UVa12569树上的机器人的规划

今天更新一篇,直奔主题。

题意:

有一棵n(4<=n<=15)个结点的树,其中一个结点有一个机器人,还有一些结点有石头。每步可以把一个机器人或石头移到一个相邻的节点。任何情况下一个结点里不能有两个东西(石头或机器人)。输入每个石头的位置和机器人的起点和终点,求最小步数的方案。如果有多解,可以输出任意解。

建议看原题。

思路:

由于(4<=n<=15)以及求最小步数的方案,所以我们可以采用bfs的方法去做,对于路径寻找问题,自然需要对不同的状态进行记录,根据n很小我们可以采用状态压缩(二进制)进行表示。
我们采用STL的set或者hash进行查找重复,但是set太慢了,也许之前做的题对时间要求不高,于是使用了set发现超时,试了试hash发现快了很多2000ms左右

下面是代码:

set超时代码

// UVa 12569
// bfs+状态压缩+剪枝 
#include <cstdio>
#include <cstring>
#include <vector>
#include <set> 
using namespace std; 
typedef long long LL; 

struct Node {
  LL tree;
  int robpos, f, t, root, dep;
  Node(LL tree=0, int robpos=0, int f=0, int t=0, int root=0, int dep=0) : 
    tree(tree), robpos(robpos), f(f), t(t), root(root), dep(dep) {}
};

int n, m, s, t, G[20][20]; 
int kase; 

int l, r;
vector<Node> v;
set<pair<LL, int> > Set;

void Move(Node& u1, int i, int j, int root) {
  if (u1.robpos == i) u1.robpos = j; 
  u1.tree ^= (1<<i); 
  u1.tree |= (1<<j);     
  u1.root = root;
  u1.f = i;
  u1.t = j; 
}

bool try_to_insert(Node& u, int i, int j) {
  LL tree = u.tree;
  int robpos = u.robpos; 
  if (robpos == i) robpos = j; 
  tree ^= (1<<i);
  tree |= (1<<j); 
  if (!Set.count(make_pair(tree, robpos))) {
    Set.insert(make_pair(tree, robpos));
    return true; 
  }
  return false; 
}

void Print(int cur) {
  if (cur != 0) { 
    Print(v[cur].root); 
    printf("%d %d\n", v[cur].f, v[cur].t);
  }
}

void bfs(Node& start) {    
  v.clear();
  Set.clear(); 
  Set.insert(make_pair(start.tree, start.robpos));
  v.push_back(start);
  l = 0; r = 1;
  
  while (l < r) {
    Node u = v[l]; l++;
    if (u.robpos == t) { printf("Case %d: %d\n", ++kase, u.dep); Print(l-1); puts(""); return; } 
    for (int i = 1; i <= n; ++i) 
      if (u.tree&(1<<i)) 
        for (int j = 1; j <= n; ++j) 
          if (G[i][j] && (u.tree&(1<<j)) == 0 && try_to_insert(u, i, j)) {
            Node u1; 
            memcpy(&u1, &u, sizeof(u1));
            u1.dep = u.dep + 1;
            Move(u1, i, j, l-1);
            v.push_back(u1); 
            r++; 
          }
  }
  printf("Case %d: -1\n\n", ++kase); 
}

int main() { 
  int T; 
  scanf("%d", &T); 
  while (T--) {
    scanf("%d%d%d%d", &n, &m, &s, &t);
    Node start(0, s, 0, 0, -1, 0);
    start.tree |= (1<<s);  
    for (int i = 0; i < m; ++i) {
      int stone; 
      scanf("%d", &stone);
      start.tree |= (1<<stone); 
    }
    memset(G, 0, sizeof(G));
    for (int i = 0; i < n-1; ++i) {
      int u, v; 
      scanf("%d%d", &u, &v); 
      G[u][v] = G[v][u] = 1;    
    } 
    bfs(start);
  }
  return 0;
}

hash快速代码

// UVa 12569
// bfs+状态压缩+剪枝(不用set用hash) 
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std; 

struct Node {
  int tree, robpos, f, t, root, dep;
  Node(int tree=0, int robpos=0, int f=0, int t=0, int root=0, int dep=0) : 
    tree(tree), robpos(robpos), f(f), t(t), root(root), dep(dep) {}
};

int n, m, s, t, G[20][20]; 
int kase; 

int l, r;
vector<Node> v;

void Move(Node& u1, int i, int j, int root) {
  if (u1.robpos == i) u1.robpos = j; 
  u1.tree ^= (1<<i); 
  u1.tree |= (1<<j);     
  u1.root = root;
  u1.f = i;
  u1.t = j; 
}

const int hashsize = 131072; // 2 ^ 16
int Head[hashsize+10], Next[100003]; // just try 

void init_lookup_table() { memset(Head, -1, sizeof(Head)); }

int try_to_insert(Node& u, int i, int j, int pos) {
  int tree = u.tree, robpos = u.robpos; 
  if (robpos == i) robpos = j; 
  tree ^= (1<<i); 
  tree |= (1<<j); 
  int u1 = Head[tree];
  while (u1 != -1) {
    if (v[u1].robpos == robpos) return 0;    
    u1 = Next[u1];
  }
  Next[pos] = Head[tree];
  Head[tree] = pos; 
  return 1;
}

void Print(int cur) {
  if (cur != 0) { 
    Print(v[cur].root); 
    printf("%d %d\n", v[cur].f, v[cur].t);
  }
}

void bfs(Node& start) {    
  init_lookup_table(); 
  v.clear();
  v.push_back(start);
  l = 0; r = 1;
  
  while (l < r) {
    Node u = v[l]; l++;
    if (u.robpos == t) { printf("Case %d: %d\n", ++kase, u.dep); Print(l-1); puts(""); return; } 
    for (int i = 1; i <= n; ++i) 
      if (u.tree&(1<<i)) 
        for (int j = 1; j <= n; ++j) 
          if (G[i][j] && (u.tree&(1<<j)) == 0 && try_to_insert(u, i, j, r)) {
            Node u1; 
            memcpy(&u1, &u, sizeof(u1));
            u1.dep = u.dep + 1;
            Move(u1, i, j, l-1);
            v.push_back(u1); 
            r++; 
          }
  }
  printf("Case %d: -1\n\n", ++kase); 
}

int main() { 
  int T; 
  scanf("%d", &T); 
  while (T--) {
    scanf("%d%d%d%d", &n, &m, &s, &t);
    Node start(0, s, 0, 0, -1, 0);
    start.tree |= (1<<s);  
    for (int i = 0; i < m; ++i) {
      int stone; 
      scanf("%d", &stone);
      start.tree |= (1<<stone); 
    }
    memset(G, 0, sizeof(G));
    for (int i = 0; i < n-1; ++i) {
      int u, v; 
      scanf("%d%d", &u, &v); 
      G[u][v] = G[v][u] = 1;    
    } 
    bfs(start);
  }
  return 0;
}

就这样,不懂的地方问我,我会及时回复的。

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