题意:给定两个四位数,一个代表密码的初始状态,一个代表密码的开锁状态。对于四位数中的任意一位,可以将其减一或者加一,数字只是1到9,没有0,如果是1,要减一的话,就变成了9,相反,9要加一的话,就变成了1.还可以将两个相邻的位交换,但是最左的位不能和最右的位交换,最右和最左亦然。每改变一次算一步操作。问给定这两种状态,从初始状态到开锁状态最少需要多少步操作?
题解:bfs求解所有状态,最先到达的点的步数肯定最少。
对每一位进行加减和交换处理,特殊情况特殊考虑。
代码:
#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std;
const int MAX_ = 10001;
bool vis[MAX_];
int d[MAX_];
int last;
char str2[10];
int change(char str[]){
int cnt = 0;
int len = strlen(str);
for(int i = 0; i < len; i++){
cnt = cnt * 10 + str[i] - '0';
}
return cnt;
}
int bfs(char str[]){
queue<int>Q;
char s[10];
memset(vis,0,sizeof(vis));
int k = change(str);
Q.push(k);
vis[k] = 1;
d[k] = 0;
while(!Q.empty()){
int N = 4,t, tmp = Q.front();
Q.pop();
t = tmp;
if(t == last)return d[t];
for(int i = N - 1; i >= 0; i--){
str[i] = '0' + t % 10;
t /= 10;
}
str[N] = '\0';
for(int i = 0; i < N; i++) {
if(str[i] > '1') {
s[i] = str[i] - 1;
} else {
s[i] = '9';
}
for(int j = 0; j < i; j++) {
s[j] = str[j];
}
for(int j = i+1; j < N; j++) {
s[j] = str[j];
}
s[N] = '\0';
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
}
for(int i = 0; i < N; i++) {
if(str[i] < '9') {
s[i] = str[i] + 1;
} else {
s[i] = '1';
}
for(int j = 0; j < i; j++) {
s[j] = str[j];
}
for(int j = i+1; j < N; j++) {
s[j] = str[j];
}
s[N] = '\0';
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
}
for(int i = 0; i < N; i++){
for(int j = 0; j < N; j++) {
s[j] = str[j];
}
s[N] = '\0';
if(i == 0){
swap(s[i],s[i+1]);
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
continue;
}
if(i == N - 1){
swap(s[i],s[i-1]);
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
continue;
}
swap(s[i],s[i+1]);
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
for(int j = 0; j < N; j++) {
s[j] = str[j];
}
s[N] = '\0';
swap(s[i],s[i-1]);
t = change(s);
if(vis[t])continue;
vis[t] = 1;
d[t] = d[tmp] + 1;
Q.push(t);
}
}
return -1;
}
int main(){
int Case;
char str1[10];
scanf("%d",&Case);
while(Case--){
scanf("%s%s",str1,str2);
last = change(str2);
printf("%d\n",bfs(str1));
}
return 0;
}