Problem UVA12569-Planning mobile robot on Tree (EASY Version)
Accept：138 Submit：686
Time Limit: 3000 mSec
Problem Description
Problem Description
Input
The first line contains the number of test cases T (T ≤ 340). Each test case begins with four integers n, m, s, t (4 ≤ n ≤ 15, 0 ≤ m ≤ n?2, 1 ≤ s,t ≤ n, s ?= t), the number of vertices, the number of obstacles and the label of the source and target. Vertices are numbered 1 to n. The next line contains m different integers not equal to s, the vertices containing obstacles. Each of the next n ? 1 lines contains two integers u and v (1 ≤ u < v ≤ n), that means there is an edge u?v in the tree.
Output
For each test case, print the minimum number of moves k in the first line. Each of the next k lines containstwointegers a and b,thatmeanstomovetherobot/obstaclefrom a to b. Ifthereisnosolution, print ‘-1’. If there are multiple solutions, any will do. Print a blank line after each test case.
Sample Input
3
4 1 1 3
2
1 2
2 3
2 4
6 2 1 4
2 3
1 2
2 3
3 4
2 5
2 6
8 3 1 5
2 3 4
1 2
2 3
3 4
4 5
1 6
1 7
2 8
Sample Ouput
Case 1: 3
2 4
1 2
2 3
Case 2: 6
2 6
3 2
2 5
1 2
2 3
3 4
Case 3: 16
1 7
2 1
1 6
7 1
1 2
2 8
3 2
2 1
1 7
4 3
3 2
2 1
8 2
2 3
3 4
4 5
题解：看到n的范围状压是比较正的思路，二进制储存，BFS搜索，vis数组稍微有一点改动，其中一维记录石头，再有一维记录机器人。水题。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 const int maxn = 16;
6 int n, m, s, t;
7 int ori;
8
9 struct Edge {
10 int to, next;
11 }edge[maxn << 1];
12
13 struct Node {
14 int sit, robot;
15 int time;
16 Node(int sit = 0, int robot = 0, int time = 0) :
17 sit(sit), robot(robot), time(time) {}
18 };
19
20 int tot, head[maxn];
21 pair<int,int> pre[40000][maxn];
22 bool vis[40000][maxn];
23
24 void init() {
25 tot = 1;
26 memset(head, -1, sizeof(head));
27 memset(pre, -1, sizeof(pre));
28 memset(vis, false, sizeof(vis));
29 }
30
31 void AddEdge(int u, int v) {
32 edge[tot].to = v;
33 edge[tot].next = head[u];
34 head[u] = tot++;
35 }
36
37 inline int get_pos(int x) {
38 return 1 << x;
39 }
40
41 int bfs(pair<int,int> &res) {
42 queue<Node> que;
43 que.push(Node(ori, s, 0));
44 vis[ori][s] = true;
45
46 while (!que.empty()) {
47 Node first = que.front();
48 que.pop();
49 if (first.robot == t) {
50 res.first = first.sit, res.second = first.robot;
51 return first.time;
52 }
53 int ssit = first.sit, rrob = first.robot;
54 //printf("%d %d\n", ssit, rrob);
55
56 for (int i = head[rrob]; i != -1; i = edge[i].next) {
57 int v = edge[i].to;
58 if (ssit&get_pos(v) || vis[ssit][v]) continue;
59 vis[ssit][v] = true;
60 que.push(Node(ssit, v, first.time + 1));
61 pre[ssit][v] = make_pair(ssit, rrob);
62 }
63
64 for (int i = 0; i < n; i++) {
65 if (ssit&(get_pos(i))) {
66 for (int j = head[i]; j != -1; j = edge[j].next) {
67 int v = edge[j].to;
68 if (v == rrob || (ssit & get_pos(v))) continue;
69 int tmp = ssit ^ get_pos(v);
70 tmp ^= get_pos(i);
71 if (vis[tmp][rrob]) continue;
72 vis[tmp][rrob] = true;
73 que.push(Node(tmp, rrob, first.time + 1));
74 pre[tmp][rrob] = make_pair(ssit, rrob);
75 }
76 }
77 }
78 }
79 return -1;
80 }
81
82 void output(pair<int,int> a) {
83 if (a.first == ori && a.second == s) return;
84 output(pre[a.first][a.second]);
85 int ppre = pre[a.first][a.second].first, now = a.first;
86
87 if (ppre^now) {
88 int b = -1, c = -1;
89 for (int i = 0; i < n; i++) {
90 if (((ppre & (1 << i)) == (1 << i)) && ((now & (1 << i)) == 0)) {
91 b = i;
92 }
93 else if (((ppre & (1 << i)) == 0) && ((now & (1 << i)) == (1 << i))) {
94 c = i;
95 }
96 }
97 printf("%d %d\n", b + 1, c + 1);
98 }
99 else {
100 printf("%d %d\n", pre[a.first][a.second].second + 1, a.second + 1);
101 }
102 }
103
104 int con = 1;
105
106 int main()
107 {
108 int iCase;
109 scanf("%d", &iCase);
110 while (iCase--) {
111 init();
112 scanf("%d%d%d%d", &n, &m, &s, &t);
113 s--, t--;
114 ori = 0;
115 int x;
116 for (int i = 1; i <= m; i++) {
117 scanf("%d", &x);
118 x--;
119 ori ^= get_pos(x);
120 }
121
122 int u, v;
123 for (int i = 1; i <= n - 1; i++) {
124 scanf("%d%d", &u, &v);
125 u--, v--;
126 AddEdge(u, v);
127 AddEdge(v, u);
128 }
129
130 pair<int, int> res;
131 int ans = bfs(res);
132 printf("Case %d: %d\n", con++, ans);
133 if (ans != -1) output(res);
134 printf("\n");
135 }
136 return 0;
137 }
原文地址：https://www.cnblogs.com/npugen/p/9600200.html