Codeup100000609问题 C: 【宽搜入门】8数码难题

题目描述：

初始状态的步数就算1，哈哈

输入：

第一个33的矩阵是原始状态，第二个33的矩阵是目标状态。

输出：

移动所用最少的步数

Input：

2 8 3
1 6 4
7 0 5
1 2 3
8 0 4
7 6 5

Output：

6

提示：

此题直接用全排列散列会造成数组过大，因此采用康托展开来求解
康托展开

实现代码1：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int temp[10] = {
      1,1,2,6,24,120,720,5040,40320,362880 };
int dis[500000]={
     0};

int kangtuo(string s) {
     	//康托展开
	int sum = 0;
	int len = s.length();
	for (int i = 0; i < len - 1; i++) {
     
		int total = 0;
		for (int j = i + 1; j < len; j++) {
     
			if (s[j] < s[i]) {
     
				total++;
			}
		}
		sum += total * temp[len - i - 1];
	}
	return sum;
}

string swap(string s, int i, int j) {
     	//交换
	char ch;
	ch = s[i];
	s[i] = s[j];
	s[j] = ch;
	return s;
}

string move(string s, int i) {
     	//进行移动操作
	int k = s.find('0');
	if (i == 1 && k >= 3) {
     
		s = swap(s, k, k - 3);
	}
	if (i == 2 && k <= 5) {
     
		s = swap(s, k, k + 3);
	}
	if (i == 3 && k % 3 != 0) {
     
		s = swap(s, k, k - 1);
	}
	if (i == 4 && k % 3 != 2) {
     
		s = swap(s, k, k + 1);
	}
	return s;
}

int BFS(string s, string t) {
     
	dis[kangtuo(s)] = 1;
	queue<string> q;
	q.push(s);
	while (!q.empty()) {
     
		string u = q.front();
		q.pop();
		if (u == t) {
     	//相等则返回值表明找到
			return dis[kangtuo(u)];
		}
		for (int i = 1; i <= 4; i++) {
     
			string v = move(u, i);
			if (dis[kangtuo(v)] == 0) {
     
				dis[kangtuo(v)] = dis[kangtuo(u)] + 1;
				q.push(v);
				if(v == t) {
     	//相等则返回值表明找到
					return dis[kangtuo(v)];
				}
			}
		}
	}
	return -1;
}

string NumberToString(int a[][3]) {
     	//整型数组转字符串
	string s;
	for (int i = 0; i < 3; i++) {
     
		for (int j = 0; j < 3; j++) {
     
			s += ('0' + a[i][j]);
		}
	}
	return s;
}

int main()
{
     
	int a[3][3];
	string s, t;
	while (1) {
     
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		s = NumberToString(a);
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		t = NumberToString(a);
		if(s == t){
     
			printf("1\n");
			break;
		}
		int ans = BFS(s, t);
		if(ans!=-1){
     
			printf("%d\n", ans);
		}
		memset(dis, 0, sizeof(dis));
	}
	return 0;
}

实现代码2(map实现)：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int temp[10] = {
      1,1,2,6,24,120,720,5040,40320,362880 };
map<int,int> dis; 

int kangtuo(string s) {
     
	int sum = 0;
	for (int i = 0; i < s.length() - 1; i++) {
     
		int total = 0;
		for (int j = i + 1; j < s.length(); j++) {
     
			if (s[j] < s[i]) {
     
				total++;
			}
		}
		sum += total * temp[s.length() - i - 1];
	}
	return sum;
}

string swap(string s, int i, int j) {
     
	char ch;
	ch = s[i];
	s[i] = s[j];
	s[j] = ch;
	return s;
}

string move(string s, int i) {
     
	int k = s.find('0');
	if (i == 1 && k >= 3) {
     
		s = swap(s, k, k - 3);
	}
	if (i == 2 && k <= 5) {
     
		s = swap(s, k, k + 3);
	}
	if (i == 3 && k % 3 != 0) {
     
		s = swap(s, k, k - 1);
	}
	if (i == 4 && k % 3 != 2) {
     
		s = swap(s, k, k + 1);
	}
	return s;
}

int BFS(string s, string t) {
     
	dis[kangtuo(s)] = 1;
	queue<string> q;
	q.push(s);
	while (!q.empty()) {
     
		string u = q.front();
		q.pop();
		if (u == t) {
     
			return dis[kangtuo(u)];
		}
		for (int i = 1; i <= 4; i++) {
     
			string v = move(u, i);
			if (dis.find(kangtuo(v))==dis.end()) {
     
				dis[kangtuo(v)] = dis[kangtuo(u)] + 1;
				q.push(v);
				if(v == t) {
     	//相等则返回值表明找到
					return dis[kangtuo(v)];
				}
			}
		}
	}
	return -1;
}

string NumberToString(int a[][3]) {
     
	string s;
	for (int i = 0; i < 3; i++) {
     
		for (int j = 0; j < 3; j++) {
     
			s += ('0' + a[i][j]);
		}
	}
	return s;
}

int main()
{
     
	int a[3][3];
	string s, t;
	while (1) {
     
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		s = NumberToString(a);
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		t = NumberToString(a);
		if(s == t){
     
			printf("1\n");
			break;
		}
		int ans = BFS(s, t);
		if(ans!=-1){
     
			printf("%d\n", ans);
		}
		dis.clear();
	}
	return 0;
}

实现代码3(map实现)：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

int temp[10] = {
      1,1,2,6,24,120,720,5040,40320,362880 };
map<string,int> dis; 

string swap(string s, int i, int j) {
     
	char ch;
	ch = s[i];
	s[i] = s[j];
	s[j] = ch;
	return s;
}

string move(string s, int i) {
     
	int k = s.find('0');
	if (i == 1 && k >= 3) {
     
		s = swap(s, k, k - 3);
	}
	if (i == 2 && k <= 5) {
     
		s = swap(s, k, k + 3);
	}
	if (i == 3 && k % 3 != 0) {
     
		s = swap(s, k, k - 1);
	}
	if (i == 4 && k % 3 != 2) {
     
		s = swap(s, k, k + 1);
	}
	return s;
}

int BFS(string s, string t) {
     
	dis[s] = 1;
	queue<string> q;
	q.push(s);
	while (!q.empty()) {
     
		string u = q.front();
		q.pop();
		if (u == t) {
     
			return dis[u];
		}
		for (int i = 1; i <= 4; i++) {
     
			string v = move(u, i);
			if (dis.find(v)==dis.end()) {
     
				dis[v] = dis[u] + 1;
				q.push(v);
				if (v == t){
     
				    return dis[v];
				}
			}
		}
	}
	return -1;
}

string NumberToString(int a[][3]) {
     
	string s;
	for (int i = 0; i < 3; i++) {
     
		for (int j = 0; j < 3; j++) {
     
			s += ('0' + a[i][j]);
		}
	}
	return s;
}

int main()
{
     
	int a[3][3];
	string s, t;
	while (1) {
     
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		s = NumberToString(a);
		for (int i = 0; i < 3; i++) {
     
			for (int j = 0; j < 3; j++) {
     
				if (scanf("%d", &a[i][j]) == EOF) {
     
					return 0;
				}
			}
		}
		t = NumberToString(a);
		if(s == t){
     
			printf("1\n");
			break;
		}
		int ans = BFS(s, t);
		if(ans!=-1){
     
			printf("%d\n", ans);
		}
		dis.clear();
	}
	return 0;
}