标签:eof deque empty users window name define str element
给你8*8的棋盘和4个棋子初始位置、最终位置,问你能否在8次操作后达到该状态。
双向BFS,起点开始正搜4步,终点倒搜4步,map标记。
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 #include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const double E=2.718281828; 45 const double PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const double ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=(int)1e6+10; 51 52 int mp[10][10]; 53 unordered_map<ull,bool>mark; 54 struct node 55 { 56 ull id; 57 int step; 58 }; 59 ull get() 60 { 61 ull now=0,temp=1; 62 int cnt=-1; 63 for(int i=1;i<=8;++i) 64 { 65 for(int j=1;j<=8;++j) 66 { 67 ++cnt; 68 if(mp[i][j]) 69 now|=temp<<cnt; 70 } 71 } 72 return now; 73 } 74 bool ok(int x,int y) 75 { 76 return x>=1&&x<=8&&y>=1&&y<=8; 77 } 78 bool ans; 79 void bfs(ull start,bool f) 80 { 81 queue<node>q; 82 q.push({start,0}); 83 while(!q.empty()) 84 { 85 node now=q.front();q.pop(); 86 if(!f) 87 { 88 if(mark[now.id])continue; 89 mark[now.id]=1; 90 } 91 else 92 { 93 if(mark[now.id]) 94 { 95 ans=1; 96 return; 97 } 98 } 99 if(now.step==4)continue; 100 int cnt=-1; 101 for(int i=1;i<=8;++i) 102 { 103 for(int j=1;j<=8;++j) 104 { 105 ++cnt; 106 mp[i][j]=(int)(now.id>>cnt)&1; 107 } 108 } 109 for(int i=1;i<=8;++i) 110 { 111 for(int j=1;j<=8;++j) 112 { 113 if(mp[i][j]&&ok(i-1,j)&&mp[i-1][j]==1&&ok(i-2,j)&&mp[i-2][j]==0) 114 { 115 mp[i-2][j]=1,mp[i][j]=0; 116 q.push({get(),now.step+1}); 117 mp[i-2][j]=0,mp[i][j]=1; 118 } 119 if(mp[i][j]&&ok(i+1,j)&&mp[i+1][j]==1&&ok(i+2,j)&&mp[i+2][j]==0) 120 { 121 mp[i+2][j]=1,mp[i][j]=0; 122 q.push({get(),now.step+1}); 123 mp[i+2][j]=0,mp[i][j]=1; 124 } 125 if(mp[i][j]&&ok(i,j-1)&&mp[i][j-1]==1&&ok(i,j-2)&&mp[i][j-2]==0) 126 { 127 mp[i][j-2]=1,mp[i][j]=0; 128 q.push({get(),now.step+1}); 129 mp[i][j-2]=0,mp[i][j]=1; 130 } 131 if(mp[i][j]&&ok(i,j+1)&&mp[i][j+1]==1&&ok(i,j+2)&&mp[i][j+2]==0) 132 { 133 mp[i][j+2]=1,mp[i][j]=0; 134 q.push({get(),now.step+1}); 135 mp[i][j+2]=0,mp[i][j]=1; 136 } 137 //-------------------------------------------------------------- 138 if(mp[i][j]==1&&ok(i-1,j)&&mp[i-1][j]==0) 139 { 140 mp[i-1][j]=1,mp[i][j]=0; 141 q.push({get(),now.step+1}); 142 mp[i-1][j]=0;mp[i][j]=1; 143 } 144 if(mp[i][j]==1&&ok(i+1,j)&&mp[i+1][j]==0) 145 { 146 mp[i+1][j]=1,mp[i][j]=0; 147 q.push({get(),now.step+1}); 148 mp[i+1][j]=0;mp[i][j]=1; 149 } 150 if(mp[i][j]==1&&ok(i,j-1)&&mp[i][j-1]==0) 151 { 152 mp[i][j-1]=1,mp[i][j]=0; 153 q.push({get(),now.step+1}); 154 mp[i][j-1]=0;mp[i][j]=1; 155 } 156 if(mp[i][j]==1&&ok(i,j+1)&&mp[i][j+1]==0) 157 { 158 mp[i][j+1]=1,mp[i][j]=0; 159 q.push({get(),now.step+1}); 160 mp[i][j+1]=0;mp[i][j]=1; 161 } 162 } 163 } 164 } 165 } 166 167 int main() 168 { 169 int x,y; 170 while(~sc("%d%d",&x,&y)) 171 { 172 ans=0; 173 mark.clear(); 174 mem(mp,0); 175 mp[x][y]=1; 176 for(int i=2;i<=4;++i) 177 { 178 sc("%d%d",&x,&y); 179 mp[x][y]=1; 180 } 181 bfs(get(),0); 182 mem(mp,0); 183 for(int i=1;i<=4;++i) 184 { 185 sc("%d%d",&x,&y); 186 mp[x][y]=1; 187 } 188 bfs(get(),1); 189 if(ans) 190 pr("YES\n"); 191 else 192 pr("NO\n"); 193 } 194 return 0; 195 } 196 197 /**************************************************************************************/
标签:eof deque empty users window name define str element
原文地址:https://www.cnblogs.com/--HPY-7m/p/11991974.html