算法提高 题目 2 密码锁 （bfs最少步数）

思路:

用bfs保存每种状态,每次从状态数组中取出队头元素，进行交换，得到新数组塞进状态数组中

代码：

#include<iostream>#include<algorithm>#include<string.h>using namespace std;int n,a[15],b[15];int step[100000]={0};int ss[100000][15]={0};int check(int c[15]){int i;for (i=0; i<n; i++){if (c[i] == 2 && i+3<n){if (c[i+1]==0 && c[i+2]==1 && c[i+3]==2){return 1;}else{return 0;} }}return 0;}int pc(int c[15],int m){for (int i=0; i<m; i++){int j;for (j=0; j<n; j++){if (ss[i][j] != c[j]){break;}}if (j==n){return 0;}}return 1;}int bfs(){int front = 0,tail = 1;int x,y,i,j,k,top=0;memcpy(ss[front],a,sizeof(a));while (front < tail){memcpy(a,ss[front],sizeof(ss[front]));if (check(a)){return step[front];}x = top-1;y = top+1;memcpy(b,a,sizeof(a));if (x>=0&&x<n){if (a[top]!=a[x]){b[x] = a[top];b[top] = a[x];if (pc(b,tail)){memcpy(ss[tail],b,sizeof(b));step[tail] = step[front]+1;if (check(b)){return step[tail];}tail++;}}}memcpy(b,a,sizeof(a));if (y>=0 && y<n){if (a[top] != a[y]){b[y] = a[top];b[top] = a[y];if (pc(b,tail)){memcpy(ss[tail],b,sizeof(b));step[tail] = step[front]+1;if (check(b)){return step[tail];}tail++;}}}top++;if (top == n){top = 0;front++;}} return -1; }int main(){int c,v[3]={0};char str[20];cin>>n;cin>>str;for (int i=0; i<n; i++){a[i] = str[i]-'0';v[a[i]]++;}cout<<bfs();return 0;}