POJ 1363 Rails （值得一记的栈好题）

Rails

Time Limit: 1000MS Memory Limit: 10000K

Description

There is a famous railway station in PopPush City. Country there is incredibly hilly. The station was built in last century. Unfortunately, funds were extremely limited that time. It was possible to establish only a surface track. Moreover, it turned out that the station could be only a dead-end one (see picture) and due to lack of available space it could have only one track.

The local tradition is that every train arriving from the direction A continues in the direction B with coaches reorganized in some way. Assume that the train arriving from the direction A has N <= 1000 coaches numbered in increasing order 1, 2, …, N. The chief for train reorganizations must know whether it is possible to marshal coaches continuing in the direction B so that their order will be a1, a2, …, aN. Help him and write a program that decides whether it is possible to get the required order of coaches. You can assume that single coaches can be disconnected from the train before they enter the station and that they can move themselves until they are on the track in the direction B. You can also suppose that at any time there can be located as many coaches as necessary in the station. But once a coach has entered the station it cannot return to the track in the direction A and also once it has left the station in the direction B it cannot return back to the station.

Input

The input consists of blocks of lines. Each block except the last describes one train and possibly more requirements for its reorganization. In the first line of the block there is the integer N described above. In each of the next lines of the block there is a permutation of 1, 2, …, N. The last line of the block contains just 0.

The last block consists of just one line containing 0.

Output

The output contains the lines corresponding to the lines with permutations in the input. A line of the output contains Yes if it is possible to marshal the coaches in the order required on the corresponding line of the input. Otherwise it contains No. In addition, there is one empty line after the lines corresponding to one block of the input. There is no line in the output corresponding to the last ``null’’ block of the input.

Sample Input

5
1 2 3 4 5
5 4 1 2 3
0
6
6 5 4 3 2 1
0
0

Sample Output

Yes
No

Yes

思路：

一开始想的是研究它的序列的顺序规律，根据升序降序把它写出来。但是对于那种先进后出的数却没办法解决。于是就调整了思路，发现数字本身的顺序一定是遵从栈的规律的，之后就从大到小寻找数字本身下手，找不到在中途遍历的数字装入栈中，最后从栈中弹出。最后判断整个顺序是否合法即可。具体见代码、

AC代码：

#include
#include
#include
#include

using namespace std;
vector<int> p;
stack<int> q;
int a[1001];
bool vis[1001];

bool check(int n)
{
	int i;
	int maxs = n;
	for (i = n; i >= 1; i--)
	{
		if (a[i] != maxs) {
			q.push(a[i]); vis[a[i]] = true;
			continue;
		}
		p.push_back(a[i]);
		maxs--;
		while (vis[maxs])
		{
			p.push_back(q.top());
			q.pop();
			maxs--;
		}
	}
	for (i = 0; i < (int)p.size(); ++i)
	{
		if (p[i] != n - i)
			return false;
	}
	return true;
}

int main()
{
	int n, x;
	int i;
	while (cin >> n, n != 0)
	{
		while (cin >> x, x != 0)
		{
			a[1] = x;
			for (i = 2; i <= n; ++i)
				cin >> a[i];
			p.clear();
			while (!q.empty())q.pop();
			memset(vis, false, sizeof vis);
			if (check(n))
				cout << "Yes" << endl;
			else
				cout << "No" << endl;
		}
		cout << endl;
	}
	return 0;
}