UVA1533-Moving Pegs（BFS+状态压缩）
Problem UVA1533-Moving Pegs
Accept：106 Submit：375
Time Limit: 3000 mSec
Problem Description
Input
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case is a single integer which means an empty hole number.
Output
For each test case, the first line of the output file contains an integer which is the number of jumps in a shortest sequence of moving pegs. In the second line of the output file, print a sequence of peg movements. Apegmovementconsistsofapairofintegersseparatedbyaspace. Thefirstintegerofthe pair denotes the hole number of the peg that is moving, and the second integer denotes a destination (empty) hole number.
Sample Input
Sample Ouput
10
12 5 3 8 15 12 6 13 7 9 1 7 10 8 7 9 11 14 14 5
题解：15个洞，二进制存储状态是比较正的思路。接下来就是水题了，只不过是把矩形地图换成了三角形地图，预处理一个临接表，存一下对于每个点能到哪些点。因为要字典序，因此顺序很重要，稍加分析就知道周围6个位置的大小关系，注意对于15个记录相邻点的数组，一定要统一顺序，除了字典序，还因为有可能要顺着一个方向走几格，这时顺序一致就很方便。
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 15, maxm = 6; 6 const int dir[maxn][maxm] = 7 { 8 {-1,-1,-1,-1, 1, 2}, {-1, 0,-1, 2, 3, 4}, { 0,-1, 1,-1, 4, 5}, {-1, 1,-1, 4, 6, 7}, 9 { 1, 2, 3, 5, 7, 8}, { 2,-1, 4,-1, 8, 9}, {-1, 3,-1, 7,10,11}, { 3, 4, 6, 8,11,12}, 10 { 4, 5, 7, 9,12,13}, { 5,-1, 8,-1,13,14}, {-1, 6,-1,11,-1,-1}, { 6, 7,10,12,-1,-1}, 11 { 7, 8,11,13,-1,-1}, { 8, 9,12,14,-1,-1}, { 9,-1,13,-1,-1,-1} 12 }; 13 14 15 int s; 16 bool vis[1 << maxn]; 17 pair<int, int> path[1 << maxn]; 18 int pre[1 << maxn]; 19 20 struct Node { 21 int sit, time; 22 int pos; 23 Node(int sit = 0, int time = 0, int pos = 0) : 24 sit(sit), time(time), pos(pos) {} 25 }; 26 27 int bfs(int &p) { 28 int cnt = 0; 29 int ori = (1 << maxn) - 1; 30 ori ^= (1 << s); 31 queue<Node> que; 32 que.push(Node(ori, 0, 0)); 33 vis[ori] = true; 34 while (!que.empty()) { 35 Node first = que.front(); 36 que.pop(); 37 if (first.sit == (1 << s)) { 38 p = first.pos; 39 return first.time; 40 } 41 42 int ssit = first.sit; 43 for (int i = 0; i < maxn; i++) { 44 if (!(ssit&(1 << i))) continue; 45 46 for (int j = 0; j < maxm; j++) { 47 int Next = dir[i][j]; 48 if (Next == -1 || !(ssit&(1 << Next))) continue; 49 50 int tmp = ssit ^ (1 << i); 51 while (Next != -1) { 52 if (!(ssit&(1 << Next))) { 53 //printf("%d %d\n",i, Next); 54 tmp ^= (1 << Next); 55 if (!vis[tmp]) { 56 Node temp(tmp, first.time + 1, ++cnt); 57 pre[cnt] = first.pos; 58 path[cnt] = make_pair(i, Next); 59 que.push(temp); 60 vis[tmp] = true; 61 } 62 break; 63 } 64 tmp ^= (1 << Next); 65 Next = dir[Next][j]; 66 } 67 } 68 } 69 } 70 return -1; 71 } 72 73 void output(int pos) { 74 if (!pre[pos]) { 75 printf("%d %d", path[pos].first + 1, path[pos].second + 1); 76 return; 77 } 78 output(pre[pos]); 79 printf(" %d %d", path[pos].first + 1, path[pos].second + 1); 80 } 81 82 int main() 83 { 84 int iCase; 85 scanf("%d", &iCase); 86 while (iCase--) { 87 scanf("%d", &s); 88 s--; 89 memset(vis, false, sizeof(vis)); 90 memset(pre, -1, sizeof(pre)); 91 int pos; 92 int ans = bfs(pos); 93 if (ans == -1) { 94 printf("IMPOSSIBLE\n"); 95 } 96 else { 97 printf("%d\n", ans); 98 output(pos); 99 printf("\n"); 100 } 101 } 102 return 0; 103 }
