JZOJ 3448.公路维护

( ext{Problem})

1.询问区间最小值是否大于 (0)
2.区间加(可正可负)
3.区间取 (max)
如果某个数经过操作后小于等于 (0),以后的操作就不会再影响这个数

( ext{Analysis})

显然要用线段树维护这个区间
区间加和 (max) 打个双标记就好了,给加法优先
然后考虑区间加的过程中一个数如果加上这个数小于等于了 (0),那么我们把这个点设成无限大,标记一下
一个区间的标记定为区间内有没有被标记了的点
设成无限大后继续操作不会影响后面的情况,这样就好极了

( ext{Code})

#include<cstdio>
#include<iostream>
#define ls (k << 1)
#define rs (ls | 1)
using namespace std;

const int N = 1e5 + 5 , INF = 0x3f3f3f3f;
int n , m , L;

struct segment{
	int mn , tag_add , tag_max, fl;
}seg[N << 2];

inline void read(int &x)
{
	x = 0; char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while (ch >= '0' && ch <= '9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
}

inline void pushup(int k)
{
	seg[k].mn = min(seg[ls].mn , seg[rs].mn);
	seg[k].fl = seg[ls].fl && seg[rs].fl;
}

inline void push_add(int k , int v)
{
	seg[k].tag_add += v, seg[k].mn += v;
	if (seg[k].tag_max > -INF) seg[k].tag_max += v;
}
inline void push_max(int k , int v)
{
	if (v <= seg[k].mn) return;
	seg[k].mn = v, seg[k].tag_max = v;
}

inline void pushdown(int k)
{
	if (seg[k].tag_add != 0)
	{
		push_add(ls , seg[k].tag_add);
		push_add(rs , seg[k].tag_add);
		seg[k].tag_add = 0;
	}
	if (seg[k].tag_max != -INF)
	{
		push_max(ls , seg[k].tag_max);
		push_max(rs , seg[k].tag_max);
		seg[k].tag_max = -INF;
	}
}

inline void build(int l , int r , int k)
{
	seg[k].tag_max = -INF;
	if (l == r)
	{
		seg[k].mn = L, seg[k].fl = 1;
		return;
	}
	int mid = (l + r) >> 1;
	build(l , mid , ls) , build(mid + 1 , r , rs);
	pushup(k);
}

inline void update_add(int l , int r , int k , int x , int y , int c)
{
	if (x <= l && r <= y) 
	{
		if (seg[k].mn + c > 0) return push_add(k , c);
		else if (l == r)
		{
			seg[k].mn = INF, seg[k].fl = 0;
			return;
		}
	}
	pushdown(k);
	int mid = (l + r) >> 1;
	if (x <= mid) update_add(l , mid , ls , x , y , c);
	if (y > mid) update_add(mid + 1 , r , rs , x , y , c);
	pushup(k);
}
inline void update_max(int l , int r , int k , int x , int y , int c)
{
	if (seg[k].mn >= c) return;
	if (x <= l && r <= y) 
	{
		push_max(k, c);
		return;
	}
	pushdown(k);
	int mid = (l + r) >> 1;
	if (x <= mid) update_max(l , mid , ls , x , y , c);
	if (y > mid) update_max(mid + 1 , r , rs , x , y , c);
	pushup(k);
}

inline int query(int l , int r , int k , int x, int y)
{
	if (x <= l && r <= y) return seg[k].fl;
	pushdown(k);
	int mid = (l + r) >> 1, res = 1;
	if (x <= mid) res = query(l , mid , ls , x, y);
	if (y > mid) res = res && query(mid + 1 , r , rs , x, y);
	return res;
}

int main()
{
	freopen("road.in", "r", stdin);
	freopen("road.out", "w", stdout);
	read(n), read(m), read(L);
	build(1 , n , 1);
	int l , r , c, op, ans = 0;
	for(int i = 1; i <= m; i++)
	{
		read(op), read(l), read(r), read(c);
		if (op == 1 && (query(1, n, 1, l, r))) ++ans, update_add(1, n, 1, l, r, -c);
		else if (op == 2) update_add(1, n, 1, l, r, c);
		else if (op == 3) update_max(1, n, 1, l, r, c);
	}
	printf("%d
", ans);
}
原文地址:https://www.cnblogs.com/leiyuanze/p/15011028.html