POJ1753-优快云博客

本文介绍了一种利用广度优先搜索算法解决棋盘状态转换问题的方法。通过将棋盘状态映射到二进制串并构建状态转换树，可以有效地找到从初始状态到达全黑或全白状态所需的最少步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

棋盘的上的每一个状态对应一个二进制串，以初始状态为跟节点建树，然后广度优先遍历这颗树，层数为所求的值。

#include<iostream>
#include<queue>
#define ALL_WHITE 0
#define ALL_BLACK 65535
using namespace std;
int states[65536];
int state,temp;
int flip(int state_current, int pos) {
	state_current ^= (1 << pos);
	if (pos - 4 >= 0) {
		state_current ^= (1 << (pos - 4));
	}
	if (pos + 4 < 16) {
		state_current ^= (1 << (pos + 4));
	}
	if (pos % 4 != 0) {
		state_current ^= (1 << (pos - 1));
	}
	if (pos % 4 != 3) {
		state_current ^= (1 << (pos + 1));
	}
	return state_current;
}
void bfs() {
	int i;
	memset(states,-1,sizeof(states));
	queue<int> q;
	states[state] = 0;
	q.push(state);
	if (state == ALL_BLACK || state == ALL_WHITE){
		cout<<"0"<<endl;
		return ;
	}
	while(!q.empty()){
		temp=q.front();
		q.pop();
		for(i=0;i<16;i++) {
			int next_state = flip(temp, i);
			if (next_state == ALL_BLACK || next_state == ALL_WHITE){
				cout<<states[temp]+1<<endl;
				return ;
			}
			if (states[next_state] == -1){
				states[next_state] = states[temp] + 1;
				q.push(next_state);
			}
		}
	}
	cout<<"Impossible"<<endl;
}
int main(){
	char c;
	int i;
	state = 0;
	for(i=0; i<16; i++){
		cin>>c;
		if(c=='b'){
			state += 1<<i;
		}
		if (c == 'w'){
			state += 0<<i;
		}
	}
	bfs();
	return 0;
}