hdu 1195 Open the Lock

挺好的一道题目，双向bfs的题目，也能单向bfs写。不过双向写的代码太恶心了......还是因为不熟悉

题目大意：给出原密码和准确密码，对于原密码每次可以进行三个操作：
1：任意一位+1 且9+1=1
2：任意一位-1 且1-1=9
3：相邻两位互换，且首位和末位不是相邻的。
（密码只有四位数，每个数字1-9）题目要求原密码变成准确密码的最少步数

单向bfs：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <climits>//形如INT_MAX一类的
#define MAX 1050
#define INF 0x7FFFFFFF
# define eps 1e-5
//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;struct node
{int num[4];int step;
}first,last;bool vis[11][11][11][11];bool ok(node t)
{int cnt = 0;for(int i=0; i<4; i++){if(t.num[i] == last.num[i])cnt++;}if(cnt == 4)return 1;return 0;
}void bfs()
{queue<node> q;vis[first.num[0]][first.num[1]][first.num[2]][first.num[3]] = 1;first.step = 0;q.push(first);while(!q.empty()){node tt = q.front();q.pop();node t;if(ok(tt)){cout << tt.step << endl;return ;}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 9)t.num[i] = 1;elset.num[i] ++;t.step = tt.step + 1;if(!vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]]){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = 1;}}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 1)t.num[i] = 9;elset.num[i] --;t.step = tt.step + 1;if(!vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]]){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = 1;}}for(int i=0; i<3; i++){t = tt;swap(t.num[i],t.num[i+1]);t.step = tt.step + 1;if(!vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]]){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = 1;}}}
}int main()
{int t;char s[10],e[10];cin >> t;while(t--){cin >> s >> e;for(int i=0; i<4; i++)first.num[i] = s[i] - '0';for(int i=0; i<4; i++)last.num[i] = e[i] - '0';memset(vis,0,sizeof(vis));bfs();}return 0;
}

双向bfs：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>//#pragma comment(linker, "/STACK:36777216") ///传说中的外挂
using namespace std;struct node
{int num[4];int step;
} first,last;int vis[11][11][11][11];
int vis2[11][11][11][11];bool ok1(node a)
{if(vis[a.num[0]][a.num[1]][a.num[2]][a.num[3]] != -1)return 1;return 0;
}
bool ok2(node a)
{if(vis2[a.num[0]][a.num[1]][a.num[2]][a.num[3]] != -1)return 1;return 0;
}void bfs()
{queue<node>q,q2;memset(vis,-1,sizeof(vis));memset(vis2,-1,sizeof(vis));int tmp1=0,tmp2=0;node t,tt;vis[first.num[0]][first.num[1]][first.num[2]][first.num[3]] = 0;first.step = 0;q.push(first);vis2[last.num[0]][last.num[1]][last.num[2]][last.num[3]] = 0;last.step = 0;q2.push(last);while(1){while(!q.empty() && q.front().step == tmp1){tt = q.front();q.pop();if(ok2(tt)){int k = vis2[tt.num[0]][tt.num[1]][tt.num[2]][tt.num[3]];cout << tt.step + k<< endl;return ;}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 9)t.num[i] = 1;elset.num[i] ++;t.step = tt.step + 1;if(ok2(t)){int k = vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok1(t)){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 1)t.num[i] = 9;elset.num[i] --;t.step = tt.step + 1;if(ok2(t)){int k = vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok1(t)){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}for(int i=0; i<3; i++){t = tt;swap(t.num[i],t.num[i+1]);t.step = tt.step + 1;if(ok2(t)){int k = vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok1(t)){q.push(t);vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}}tmp1 = q.front().step;//cout << "tmp1" <<' ' << tmp1 << endl;while(!q2.empty() && q2.front().step == tmp2){tt = q2.front();q2.pop();if(ok1(tt)){int k = vis[tt.num[0]][tt.num[1]][tt.num[2]][tt.num[3]];cout << tt.step + k<< endl;return ;}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 9)t.num[i] = 1;elset.num[i] ++;t.step = tt.step + 1;if(ok1(t)){int k = vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok2(t)){q2.push(t);vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}for(int i=0; i<4; i++){t = tt;if(t.num[i] == 1)t.num[i] = 9;elset.num[i] --;t.step = tt.step + 1;if(ok1(t)){int k = vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok2(t)){q2.push(t);vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}for(int i=0; i<3; i++){t = tt;swap(t.num[i],t.num[i+1]);t.step = tt.step + 1;if(ok1(t)){int k = vis[t.num[0]][t.num[1]][t.num[2]][t.num[3]];cout << t.step + k<< endl;return ;}if(!ok2(t)){q2.push(t);vis2[t.num[0]][t.num[1]][t.num[2]][t.num[3]] = t.step;}}}tmp2 = q2.front().step;}}int main()
{int t;char s[10],e[10];cin >> t;while(t--){cin >> s >> e;for(int i=0; i<4; i++)first.num[i] = s[i] - '0';for(int i=0; i<4; i++)last.num[i] = e[i] - '0';bfs();}return 0;
}

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