UVA12569-Planning mobile robot on Tree (EASY Version)（BFS+状态压缩）

Problem UVA12569-Planning mobile robot on Tree (EASY Version)

Accept：138  Submit：686

Time Limit: 3000 mSec

 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 }

 

posted on 2018-09-06 18:32  随缘&不屈  阅读(217)  评论(0)  编辑  收藏  举报

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