HDU 1540 Tunnel Warfare （线段树，区间合并）

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#线段树 #区间合并

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本文介绍了一种基于线段树的数据结构解决隧道战中村庄连接状态查询的问题。通过D操作摧毁村庄，R操作恢复村庄，Q操作查询指定村庄的最大连续区间。利用线段树维护每个村庄的连接状态，实现快速更新和查询。

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Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9270 Accepted Submission(s): 3609

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

题意：D表示摧毁某个村庄，R表示修复最后一个被摧毁的村庄，Q表示询问包含x在内的最大连续区间

思路：利用线段树求最大连续区间，我最近刚学线段树，这种题目对我来说还是有难度的，参考了别人的博客，才理解

上代码

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x3f3f3f3f
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define maxn 50010
int len, R, L;
struct node
{
	int llen, rlen;//左右最大连续区间
}tree[maxn << 2];
void build(int l, int r, int rt)
{
	tree[rt].llen = tree[rt].rlen = r - l + 1;//一开始左右最大连续区间都是满的
	if (l == r)
		return;
	int m = (l + r) / 2;
	build(ls);
	build(rs);
}
void updata(int l, int r, int rt, int id, int a)
{
	if (l == r)
	{
		tree[rt].llen = tree[rt].rlen = a;//摧毁或者回复
		return;
	}
	int m = (l + r) / 2;
	if (id <= m)
		updata(ls, id, a);
	else
		updata(rs, id, a);
	//以下是代码的核心
	//父区间的最大左右区间
	tree[rt].llen = tree[rt << 1].llen;
	tree[rt].rlen = tree[rt << 1 | 1].rlen;
	if (tree[rt].llen == m - l + 1)//如果左子树区间满了，父左区间要加上右子树的左区间
		tree[rt].llen += tree[rt << 1 | 1].llen;
	if (tree[rt].rlen == r - m)//如果右子树右区间满了，父右区间要加上左子树的右区间
		tree[rt].rlen += tree[rt << 1].rlen;
}
//查询的话，就从那个数开始，看能不能和自己左（右）区间的合并
void query(int l, int r, int rt, int id)
{
	if (l == r)
	{
		if (tree[rt].llen)
			R = L = 1;
		else
			R = L = 0;
		len = tree[rt].llen;
		return;
	}
	int m = (l + r) / 2;
	//左区间满向右，右区间满向左
	if (id <= m)
	{
		query(ls, id);
		if (L)
			len += tree[rt << 1 | 1].llen;//能和右区间的合并
		if (tree[rt << 1 | 1].llen<r - m)//父区间不满，不能再和右区间合并了
			L = 0;
	}
	else
	{
		query(rs, id);
		if (R)
			len += tree[rt << 1].rlen;//能和左区间合并
		if (tree[rt << 1].rlen<m - l + 1)//父区间不满，不能再和左区间合并了
			R = 0;
	}
}
int main()
{
	//freopen("Text.txt", "r", stdin);
	int n, m, s[maxn], sn, id;
	char ch[2];
	while (scanf("%d%d", &n, &m) != EOF)
	{
		build(1, n, 1);
		sn = 0;
		while (m--)
		{
			scanf("%s", ch);
			if (ch[0] == 'D')
			{
				scanf("%d", &id);
				s[++sn] = id;
				updata(1, n, 1, id, 0);
			}
			else if (ch[0] == 'R')
			{
				if (sn)
					updata(1, n, 1, s[sn], 1), sn--;//修复最后一个被破坏的村庄
			}
			else
			{
				scanf("%d", &id);
				query(1, n, 1, id);
				printf("%d\n", len);
			}
		}
	}
}