cf600E. Lomsat gelral(dsu on tree)

最新推荐文章于 2020-11-11 13:58:30 发布

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最新推荐文章于 2020-11-11 13:58:30 发布

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本文深入探讨了DSU on Tree算法，一种用于解决树上动态连通性问题的高效算法。通过具体实例，详细解释了如何在树的每个节点上应用该算法，以找出子树中出现次数最多的颜色编号及其总和。代码示例展示了从输入读取到最终输出的完整过程，包括树的深度优先搜索、子树信息更新和答案计算。

题意

题目链接

给出一个树，求出每个节点的子树中出现次数最多的颜色的编号和

Sol

dsu on tree的裸题。

一会儿好好总结总结qwq

#include<bits/stdc++.h>
#define LL long long 
using namespace std;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, col[MAXN], son[MAXN], siz[MAXN], cnt[MAXN], Mx, Son;
LL sum = 0, ans[MAXN];
vector<int> v[MAXN];
void dfs(int x, int fa) {
    siz[x] = 1;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa) continue;
        dfs(to, x);
        siz[x] += siz[to];
        if(siz[to] > siz[son[x]]) son[x] = to;
    }
}
void add(int x, int fa, int val) {
    cnt[col[x]] += val;
    if(cnt[col[x]] > Mx) Mx = cnt[col[x]], sum = col[x];
    else if(cnt[col[x]] == Mx) sum += (LL)col[x];
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa || to == Son) continue;
        add(to, x, val);
    }
}
void dfs2(int x, int fa, int opt) {
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i];
        if(to == fa) continue;
        if(to != son[x]) dfs2(to, x, 0);
    }
    if(son[x]) dfs2(son[x], x, 1), Son = son[x];

    add(x, fa, 1); Son = 0;
    ans[x] = sum;
    if(!opt) add(x, fa, -1), sum = 0, Mx = 0;
}
int main() {
    N = read();
    for(int i = 1; i <= N; i++) col[i] = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }
    dfs(1, 0);
    dfs2(1, 0, 0);
    for(int i = 1; i <= N; i++) printf("%I64d ", ans[i]);
    return 0;
}