【CF600E】Lomsat gelral——树上启发式合并

（题面来自luogu）

题意翻译

一棵树有n个结点，每个结点都是一种颜色，每个颜色有一个编号，求树中每个子树的最多的颜色编号的和。

ci <= n <= 1e5

　　裸题。统计时先扫一遍得到出现次数最大值，然后再扫一遍看哪个颜色的出现次数与mxCnt相等。注意用一个bool数组判重，清空轻儿子贡献时要顺手把bool数组也打成false。（这是错的，请看下一份代码）

代码：

#include <iostream>
#include <cstdio>
#include <cctype>
#define maxn 100010
using namespace std;
template <typename T>
void read(T &x) {
    x = 0;
    char ch = getchar();
    while (!isdigit(ch))
        ch = getchar();
    while (isdigit(ch))
        x = x * 10 + (ch ^ 48), ch = getchar();
}
int n;
int head[maxn], top;
struct E {
    int to, nxt;
} edge[maxn << 1];
inline void insert(int u, int v) {
    edge[++top] = (E) {v, head[u]};
    head[u] = top;
}
int son[maxn], size[maxn];
long long ans[maxn];
void dfs1(int u, int pre) {
    size[u] = 1;
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre) continue;
        dfs1(v, u);
        size[u] += size[v];
        if (size[v] > size[son[u]])
            son[u] = v;
    }
    return;
}
int color[maxn], cnt[maxn], sum;
int curSon, mxCnt;
bool vis[maxn];
void calc(int u, int pre, bool op) {
    op ? (++cnt[color[u]], mxCnt = max(mxCnt, cnt[color[u]])) : (--cnt[color[u]], vis[color[u]] = false);
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre || v == curSon)
            continue;
        calc(v, u, op);
    }
}
void stats(int u, int pre) {
    if (cnt[color[u]] == mxCnt && !vis[color[u]])
        sum += color[u], vis[color[u]] = true;
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v != pre)
            stats(v, u);
    }
    return;
}
void dsu(int u, int pre, bool op) {
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre || v == son[u])
            continue;
        dsu(v, u, 0);
    }
    if (son[u])
        dsu(son[u], u, 1), curSon = son[u];
    mxCnt = 0;
    calc(u, pre, 1);
    stats(u, pre);
    ans[u] = sum;
    curSon = 0;
    if (!op) {
        sum = 0;
        calc(u, pre, 0);
    }
}
int main() {
    read(n);
    for (int i = 1; i <= n; ++i)
        read(color[i]);
    int u, v;
    for (int i = 1; i < n; ++i) {
        read(u), read(v);
        insert(v, u), insert(u, v);
    }
    dfs1(1, 0);
    dsu(1, 0, 1);
    for (int i = 1; i <= n; ++i)
        printf("%I64d ", ans[i]);
    return 0;
}

　　 （交这个题的时候cf炸了……才不是因为我TLE呢

--------------------------

　　然而果然出了锅，昨天CF评测机大概是被这个弱智错误笑死的……

　　重测了下，上面那份代码是错的，原因是没有考虑重儿子内的最大出现次数。然而原思路越改越复杂，后来想了想，每遍历到重复的颜色其出现次数都会+1，因此不用判重，从每个点统计时扫一遍就好了。

代码：

#include <iostream>
#include <cstdio>
#include <cctype>
#define maxn 100010
using namespace std;
template <typename T>
void read(T &x) {
    x = 0;
    char ch = getchar();
    while (!isdigit(ch))
        ch = getchar();
    while (isdigit(ch))
        x = x * 10 + (ch ^ 48), ch = getchar();
}
int n;
int head[maxn], top;
struct E {
    int to, nxt;
} edge[maxn << 1];
inline void insert(int u, int v) {
    edge[++top] = (E) {v, head[u]};
    head[u] = top;
}
int son[maxn], size[maxn];
long long ans[maxn];
void dfs1(int u, int pre) {
    size[u] = 1;
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre) continue;
        dfs1(v, u);
        size[u] += size[v];
        if (size[v] > size[son[u]])
            son[u] = v;
    }
    return;
}
int color[maxn], cnt[maxn];
long long sum;
int curSon, mxCnt;
void calc(int u, int pre, int val) {
    cnt[color[u]] += val;
    if (val == 1) {
        if (cnt[color[u]] > mxCnt)
            mxCnt = cnt[color[u]], sum = color[u];
        else if (cnt[color[u]] == mxCnt)
            sum += color[u];
    }
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre || v == curSon)
            continue;
        calc(v, u, val);
    }
}
void dsu(int u, int pre, bool op) {
    for (int i = head[u]; i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == pre || v == son[u])
            continue;
        dsu(v, u, 0);
    }
    if (son[u])
        dsu(son[u], u, 1), curSon = son[u];
    calc(u, pre, 1);
    ans[u] = sum;
    curSon = 0;
    if (!op) {
        mxCnt = sum = 0;
        calc(u, pre, -1);
    }
}
int main() {
    read(n);
    for (int i = 1; i <= n; ++i)
        read(color[i]);
    int u, v;
    for (int i = 1; i < n; ++i) {
        read(u), read(v);
        insert(v, u), insert(u, v);
    }
    dfs1(1, 0);
    dsu(1, 0, 1);
    for (int i = 1; i <= n; ++i)
        printf("%I64d ", ans[i]);
    return 0;
}