分治线段树,其实就是将标记永久化,到最后再统一下传所有标记。
至于先后顺序,可以给每个节点开一个时间戳。
一般地,分治线段树用于离线,只查询一次答案的题目。
本题中,标记要被下传 222 次。
Code:
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 100000 + 10;
int n,m, tag[maxn << 2], lazy[maxn << 2], siz[maxn], root[maxn], p[maxn], tag_root[maxn], ans[maxn];
int find(int x){ return p[x] == x ? x : p[x] = find(p[x]); }
struct Segment_Tree
{
# define lson (o << 1)
# define rson (o << 1)|1
void update(int l, int r, int L, int R, int col, int times, int o)
{
if(l > r || r < L || l > R) return ;
if(l >= L && r <= R)
{
lazy[o] = col, tag[o] = times;
return;
}
int mid = (l + r) >> 1;
update(l, mid, L, R,col, times, lson);
update(mid + 1, r, L, R, col , times, rson);
}
inline void pushdown(int o)
{
if(tag[o] > tag[lson])
lazy[lson] = lazy[o], tag[lson] = tag[o];
if(tag[o] > tag[rson])
lazy[rson] = lazy[o], tag[rson] = tag[o];
}
void release(int o, int l, int r)
{
if(l > r) return ;
if(l == r)
{
int x = find(l);
if(tag_root[x] > tag[o])
ans[l] = root[x];
else
ans[l] = lazy[o], tag_root[x] = tag[o], root[x] = lazy[o];
return ;
}
int mid = (l + r) >> 1;
pushdown(o);
release(lson, l, mid);
release(rson, mid + 1, r);
}
}T;
int main()
{
scanf("%d%d",&n,&m);
for(int i = 1;i <= n; ++i)
{
int c; scanf("%d",&c);
p[i] = i, root[i] = c, siz[i] = 1, tag_root[i] = 1;
}
for(int i = 2;i <= m + 1; ++i)
{
int opt, l, r, x;
scanf("%d%d%d%d",&opt,&l,&r,&x);
switch(opt)
{
case 1:
T.update(1, n, l, r, x, i, 1);
break;
case 2:
int a = find(l), b = find(r);
root[b] = x, tag_root[b] = i;
if(a != b) { p[a] = b, siz[b] += siz[a]; }
break;
}
}
T.release(1, 1, n);
T.release(1, 1, n);
for(int i = 1;i <= n; ++i) printf("%d ",ans[i]);
printf("\n");
for(int i = 1;i <= n; ++i)
{
int x = find(i);
printf("%d ",siz[x] - 1);
}
return 0;
}