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Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village
was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration
of connection must be done immediately!

Input

The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.

Output

Output the answer to each of the Army commanders’ request in order on a separate line.

Sample Input

7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4

Sample Output

1
0
2
4

Source

POJ Monthly

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这道题也是典型的线段树区间合并的题, 主要是要记录一段区间左边连续的长度, 右边连续长度, 单点更新,查询时只要判断要查的点是不是落在中间连续处,否则就继续递归查询

/*************************************************************************
    > File Name: hdu1540.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年01月11日 星期日 12时24分34秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 500010;
stack <int> re;

struct node
{
	int l, r;
	int l_sum;
	int r_sum;
}tree[N << 2];

void pushup (int p)
{
	tree[p].l_sum = tree[p << 1].l_sum;
	tree[p].r_sum = tree[p << 1 | 1].r_sum;
	if (tree[p << 1].l_sum == tree[p << 1].r - tree[p << 1].l + 1)
	{
		tree[p].l_sum += tree[p << 1 | 1].l_sum;
	}
	if (tree[p << 1 | 1].r_sum == tree[p << 1 | 1].r - tree[p << 1 | 1].l + 1)
	{
		tree[p].r_sum += tree[p << 1].r_sum;
	}
}

void build (int p, int l, int r)
{
	tree[p].l = l;
	tree[p].r = r;
	tree[p].l_sum = tree[p].r_sum = (r - l + 1);
	if (l == r)
	{
		return;
	}
	int mid = (l + r) >> 1;
	build (p << 1, l, mid);
	build (p << 1 | 1, mid + 1, r);
}

void update (int p, int pos, int sta)
{
	if (tree[p].l == tree[p].r)
	{
		tree[p].l_sum = tree[p].r_sum = sta;
		return;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (pos <= mid)
	{
		update (p << 1, pos, sta);
	}
	else
	{
		update (p << 1 | 1, pos, sta);
	}
	pushup (p);
}

int query (int p, int pos)
{
	if (tree[p].l == tree[p].r)
	{
		if (tree[p].l_sum == 1)
		{
			return 1;
		}
		return 0;
	}
	int mid = (tree[p].l + tree[p].r) >> 1;
	if (pos <= mid)
	{
		if (pos >= mid - tree[p << 1].r_sum + 1)
		{
			return tree[p << 1].r_sum + tree[p << 1 | 1].l_sum;
		}
		return query (p << 1, pos);
	}
	else
	{
		if (pos <= mid + tree[p << 1 | 1].l_sum)
		{
			return tree[p << 1 | 1].l_sum + tree[p << 1].r_sum;
		}
		return query (p << 1 | 1, pos);
	}
}

int main()
{
	int n, m;
	int x;
	char op[10];
	while (~scanf("%d%d", &n, &m))
	{
		build (1, 1, n);
		while (!re.empty())
		{
			re.pop();
		}
		while (m--)
		{
			scanf("%s", op);
			if(op[0] == 'R')
			{
				if (re.empty())
				{
					continue;
				}
				update (1, re.top(), 1);
				re.pop();
			}
			else
			{
				scanf("%d", &x);
				if (op[0] == 'D')
				{
					update (1, x, 0);
					re.push(x);
				}
				else
				{
					printf("%d\n", query (1, x));
				}
			}
		}
	}
	return 0;
}