Ice and Fire

题目链接

题目描述：

Little09 and his friends are playing a game. There are \(n\) players, and the temperature value of the player \(i\) is \(i\).

The types of environment are expressed as \(0\) or \(1\). When two players fight in a specific environment, if its type is \(0\), the player with a lower temperature value in this environment always wins; if it is \(1\), the player with a higher temperature value in this environment always wins. The types of the \(n−1\) environments form a binary string s with a length of \(n−1\).

If there are \(x\) players participating in the game, there will be a total of \(x−1\) battles, and the types of the \(x−1\) environments will be the first \(x−1\) characters of \(s\). While there is more than one player left in the tournament, choose any two remaining players to fight. The player who loses will be eliminated from the tournament. The type of the environment of battle \(i\) is \(s_i\).

For each \(x\) from \(2\) to \(n\), answer the following question: if all players whose temperature value does not exceed \(x\) participate in the game, how many players have a chance to win?

输入描述：

Each test contains multiple test cases. The first line contains a single integer \(t\) \((1≤t≤10^3)\) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer \(n (2≤n≤2⋅10^5)\) — the number of players.

The second line of each test case contains a binary string \(s\) with a length \(n−1\).

It is guaranteed that the sum of \(n\) over all test cases does not exceed \(3⋅10^5\).

输出描述：

For each test case output \(n−1\) integers — for each \(x\) from \(2\) to \(n\), output the number of players that have a chance to win.

样例：

input:

2
4
001
4
101

output:

1 1 3
1 2 3

Note:

In the first test case, for \(x=2\) and \(x=3\), only the player whose temperature value is \(1\) can be the winner. For \(x=4\), the player whose temperature value is \(2,3,4\) can be the winner.

AC代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;

int T;
int n;
int a[2];

// 需要判断一下后缀有多少个连续的1或者0
// 假设有n个人，需要n - 1场战斗，s字符串最后有x个连续的1
// 1 ~ x中没有人可以活，因为最后x场战斗必须要有编号比他小人
// 而最好的情况就是第x个人，而第x个人也只有x - 1个编号比他小的人，x - 1比x小所以最后x肯定会死掉
// 而x + 1 ~ n中都有机会活下去，在x + 1 ~ n中选择两个人活下去，而x + 1 ~ n的其他人都在n - x - 2场战斗中死掉
// 因为第x - 1个字符必定是0，则所以在第n - x - 1次战斗中选择让两个人中的一个人和1战斗，1必获胜
// 那么最后剩下的那一个人成为了编号最大的一个人，那么在剩下的x场战斗中那个人必然获胜，而显然在x - 1 ~ n中每个人都可以成为最后剩下的那个人
// 后缀连续为0的情况同理
// 假设f是连续后缀的最大长度
// 所以综上所述i场战斗最后有机会获胜的人是i - fi + 1
// 而i - fi实际上也就是连续后缀前面的字符的个数
void solve()
{
	cin >> n;

	string s;
	cin >> s;

	a[0] = a[1] = 0;

	for(int i = 0; i < n - 1; i ++)
	{
		a[s[i] - '0'] = i + 1;

		if(s[i] == '0')
		{
			cout << a[1] + 1 << ' ';
		}
		else
		{
			cout << a[0] + 1 << ' ';
		}
	}

	cout << '\n';
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);

	cin >> T;

	while(T --)
		solve();

	return 0;
}