UVa 1533 - Moving Pegs (隐式图搜索 + 状态压缩)

题意

15个格子组成的三角形，一开始给出一个空位置，问能不能最后剩下一个棋子并且正好在这个位置上。

思路

一开始想用回溯，不过这跳来跳去的实在没办法表示。后来看到题目说固定的15个位置，想到每个棋子能走的地方都可以列出来。那就干脆都列出来一一判断呗。

这题是一个隐式图遍历问题。

15个格子，可以状态压缩一下。

先把原始的状态图作为基准，然后就开始跳吧，如果能达到新的状态，入队，直到找到结果或者队伍为空

我知道位运算的优先级很低，没想到比等号还低，漏了一个括号从早上调到现在TAT

代码

  
  
  
  

   
   
   
   
#include <cstdio>
   
   
   
   
#include <stack>
   
   
   
   
#include <set>
   
   
   
   
#include <iostream>
   
   
   
   
#include <string>
   
   
   
   
#include <vector>
   
   
   
   
#include <queue>
   
   
   
   
#include <functional>
   
   
   
   
#include <cstring>
   
   
   
   
#include <algorithm>
   
   
   
   
#include <cctype>
   
   
   
   
#include <string>
   
   
   
   
#include <map>
   
   
   
   
#include <cmath>
   
   
   
   
#define LL long long
   
   
   
   
#define SZ(x) (int)x.size()
   
   
   
   
#define Lowbit(x) ((x) & (-x))
   
   
   
   
#define MP(a, b) make_pair(a, b)
   
   
   
   
#define MS(arr, num) memset(arr, num, sizeof(arr))
   
   
   
   
#define PB push_back
   
   
   
   
#define F first
   
   
   
   
#define S second
   
   
   
   
#define ROP freopen("input.txt", "r", stdin);
   
   
   
   
#define MID(a, b) (a + ((b - a) >> 1))
   
   
   
   
#define LC rt << 1, l, mid
   
   
   
   
#define RC rt << 1|1, mid + 1, r
   
   
   
   
#define LRT rt << 1
   
   
   
   
#define RRT rt << 1|1
   
   
   
   
#define BitCount(x) __builtin_popcount(x)
   
   
   
   
const double PI = acos(-1.0);
   
   
   
   
const int INF = 0x3f3f3f3f;
   
   
   
   
using namespace std;
   
   
   
   
const int MAXN = 100 + 10;
   
   
   
   
const int MOD = 1e9 + 7;
   
   
   
   
const int dir[][6] = { {-1,-1,-1,-1,-1,-1}, {-1,-1,-1,-1,2,3}, {-1,1,-1,3,4,5}, {1,-1,2,-1,5,6},
   
   
   
   
{-1,2,-1,5,7,8}, {2,3,4,6,8,9}, {3,-1,5,-1,9,10},
   
   
   
   
{-1,4,-1,8,11,12}, {4,5,7,9,12,13}, {5,6,8,10,13,14}
   
   
   
   
, {6,-1,9,-1,14,15}, {-1,7,-1,12,-1,-1}, {7,8,11,13,-1,-1}
   
   
   
   
, {8,9,12,14,-1,-1}, {9,10,13,15,-1,-1}, {10,-1,14,-1,-1,-1} };
   
   
   
   

   
   
   
   
typedef pair<int, int> pii;
   
   
   
   
typedef vector<int>::iterator viti;
   
   
   
   
typedef vector<pii>::iterator vitii;
   
   
   
   

   
   
   
   
struct STATE
   
   
   
   
{
   
   
   
   
 int state, rem, path[MAXN][2];
   
   
   
   
 int pos;
   
   
   
   
};
   
   
   
   

   
   
   
   
set<int> mp;
   
   
   
   
int emp;
   
   
   
   
queue<STATE> Q;
   
   
   
   
STATE ans;
   
   
   
   

   
   
   
   
void Solve()
   
   
   
   
{
   
   
   
   
 int iniState = 0;
   
   
   
   
 STATE st;
   
   
   
   
 for (int i = 1; i <= 15; i++)
   
   
   
   
 if (i != emp) iniState |= (1 << i);
   
   
   
   
 st.state = iniState; st.rem = 14; st.pos = 0;
   
   
   
   
 Q.push(st);
   
   
   
   
 while (!Q.empty())
   
   
   
   
 {
   
   
   
   
 STATE cur = Q.front(); Q.pop();
   
   
   
   
 const int tmpState = cur.state;
   
   
   
   
 for (int i = 1; i <= 15; i++)
   
   
   
   
 {
   
   
   
   
 if ((tmpState & (1 << i)) == 0) continue; //如果这个位置是黑色
   
   
   
   
 for (int j = 0; j < 6; j++)
   
   
   
   
 {
   
   
   
   
 STATE newState = cur;
   
   
   
   
 newState.state ^= (1 << i);
   
   
   
   
 if (dir[i][j] != -1 && (tmpState & (1 << dir[i][j])))
   
   
   
   
 {
   
   
   
   
 int t = dir[i][j]; //接下来找这个方向白色的
   
   
   
   
 newState.state -= (1 << t);
   
   
   
   
 while (dir[t][j] != -1 && (tmpState & (1 << dir[t][j])))
   
   
   
   
 {
   
   
   
   
 newState.state -= (1 << dir[t][j]); newState.rem--;
   
   
   
   
 t = dir[t][j];
   
   
   
   
 }
   
   
   
   
 if (dir[t][j] == -1) continue;
   
   
   
   
 newState.state |= (1 << dir[t][j]); newState.rem--;
   
   
   
   
 newState.path[newState.pos][0] = i; newState.path[newState.pos++][1] = dir[t][j];
   
   
   
   
 if (mp.count(newState.state)) continue;
   
   
   
   
 mp.insert(newState.state);
   
   
   
   
 Q.push(newState);
   
   
   
   
 //if (newState.rem == 1) printf("%d\n", dir[t][j]);
   
   
   
   
 if (newState.rem == 1 && newState.state & (1 << emp)) //如果正好跳到空白点而且只剩一个
   
   
   
   
 {
   
   
   
   
 ans = newState;
   
   
   
   
 return;
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
}
   
   
   
   

   
   
   
   
void Init()
   
   
   
   
{
   
   
   
   
 while (!Q.empty()) Q.pop();
   
   
   
   
 mp.clear();
   
   
   
   
 ans.pos = INF;
   
   
   
   
}
   
   
   
   

   
   
   
   
int main()
   
   
   
   
{
   
   
   
   
 //ROP;
   
   
   
   
 int T, i, j;
   
   
   
   
 scanf("%d", &T);
   
   
   
   
 while (T--)
   
   
   
   
 {
   
   
   
   
 Init();
   
   
   
   
 scanf("%d", &emp);
   
   
   
   
 if (emp < 1 || emp > 15) puts("IMPOSSIBLE");
   
   
   
   
 Solve();
   
   
   
   
 if (ans.pos == INF) puts("IMPOSSIBLE");
   
   
   
   
 else
   
   
   
   
 {
   
   
   
   
 printf("%d\n", ans.pos);
   
   
   
   
 for (i = 0; i < ans.pos; i++)
   
   
   
   
 {
   
   
   
   
 if (i) printf(" %d %d", ans.path[i][0], ans.path[i][1]);
   
   
   
   
 else printf("%d %d", ans.path[i][0], ans.path[i][1]);
   
   
   
   
 }
   
   
   
   
 }
   
   
   
   
 puts("");
   
   
   
   
 }
   
   
   
   
 return 0;
   
   
   
   
}