初始状态的步数就算1,哈哈
第一个33的矩阵是原始状态,第二个33的矩阵是目标状态。
移动所用最少的步数
2 8 3
1 6 4
7 0 5
1 2 3
8 0 4
7 6 5
6
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
#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;
}
#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;
}
#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;
}