标签:sha 答案 algo har can span const bsp stream
DUS on tree
难得都不会,会的都是板子,可悲,可悲
题意:略
先想一个O(n^2)的写法,然后想办法去掉重复计算。究竟哪里重复 了呢?
假设p是x的儿子,p有很多个。每次计算答案的时候,如果“重儿子”(子孙最多的p)的答案可以直接用的话,
就可以省去很多的重复计算,这就是书上启发式合并
DUS ON TREE写法类似板子(恕我无知,没见过其他整法)
#include<cstdio>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 7;
ll list[maxn];
ll ans, mx;
int siz[maxn];//根
int son[maxn];//重儿子
ll cns[maxn];
vector<int>G[maxn];
void insert(int be, int en) {
G[be].push_back(en);
}
int dfs(int x, int fa) {
siz[x] = 1;
int c = 0;
for(int p : G[x]) {
if (p == fa) continue;
dfs(p, x);
siz[x] += siz[p];
if (c < siz[p]) {
c = siz[p];
son[x] = p;
}
}
return 0;
}
ll cnt[maxn];//出现次数
int cal(int x, int fa, int cn, int ban) {
cnt[list[x]] += cn;
if (cnt[list[x]] > mx) {
mx = cnt[list[x]];//times
ans = list[x];
}
else if (cnt[list[x]] == mx) {
ans += list[x];
}
for (int p : G[x]) {
if (p == fa || p == ban) continue;
cal(p, x, cn, ban);
}
return 0;
}
int dfs2(int x, int fa,int f) {
for (int p : G[x]) {
if (p == fa || p == son[x]) continue;
dfs2(p, x, 1);
}
if (son[x] ) {
dfs2(son[x], x, 0);
}
cal(x, fa, 1, son[x]);
cns[x] = ans;
if (f == 1) {
cal(x, fa, -1, 0);//把树清理干净
ans = 0;
mx = 0;
}
return 0;
}
int main() {
int n;
int be, en;
ans = 0;
mx = 0;
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
scanf("%lld", &list[i]);
}
for (int i = 1; i < n; i++) {
scanf("%d %d", &be, &en);
insert(be, en);
insert(en, be);
}
dfs(1, -1);
dfs2(1, -1, 0);
for (int i = 1; i <= n; i++) {
if (i == 1) printf("%lld", cns[i]);
else printf(" %lld", cns[i]);
}
printf("\n");
return 0;
}
标签:sha 答案 algo har can span const bsp stream
原文地址:https://www.cnblogs.com/lesning/p/12249955.html