POJ_3087_(BFS)

POJ 3087

• 题目意思读懂:
• 给出字符串s1,s2,ans,创建一个空字符串s12,将s2第一个放在s12第一个,s1第一个放在s12第二个,然后依次类推,s12前一半是新的s1,后一半是新的s2,输出模拟后s12 == ans的步数(BFS),否则输出-1,
• map<string,bool>标记就行了

#include <iostream>
#include <string>
#include <map>
#include <queue>

using namespace std;


map<string,bool> vis;
int len;
string a,b,c;

struct node {
	string sa;
	string sb;
	int step;
	node();
	node(string s1,string s2,int step) {
		this->sa = s1;
		this->sb = s2;
		this->step = step;
	}
};

int bfs() {
	vis.clear();
	queue< node > que;
	que.push( node(a,b,0) );

	while ( !que.empty() ) {
		node cur = que.front();
		que.pop();

		string nes;
		int index1 = 0,index2 = 0;
		for (int i=0; i<2*len; i++) {
			if (i % 2 == 0) {
				nes += cur.sb[index2++];
			}
			else {
				nes += cur.sa[index1++];
			}
		}
		
		if (nes == c) { // 等于c 肯定没出现
			return cur.step+1;
		}
		else if (vis[nes] && nes != c) { // 出现了一次,并且不是结果
			return -1;
		}
		else {
			vis[nes] = 1;
			string s1 = nes.substr(0,len);
			
			string s2 = nes.substr(len,len);
			node nex(s1,s2,cur.step+1);
			que.push(nex);
		}
	}
	return -1;
}

int main() {
//	freopen("a.txt","r",stdin);
	int times;
	cin >> times;
	int cnt = 0;
	while (++cnt <= times) {
		cin >> len;
		cin >> a >> b >> c;
		vis[c] = 1;
		cout << cnt <<" "<< bfs() << endl;
	}
	return 0;
}