双向广搜-HDU1401 Solitaire

Posted 2023-01-12 吾仄lo咚锵

文章目录

• 双向广搜
• 例题
• • 题意
  • 分析
  • 代码
• 小结

双向广搜

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什么是双向广搜？
如果把bfs想象成在平静的池塘丢一颗石头，激起的波浪一层层扩散到整个空间直到到达目标，就得到起点到终点的最优路径。那么双向广搜就是在起点和终点同时丢石头，两个波浪将在中间某个位置相遇，即得到最优路径。

例题

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传送门: HDU-1401

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.
There are four identical pieces on the board. In one move it is allowed to:
move a piece to an empty neighboring field (up, down, left or right),
jump over one neighboring piece to an empty field (up, down, left or right).
>There are 4 moves allowed for each piece in the configuration shown above. As an example let’s consider a piece placed in the row 4, column 4. It can be moved one row up, two rows down, one column left or two columns right.
Write a program that:
reads two chessboard configurations from the standard input,
verifies whether the second one is reachable from the first one in at most 8 moves,
writes the result to the standard output.

input:

Each of two input lines contains 8 integers a1, a2, …, a8 separated by single spaces and describes one configuration of pieces on the chessboard. Integers a2j-1 and a2j (1 <= j <= 4) describe the position of one piece - the row number and the column number respectively. Process to the end of file.

output:

The output should contain one word for each test case - YES if a configuration described in the second input line is reachable from the configuration described in the first input line in at most 8 moves, or one word NO otherwise.

Sample Input:

4 4 4 5 5 4 6 5
2 4 3 3 3 6 4 6

Sample Output:

YES

题意

在8*8棋盘有四颗棋子，输入开始状态和目标状态的棋子坐标，每次可以移动一个棋子也可以跳过另一个棋子，问能否在8步内走到目标状态。

分析

用双向广搜求解，8个坐标值位压缩成为10进制作为hash值（或者8维数组？）并用unordered_set判重，当hash值出现在另一个分支即相遇。注意剪枝：每个棋子最多只能走4步，因为如果多于4步，那么他们步数之和就会大于8。

代码

#include <bits/stdc++.h>
#include <unordered_set>
using namespace std;
typedef long long ll;
int stepx[4] =  -1,1,0,0 ;
int stepy[4] =  0,0,-1,1 ;
struct node 
	int p[4], vis;
	bool check(int v)  //查重
		for (int i : p)
			if (i == v)
				return 1;
		return 0;
	
	int hash() 
		int a[4];
		memcpy(a, p, sizeof(a));
		sort(a, a + 4);
		int res = 0;
		for (int i : a)
			res = res * 100 + i;
		return res;
	
;

unordered_set<int> st[2]; //存hashc
queue<node> q[2];

void clear() 
	for (int i = 0; i < 2; i++) 
		queue<node>c;
		swap(q[i], c);
		st[i].clear();
	

bool ans(int i, int hash) //队列下标和待查hash
	if (st[i].find(hash) != st[i].end()) //相遇
		return true;
	return false;

bool dbfs() 
	while (!q[0].empty() || !q[1].empty()) 
		for (int i = 0; i < 2; ++i)  //2个队列
			if (!q[i].empty()) 
				node cur = q[i].front(); q[i].pop();
				cur.vis++;
				for (int j = 0; j < 4; ++j)  //4个点
					int x = cur.p[j] / 10, y = cur.p[j] % 10;
					for (int k = 0; k < 4; ++k)  //4个方向
						int nx = x + stepx[k], ny = y + stepy[k];
						if (nx >= 1 && nx <= 8 && ny >= 1 && ny <= 8 && cur.check(nx * 10 + ny))
							nx += stepx[k], ny += stepy[k];
						if (nx >= 1 && nx <= 8 && ny >= 1 && ny <= 8 && cur.vis <= 4 && //小于四步
							!cur.check(nx * 10 + ny)) 
							cur.p[j] = nx * 10 + ny;
							if (st[i].find(cur.hash()) == st[i].end()) 
								if (ans(!i, cur.hash()))return true;
								q[i].push(cur), st[i].insert(cur.hash()); //入队
							
						
					
					cur.p[j] = x * 10 + y; //回溯
				
			
		
	
	return false;

int main() 
	while (true) 
		clear();
		node t;
		int a;
		t.vis = 0;
		for (int k = 0; k < 2; ++k) 
			for (int i = 0; i < 4; ++i) 
				t.p[i] = 0;
				for (int j = 0; j < 2; ++j) 
					if (scanf("%d", &a) == EOF)exit(0);
					t.p[i] = t.p[i] * 10 + a;
				
			
			q[k].push(t);
			st[k].insert(t.hash());
		
		if (ans(0, q[1].front().hash()) || dbfs())
			printf("YES\\n");
		else printf("NO\\n");
	
	return 0;

小结

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1.如果爆栈：考虑在入队前做答案检查，出队才检查会多一轮消耗。
2.如何保存中间状态？位压缩.位运算.unordered_set…数组？
3.String判重TLE让你怀疑人生，string复制会超时,还是老老实实哈希或者康托展开什么的。
4.神奇剪枝起死回生，一定要注意考虑剪枝，因题制宜。

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