大家好,又见面了,我是全栈君。
其核心思想是记录最大的节点值和最低值,假设max<p要么min>=p时间,在节点只变化add值,不要子树遍历;否则,就往子树递归。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn = 2e5+50;
int N, P;
struct node{
int l, r, Min, Max, add;
int mid() { return (l+r)/2; }
}tree[maxn<<2];
int p, n;
void buildTree(int l, int r, int rt)
{
tree[rt].l = l;
tree[rt].r = r;
tree[rt].add = 0;
tree[rt].Min = 0;
tree[rt].Max = 0;
if(l == r) return ;
int mid = tree[rt].mid();
buildTree(l, mid, rt<<1);
buildTree(mid+1, r, rt<<1|1);
}
void update(int l, int r, int rt, int L, int R, int add)
{
if(L <= l && R >= r)
{
if(tree[rt].Max < P)
{
tree[rt].add += add;
tree[rt].Min += add;
tree[rt].Max += add;
return ;
}
else if(tree[rt].Min >= P)
{
tree[rt].add += 2*add;
tree[rt].Min += 2*add;
tree[rt].Max += 2*add;
return ;
}
}
if(tree[rt].add){
tree[rt<<1].add += tree[rt].add;
tree[rt<<1].Min += tree[rt].add;
tree[rt<<1].Max += tree[rt].add;
tree[rt<<1|1].add += tree[rt].add;
tree[rt<<1|1].Min += tree[rt].add;
tree[rt<<1|1].Max += tree[rt].add;
tree[rt].add = 0;
}
if(l == r) return ;
int mid = tree[rt].mid();
if(L <= mid) update(l, mid, rt<<1, L, R, add);
if(R > mid) update(mid+1, r, rt<<1|1, L, R, add);
tree[rt].Min = min(tree[rt<<1].Min, tree[rt<<1|1].Min);
tree[rt].Max = max(tree[rt<<1].Max, tree[rt<<1|1].Max);
}
void query(int l, int r, int rt)
{
if(tree[rt].Min == tree[rt].Max){
for(int i = l; i <= r; i ++)
printf( i == N ? "%d\n" : "%d ", tree[rt].Min );
return ;
}
if(tree[rt].add)
{
tree[rt<<1].add += tree[rt].add;
tree[rt<<1].Min += tree[rt].add;
tree[rt<<1].Max += tree[rt].add;
tree[rt<<1|1].add += tree[rt].add;
tree[rt<<1|1].Min += tree[rt].add;
tree[rt<<1|1].Max += tree[rt].add;
tree[rt].add = 0;
}
if(l == r) return ;
int mid = tree[rt].mid();
query(l, mid, rt<<1);
query(mid+1, r, rt<<1|1);
}
int main()
{
int n, m, p;
int a, b, c;
while(~scanf("%d%d%d", &n, &m, &p))
{
N = n;
P = p;
buildTree(1, n, 1);
for(int i = 0; i < m; i ++)
{
scanf("%d%d%d", &a, &b, &c);
update(1, n, 1, a, b, c);
}
query(1, n, 1);
}
}