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;

}