POJ 1753

BFS可以延伸的应用变化繁多，这道题配合状态压缩共同解决，不过时空复杂度还可以进一步优化。

#include <iostream>
#include <algorithm>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <stack>
#include <map>
#include <set>
using namespace std;

const int maxl= 16;
const int maxs= (1<<maxl)+5;
const int maxn= 6;
const int WHITE= 0;
const int BLACK= 1;
const int END0= 0;
const int ENDI= 0xffff;

int bod[maxl+5];
int vis[maxs], dis[maxs];
int flip[maxs];

int Encode(int *code, int m)
{
	int st= 0;
	for (int i= m-1; i>= 0; --i){
		st<<= 1;
		st|= code[i];
	}

	return st;
}
inline void FlipInit(const int m)
{
	int x;
	for (int i= 0; i< m; ++i){
		x= 1<<i;
		if (i-4>= 0 && i-4< m){
			x|= 1<<(i-4);
		}
		if (i+4>= 0 && i+4< m){
			x|= 1<<(i+4);
		}
		if (i & 3){
			x|= 1<<(i-1);
		}
		if ((i+1) & 3){
			x|= 1<<(i+1);
		}
		flip[i]= x;
	}
}
int BFS()
{
	int st= Encode(bod, maxl), n_st;
	if (END0== st || ENDI== st){
		return 0;
	}
	queue<int> Q;
	memset(vis, 0, sizeof(vis));
	FlipInit(maxl);
	int ans= 0;
	vis[st]= 1;
	dis[st]= 0;
	Q.push(st);

	while (!Q.empty()){
		st= Q.front();
		Q.pop();
		ans= dis[st]+1;
		for (int i= 0; i< maxl; ++i){
			n_st= st ^ flip[i];
			if (END0== n_st || ENDI== n_st){
				return ans;
			}
			if (!vis[n_st]){
				vis[n_st]= 1;
				dis[n_st]= ans;
				Q.push(n_st);
			}
		}
	}

	return -1;
}

int main()
{
	char in_s[maxn];
	int k= 0, ans;
	for (int i= 1; i<= 4; ++i){
		scanf("%s", in_s+1);
		for (int j= 1; j<= 4; ++j){
			bod[k++]= 'b'== in_s[j] ? BLACK : WHITE;
		}
	}

	ans= BFS();

	if (-1== ans){
		printf("Impossible");
	}
	else{
		printf("%d", ans);
	}
	return 0;
}