(HashNode *hashTable[])
{
memset(hashTable, 0, sizeof(HashNode*) * HASH_TABLE_MAX_SIZE);
}
//string hash function
unsigned int hash_table_hash_str(const char* skey)
{
const signed char *p = (const signed char*)skey;
unsigned int h = *p;
if(h)
{
for(p += 1; *p != '\0'; ++p)
h = (h << 5) - h + *p;
}
return h;
}
//insert key-value into hash table
void hash_table_insert(HashNode *hashTable[], char* skey, int nvalue)
{
unsigned int pos = hash_table_hash_str(skey) % HASH_TABLE_MAX_SIZE;
HashNode* pHead = hashTable[pos];
while(pHead)
{
if(strcmp(pHead->sKey, skey) == 0)
{
printf("%s already exists!\n", skey);
return ;
}
pHead = pHead->pNext;
}
HashNode* pNewNode = (HashNode*)malloc(sizeof(HashNode));
memset(pNewNode, 0, sizeof(HashNode));
pNewNode->sKey = skey;
pNewNode->nValue = nvalue;
pNewNode->pNext = hashTable[pos];
hashTable[pos] = pNewNode;
}
//lookup a key in the hash table
HashNode* hash_table_lookup(HashNode *hashTable[], const char* skey)
{
unsigned int pos = hash_table_hash_str(skey) % HASH_TABLE_MAX_SIZE;
if(hashTable[pos])
{
HashNode* pHead = hashTable[pos];
while(pHead)
{
if(strcmp(skey, pHead->sKey) == 0)
return pHead;
pHead = pHead->pNext;
}
}
return NULL;
}
//free the memory of the hash table
void hash_table_release(HashNode *hashTable[])
{
int i;
for(i = 0; i < HASH_TABLE_MAX_SIZE; ++i)
{
if(hashTable[i])
{
HashNode* pHead = hashTable[i];
while(pHead)
{
HashNode* pTemp = pHead;
pHead = pHead->pNext;
if(pTemp)
{
free(pTemp);
}
}
}
}
}
/* ===============================hash table end=========================*/
//read a tile 3 by 3
void readtile()
{
int i;
char temp[10];
for(i = 0; i < 9; ++i)
{
scanf("%s", temp);
start[i] = temp[0];
}
start[9] = '\0';
}
//print result
void print_backward(int i)
{
if(queue[0][i].parent != -1)
{
print_backward(queue[0][i].parent);
printf("%c", queue[0][i].dir);
}
}
void print_forward(int j)
{
if(queue[1][j].parent != -1)
{
printf("%c", queue[1][j].dir);
print_forward(queue[1][j].parent);
}
}
void print_result(int i, int j)
{
//printf("%d,%d\n", i, j);
print_backward(i);
print_forward(j);
printf("\n");
}
//init the queue
void init(int qi, const char* state)
{
strcpy(queue[qi][0].tile, state);
queue[qi][0].pos = strchr(state, 'x') - state;
queue[qi][0].parent = -1;
head[qi] = tail[qi] = 0;
}
//check if there are duplicates in the queue
//time comlexity:O(n)
//We can optimise this function using HashTable
int isduplicate(int qi)
{
if(hash_table_lookup(ht[qi], queue[qi][tail[qi]].tile))
{
return 1;
}
return 0;
}
//check if the current node is in another queue!
//time comlexity:O(n)
//We can optimise this function using HashTable
int isintersect(int qi)
{
HashNode* hn = hash_table_lookup(ht[1 - qi], queue[qi][tail[qi]].tile);
if(hn)
{
return hn->nValue;
}
return -1;
}
//expand nodes
int expand(int qi)
{
int i, x, y, r;
Node* p = &(queue[qi][head[qi]]);
head[qi]++;
for(i = 0; i < 4; ++i)
{
x = p->pos / 3 + shift[i][0];
y = p->pos % 3 + shift[i][1];
if(x >= 0 && x <= 2 && y >= 0 && y <= 2)
{
tail[qi]++;
Node* pNew = &(queue[qi][tail[qi]]);
strcpy(pNew->tile, p->tile);
SWAP(pNew->tile[ 3 * x + y], pNew->tile[p->pos]);
pNew->pos = 3 * x + y;
pNew->parent = head[qi] - 1;
pNew->dir = dir[i][qi];
if(isduplicate(qi))
{
tail[qi]--;
}
else
{
if((r = isintersect(qi)) != -1)
{
if(qi == 1)
{
print_result(r, tail[qi]);
}
else
{
print_result(tail[qi], r);
}
return 1;
}
hash_table_insert(ht[qi], pNew->tile, tail[qi]);
}
}
}
return 0;
}
//call expand to generate queues
int solve()
{
init(0, start);
init(1, end);
while(head[0] <= tail[0] && head[1] <= tail[1])
{
//expand the shorter queue firstly
if(tail[0] - head[0] >= tail[1] - head[1])
{
if(expand(1)) return 1;
}
else
{
if(expand(0)) return 1;
}
}
while(head[0] <= tail[0]) if(expand(0)) return 1;
while(head[1] <= tail[1]) if(expand(1)) return 1;
return 0;
}
int main(int argc, char** argv)
{
hash_table_init(ht[0]);
hash_table_init(ht[1]);
readtile();
if(!solve())
{
printf("unsolvable\n");
}
hash_table_release(ht[0]);
hash_table_release(ht[1]);
//system("pause"); //pause
return 0;
}
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