题目大意:给以1为根的一棵树,求树上每个点子树中出现次数最多的权值(如果有多个就求他们的和)
对每个点开一个线段树维护子树内权值的桶,dfs时候线段树合并就行了。
因为最后线段树一共插入最多 \(O(n\log n)\) 个节点,每个节点最多会被合并一次,所以复杂度是 \(O(n\log n)\) 的。
#include <cstdio>
#include <vector>
using namespace std;
struct fuck { int maxval; long long sum; } tree[2333333];
int lc[2333333], rc[2333333], tot;
int n, val[100010], fa[100010], rt[100010];
vector<int> out[100010];
long long ans[100010];
fuck operator*(const fuck &a, const fuck &b)
{
if (a.maxval > b.maxval) return a;
if (a.maxval < b.maxval) return b;
return (fuck){a.maxval, a.sum + b.sum};
}
int mg(int x1, int x2, int cl, int cr)
{
if (x1 == 0) return x2;
if (x2 == 0) return x1;
if (cl == cr)
{
tree[x1] = (fuck){tree[x1].maxval + tree[x2].maxval, tree[x1].sum};
return x1;
}
int mid = (cl + cr) / 2;
lc[x1] = mg(lc[x1], lc[x2], cl, mid);
rc[x1] = mg(rc[x1], rc[x2], mid + 1, cr);
tree[x1] = tree[lc[x1]] * tree[rc[x1]];
return x1;
}
void chenge(int &x, int cl, int cr, int pos, int val)
{
if (x == 0) x = ++tot;
if (cl == cr) { tree[x].maxval++; tree[x].sum = cl; return; }
int mid = (cl + cr) / 2;
if (pos > mid) chenge(rc[x], mid + 1, cr, pos, val);
else chenge(lc[x], cl, mid, pos, val);
tree[x] = tree[lc[x]] * tree[rc[x]];
}
void dfs(int x)
{
for (int i : out[x]) if (fa[x] != i)
{
fa[i] = x; dfs(i), rt[x] = mg(rt[x], rt[i], 1, n);
}
chenge(rt[x], 1, n, val[x], 1);
ans[x] = tree[rt[x]].sum;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
for (int x, y, i = 1; i < n; i++)
{
scanf("%d%d", &x, &y);
out[x].push_back(y), out[y].push_back(x);
}
dfs(1);
for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
return 0;
}
原文地址:https://www.cnblogs.com/oier/p/10586807.html
时间: 2024-10-03 21:34:34