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HDU-Tunnel Warfare

2022-04-23 05:51:00 【Round moon】

Tunnel Warfare

题目描述

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!

输入描述

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.

输出描述

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

题目大意

一段列表，选中的点被摧毁D。查询Q选中的点所在连续区间的长度。R恢复最后一个被摧毁的点。
不需要考虑重复摧毁
样例解释
经过三轮摧毁，3，6，5被摧毁。4号位置连续长度为1

在这里插入图片描述
5号位置被摧毁，长度为0
恢复最后一个被摧毁的5点。4号位置的连续长度为2
在这里插入图片描述
恢复最后一个被摧毁的点6。4号位置的连续长度为4
在这里插入图片描述

思路解析（思路一，思路二）

本题用两种思路去解题，第一种思路，维护区间最大值和最小值的方法去解决。第二种思路，线段树的区间合并方法去解决。

思路一

我们把被摧毁的点设置为自己的id值，如果没被摧毁就设置为很大的一个数字。这样我们就能知道从自己包括自己的点到自己最右侧的最小值就是自己的右边界。
在这里插入图片描述
这里发现最小值是5
如果把没被摧毁的设置为一个很小的值，那么从自己包括自己到最左侧的最大值就是自己的左边界。
在这里插入图片描述
这里发现最大值是3
那么中间的长度就是5-3-1
这里需要特殊的判定一下，如果这两个值的值重复了，那么就是0
我们这里的很大的值设为长度+1，很小的值就是0，这样我们就能把整个区间给包括起来了。

AC代码

//#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<string.h>
#include <iomanip>
#include<stdio.h>
#include<vector>
#include<string>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll ll_inf = 9223372036854775807;
const int int_inf = 2147483647;
const short short_inf = 32767;
const ll less_inf = 0x3f3f3f3f;
const char char_inf = 127;
#pragma GCC optimize(2)
#define accelerate cin.tie(NULL);cout.tie(NULL);ios::sync_with_stdio(false);
#define PI 3.141592653589793
#define EPS 1.0e-8
ll gcd(ll a, ll b) {
    
	return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
    
	return a * b / gcd(a, b);
}
inline ll read() {
    
	ll c = getchar(), Nig = 1, x = 0;
	while (!isdigit(c) && c != '-')c = getchar();
	if (c == '-')Nig = -1, c = getchar();
	while (isdigit(c))x = ((x << 1) + (x << 3)) + (c ^ '0'), c = getchar();
	return Nig * x;
}
inline void out(ll a) {
    
	if (a < 0)putchar('-'), a = -a;
	if (a > 9)out(a / 10);
	putchar(a % 10 + '0');
}
ll qpow(ll x, ll n, ll mod) {
    
	ll res = 1;
	while (n > 0) {
    
		if (n & 1)res = (res * x) % mod;
		x = (x * x) % mod;
		n >>= 1;
	}
	return res;
}
#define read read()
const int N = 50000;
struct node
{
    
	int maxn, minn;
}tree[N * 4 + 7];
int n, m;
void push_up(int rt)
{
    
	tree[rt].maxn = max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn);
	tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
void creat(int l, int r, int rt)
{
    
	if (l == r)
	{
    
		tree[rt].maxn =0;
		tree[rt].minn = n+1;
		return;
	}
	int mid = l + r >> 1;
	creat(l, mid, rt << 1);
	creat(mid + 1, r, rt << 1 | 1);
	push_up(rt);
}
void updata(int pos, int maxn, int minn, int l, int r, int rt)
{
    
	if (l == r)
	{
    
		tree[rt].maxn = maxn;
		tree[rt].minn = minn;
		return;
	}
	int mid = l + r >> 1;
	if (pos <= mid)updata(pos, maxn, minn, l, mid, rt << 1);
	else updata(pos, maxn, minn, mid + 1, r, rt << 1 | 1);
	push_up(rt);
}
node query(int L, int R, int l, int r, int rt)
{
    
	node temp;
	int maxn = -int_inf;
	int minn = int_inf;
	if (L <= l && r <= R)
	{
    
		temp.maxn = tree[rt].maxn;
		temp.minn = tree[rt].minn;
		return temp;
	}
	int mid = l + r >> 1;
	if (L <= mid) {
    
		temp = query(L, R, l, mid, rt << 1);
		maxn = max(maxn, temp.maxn);
		minn = min(minn, temp.minn);
	}
	if (R > mid) {
    
		temp = query(L, R, mid + 1, r, rt << 1 | 1);
		maxn = max(maxn, temp.maxn);
		minn = min(minn, temp.minn);
	}
	return node{
     maxn,minn };
}
stack<int>s;
int main()
{
    
	while (scanf("%d%d", &n, &m) != EOF)
	{
    
		while (!s.empty())s.pop();
		creat(1, n, 1);

		while (m--)
		{
    
			char mark[3];
			scanf("%s", mark);
			if (mark[0] == 'D')
			{
    
				int id;
				scanf("%d", &id);
				updata(id, id, id, 1, n, 1);
				s.push(id);
			}
			else if (mark[0] == 'Q')
			{
    
				int id;
				scanf("%d", &id);
				node left = query(1, id, 1, n, 1);
				node right = query(id, n, 1, n, 1);
				if (right.minn == left.maxn)puts("0");
				else printf("%d\n", right.minn - left.maxn - 1);
			}
			else
			{
    
				int id = s.top();
				s.pop();
				updata(id, 0, n + 1, 1, n, 1);
			}
		}
	}
}

思路二

我们利用区间合并的方法，去单点修改每一个点。在查询的时候，我们判断当前要查寻的点是不是位于组合成的区间内，如果是，在递归一半区间的时候要注意把另一半的区间也要递归下去。详情见代码

AC代码

//#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<string.h>
#include <iomanip>
#include<stdio.h>
#include<vector>
#include<string>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll ll_inf = 9223372036854775807;
const int int_inf = 2147483647;
const short short_inf = 32767;
const ll less_inf = 0x3f3f3f3f;
const char char_inf = 127;
#pragma GCC optimize(2)
#define accelerate cin.tie(NULL);cout.tie(NULL);ios::sync_with_stdio(false);
#define PI 3.141592653589793
#define EPS 1.0e-8
ll gcd(ll a, ll b) {
    
	return b ? gcd(b, a % b) : a;
}
ll lcm(ll a, ll b) {
    
	return a * b / gcd(a, b);
}
inline ll read() {
    
	ll c = getchar(), Nig = 1, x = 0;
	while (!isdigit(c) && c != '-')c = getchar();
	if (c == '-')Nig = -1, c = getchar();
	while (isdigit(c))x = ((x << 1) + (x << 3)) + (c ^ '0'), c = getchar();
	return Nig * x;
}
inline void out(ll a) {
    
	if (a < 0)putchar('-'), a = -a;
	if (a > 9)out(a / 10);
	putchar(a % 10 + '0');
}
ll qpow(ll x, ll n, ll mod) {
    
	ll res = 1;
	while (n > 0) {
    
		if (n & 1)res = (res * x) % mod;
		x = (x * x) % mod;
		n >>= 1;
	}
	return res;
}
#define read read()
const int N = 50000;
struct node
{
    
	int len, maxn, lmaxn, rmaxn;
}tree[N * 4 + 7];
void push_up(int rt)
{
    
	if (tree[rt << 1].lmaxn == tree[rt << 1].len)
		tree[rt].lmaxn = tree[rt << 1].maxn + tree[rt << 1 | 1].lmaxn;
	else
		tree[rt].lmaxn = tree[rt << 1].lmaxn;
	if (tree[rt << 1 | 1].rmaxn == tree[rt << 1 | 1].len)
		tree[rt].rmaxn = tree[rt << 1].rmaxn + tree[rt << 1 | 1].maxn;
	else
		tree[rt].rmaxn = tree[rt << 1 | 1].rmaxn;
	tree[rt].maxn = max(tree[rt << 1].rmaxn + tree[rt << 1 | 1].lmaxn, max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn));
}
void creat(int l, int r, int rt)
{
    
	tree[rt].len = r - l + 1;
	if (l == r)
	{
    
		tree[rt].rmaxn = tree[rt].lmaxn = tree[rt].maxn = 1;
		return;
	}
	int mid = l + r >> 1;
	creat(l, mid, rt << 1);
	creat(mid + 1, r, rt << 1 | 1);
	push_up(rt);
}
void update(int pos, int mark, int l, int r, int rt)
{
    
	if (l == r)
	{
    
		tree[rt].lmaxn = tree[rt].rmaxn = tree[rt].maxn = mark;
		return;
	}
	int mid = l + r >> 1;
	if (pos <= mid)update(pos, mark, l, mid, rt << 1);
	else update(pos, mark, mid + 1, r, rt << 1 | 1);
	push_up(rt);
}
int query(int pos, int l, int r, int rt)
{
    
	int mid = l + r >> 1;
	if (l == r || tree[rt].maxn == 0 || tree[rt].maxn == tree[rt].len)
		return tree[rt].maxn;
	if (pos <= mid)
	{
    
		if (pos >= mid - tree[rt << 1].rmaxn + 1)
			return query(pos, l, mid, rt << 1) + query(mid + 1, mid + 1, r, rt << 1 | 1);
		else
			return query(pos, l, mid, rt << 1);
	}
	else
	{
    
		if (pos <= mid + tree[rt << 1 | 1].lmaxn)
			return query(mid, l, mid, rt << 1) + query(pos, mid + 1, r, rt << 1 | 1);
		else return query(pos, mid + 1, r, rt << 1 | 1);
	}
}
stack<int>s;
int main()
{
    
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
    
		while (!s.empty())s.pop();
		memset(tree, 0, sizeof(tree));
		creat(1, n, 1);
		while (m--)
		{
    
			char Str[5];
			scanf("%s", Str);
			if (Str[0] == 'D')
			{
    
				int id = read;
				s.push(id);
				update(id, 0, 1, n, 1);
			}
			else if (Str[0] == 'R')
			{
    
				int id = s.top();
				s.pop();
				update(id, 1, 1, n, 1);
			}
			else
			{
    
				int id = read;
				int len = query(id, 1, n, 1);
				out(len);
				puts("");
			}
		}
	}
}

By-Round Moon

版权声明
本文为[Round moon]所创，转载请带上原文链接，感谢
https://blog.csdn.net/qq_35339563/article/details/119970721

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