题意：http://acm.hdu.edu.cn/showproblem.php?pid=1401
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。
思路：
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 /**************************************************************************************/
原文地址：https://www.cnblogs.com/--HPY-7m/p/11991974.html
时间： 2024-10-08 03:31:14