hdu1540 Tunnel Warfare 线段树区间合并

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6565    Accepted Submission(s): 2527

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
   
   
   
   

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <stack>
#include <algorithm>

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1

using namespace std;

const int maxn = 50000 + 10;

int msum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];

void PushUp(int rt, int m) {
	lsum[rt] = lsum[rt << 1];
	rsum[rt] = rsum[rt << 1 | 1];

	//若左区间左连续满了，就还要考虑右区间的
	if (lsum[rt << 1] == m - (m >> 1)) {
		lsum[rt] += lsum[rt << 1 | 1];
	}

	//同上
	if (rsum[rt << 1 | 1] == (m >> 1)) {
		rsum[rt] += rsum[rt << 1];
	}

	//当前区间的最大连续长度在左右区间和跨区间的区间三个中找
	msum[rt] = max(rsum[rt << 1] + lsum[rt << 1 | 1], max(msum[rt << 1], msum[rt << 1 | 1]));
}

void build(int l, int r, int rt) {
	msum[rt] = lsum[rt] = rsum[rt] = r - l + 1;
	if (l == r) return;
	int m = (l + r) >> 1;
	build(lson);
	build(rson);
}

//c为1的时候是恢复操作，c为0的时候是毁灭
void update(int x, int c, int l, int r, int rt) {
	if (l == r) {
		msum[rt] = lsum[rt] = rsum[rt] = c;
		return;
	}
	int m = (l + r) >> 1;
	if (x <= m) {
		update(x, c, lson);
	}
	else {
		update(x, c, rson);
	}
	PushUp(rt, r - l + 1);
}

int query(int x, int l, int r, int rt) {
	//到根节点或者区间最大连续长度为0，或者区间满直接返回
	if (l == r || msum[rt] == 0 || msum[rt] == r - l + 1) {
		return msum[rt];
	}
	int m = (l + r) >> 1;
	if (x <= m) {
		//查询的点在左区间的右连续区间范围内，就还要考虑右区间
		if (x >= m - rsum[rt << 1] + 1) {
			return query(x, lson) + query(m + 1, rson); //注意是m + 1，即右区间左端点
		}
		else {
			//否则只考虑左区间就行了
			return query(x, lson);
		}
	}
	else {
		//同上
		if (x <= (m + 1) + lsum[rt << 1 | 1] - 1) {
			return query(x, rson) + query(m, lson);  //注意是m，即左区间右端点
		}
		else {
			return query(x, rson);
		}
	}
}

int main()
{
	int n, m, x;
	char op[2];
	while (~scanf("%d%d", &n, &m)) {
		build(1, n, 1);
		stack<int> s;
		while (m--) {
			scanf("%s", op);
			if (op[0] == 'D') {
				scanf("%d", &x);
				s.push(x);
				update(x, 0, 1, n, 1);
			}
			else if (op[0] == 'Q') {
				scanf("%d", &x);
				printf("%d\n", query(x, 1, n, 1));
			}
			else {
				if (!s.empty()) {
					x = s.top(); s.pop();
					update(x, 1, 1, n, 1);
				}
			}
		}
	}
	return 0;
}