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即使愿望永远无法实现
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翻车笔记 BZOJ4765 普通计算姬 [分块, 清奇脑回路, 大分块, 翻车]

合理平衡复杂度是手筋

懵逼的 题目

qwq

扯淡的 题解

不会, 抄的题解

来...让我们想一下 树链剖分一次修改的复杂度是\(\mathrm O(\lg n)\), 但是查询是\(\mathrm O(n)\)

好...那我们就可以平衡复杂度了

考虑一下修改一个点对答案的贡献, 修改一个节点的权值只会对他的祖先造成影响, 我们就可以搞一个区间求和的大分块, 记录最后的ans, 然后搞一个数组g[i][j], 表示第\(i\)个节点在第\(j\)个块有几个祖先(包括自己), 这样修改的时候就可以直接sum[i] += g[dq][i] * zh

那么零散的块呢? 总不能搞一个树链剖分吧...

Emmmmm...如果我们不记录子树和, 而是每一次都重新计算呢?

\(dfs\)序上搞一个树状数组, 然后每一次都重新计算子树和!

完美!

感觉再给我三辈子我也想不出来_(:з」∠)_

好...让我们想想我们需要些什么...

  • 一个区间修改, 区间查询的块qwq, 用来存储最终的答案
    • 一个数组sum[MAXK], 记录当前块的数字和; 一个单点修改, 区间查询的树状数组b, 建在\(dfs\)序上, 用来存储每一个节点的子树和
    • ULL cx(int le, int ri):
      • 找到\([le, ri]\)间的整块, re += sum[i]
      • 对于所有的零散点, re += cxsum(i)
    • void xg(int wz, ULL zh): 遍历所有的块, sum[i] += zh * g[wz][i], xgsum(ta[wz].begin, zh)
  • 一个数组g[MAXN][MAXK]: g[i][j]用来存储第i个节点在第j个块中的祖先个数(用来记录贡献)
  • 一个结构体ta[MAXN] = {begin, end}, 来搞\(dfs\)

沙茶的 代码

注意几个细节:

  • 分块的时候一定要在最后一个块的右边界上加一个和\(n\)的取\(\min\)...否则\(RE\)死你...
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#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MAXN (100000 + 5)
#define MAXK (350 + 5)
#define INF (0x7ffffff)
#define lowbit(x) ((x) & (-(x)))
#define ULL unsigned long long
#define LL long long
#define rint register int
#define cint const int
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) < (b) ? (a) : (b))
using namespace std;
struct edg
{
int from, to, next;
}tb[MAXN << 1];
struct tnode
{
int begin, end, fa;
}ta[MAXN];
int n, root, tg[MAXN], g[MAXN][MAXK], cnta, cntb, block, le[MAXK], ri[MAXK];
ULL sum[MAXK], b[MAXN];
LL a[MAXN];
inline int belong(int x)
{ return ((x - 1) / block + 1); }
inline ULL cxsum(int x)
{
ULL re = 0;
for (int i = ta[x].end; i; i -= lowbit(i))
re += b[i];
for (int i = ta[x].begin - 1; i; i -= lowbit(i))
re -= b[i];
return re;
}
inline void xgsum(int wz, LL zh)
{
for (int i = wz; i <= n; i += lowbit(i))
b[i] += zh;
}
void dfs(int dq)
{
ta[dq].begin = ++cnta;
xgsum(cnta, a[dq]);
for (int i = 1; i <= le[0]; i++)
g[dq][i] = g[ta[dq].fa][i];
++g[dq][belong(dq)];
for (int i = tg[dq]; i; i = tb[i].next)
if (tb[i].to != ta[dq].fa)
{
ta[tb[i].to].fa = dq;
dfs(tb[i].to);
}
ta[dq].end = cnta;
}
void xg(int wz, LL zh)
{
a[wz] += zh;
for (int i = 1; i <= le[0]; i++)
sum[i] += zh * g[wz][i];
xgsum(ta[wz].begin, zh);
}
ULL cx(int l, int r)
{
ULL re = 0;
int lerk = INF, rigk = -INF;
for (int i = 1; i <= le[0]; i++)
if (le[i] >= l && ri[i] <= r)
{
re += sum[i];
lerk = min(lerk, i), rigk = max(rigk, i);
}
if (lerk >= INF || rigk < 0)
{
for (int i = l; i <= r; i++)
re += cxsum(i);
return re;
}
if (lerk > 1)
for (int i = l; i <= ri[lerk - 1]; i++)
re += cxsum(i);
if (rigk < le[0])
for (int i = le[rigk + 1]; i <= r; i++)
re += cxsum(i);
return re;
}
void init()
{
block = sqrt(n) + 1;
for (int i = 1; i <= n; i += block)
le[++le[0]] = i, ri[++ri[0]] = min(n, i + block - 1);
}
void adn(int from, int to)
{
tb[++cntb].next = tg[from];
tb[cntb].from = from;
tb[cntb].to = to;
tg[from] = cntb;
}
int main()
{
freopen("in.in", "r", stdin);
int q;
scanf("%d%d", &n, &q);
init();
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]);
for (int i = 1, srx, sry; i <= n; i++)
{
scanf("%d%d", &srx, &sry);
if (!srx)
{
root = sry;
continue;
}
adn(srx, sry);
adn(sry, srx);
}
ta[root].fa = root;
dfs(root);
for (int i = 1; i <= le[0]; i++)
for (int j = le[i]; j <= ri[i]; j++)
sum[i] += cxsum(j);
for (int i = 1, sre, srx, sry; i <= q; i++)
{
scanf("%d%d%d", &sre, &srx, &sry);
if (sre == 1)
xg(srx, sry - a[srx]);
else
printf("%llu\n", cx(srx, sry));
}
return 0;
}

By 大傻叉 Cansult