【宽搜】Vijos P1360 八数码问题

来源:互联网 发布:人工智能 仿真机器人 编辑:程序博客网 时间:2024/06/16 06:21

题目链接:

  https://vijos.org/p/1360

题目大意:

  3x3格子上放1~8数字,一个空位,每次空位可与上下左右交换,固定终止布局,求输入的起始布局需要几步到达终止布局

题目思路:

  【搜索】

  一眼题BFS,宽搜即可,判重我比较暴力直接把状态记下,没hash、cantor什么的。


////by coolxxx//#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define sqr(a) (a)*(a)#define swap(a,b) (a)^=(b),(b)^=(a),(a)^=(b)#define eps 1e-8#define S 10000#define MAX 0x7f7f7f7f#define PI 3.1415926535897#define N 14#define M 100000using namespace std;int n,m,cas,lll,ans;int r[N]={1,10,100,1000,10000,100000,1000000,10000000,100000000};int d[4]={-3,-1,1,3};int q[M][N]={1,2,3,8,0,4,7,6,5};int v[M];bool u[87654321];bool ok(int x,int y){if(y==-3)return (x/3>0);if(y==-1)return (x%3>0);if(y==1) return (x%3<2);if(y==3) return (x/3<2);}int spfa(){int i,j,now,h=-1,t=0;int x;u[m%r[8]]=1;while(h++<t){for(x=0,i=0;i<9;i++){x+=q[h][i]*r[8-i];if(q[h][i]==0)now=i;}if(x==n)return v[h];for(i=0;i<4;i++){if(ok(now,d[i])){memcpy(q[t+1],q[h],sizeof(q[h]));swap(q[t+1][now],q[t+1][now+d[i]]);for(x=0,j=1;j<9;j++)x+=q[t+1][j]*r[8-j];if(!u[x]){v[++t]=v[h]+1;u[x]=1;}}}}}int main(){#ifndef ONLINE_JUDGE//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;//while(~scanf("%s%s",sn,sm))while(~scanf("%d",&n) && n){m=123804765;printf("%d\n",spfa());}return 0;}/*////*/


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