[ZJOI2012]灾难

Description

[ZJOI2012]灾难

Solution

仿佛叫做毁灭树
如果一个点死了，只有它的每个儿子的祖宗都死了
但是这是一个图，不好处理
先跑一边\(TopSort\)，将原图变成一棵树，将每个点直接连到所有的儿子的LCA上，并且加上贡献
最后输出的时候要\(-1\)，要把自己对自己的贡献减掉

Code

#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
#define MAXN 1000010
struct rec {
	int nxt, ver;
} t[MAXN];
int n, cnt, s;
int head[MAXN], in[MAXN], size[MAXN], d[MAXN], lca[MAXN][21], h[MAXN];
queue <int> q;
inline int read() {
	int s = 0, w = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
	for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
	return s * w;
}
inline void add(int u, int v) {
	t[++cnt].nxt = head[u], t[cnt].ver = v, head[u] = cnt;
}
inline void TopSort() {
	for (register int i = 1; i <= n; i++)
		if (!in[i]) q.push(i);
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		h[++s] = u;
		for (register int i = head[u]; i; i = t[i].nxt) {
			int v = t[i].ver;
			if (!(--in[v])) q.push(v);
		}
	}
}
inline int LCA(int u, int v) {
	if (d[u] < d[v]) swap(u, v);
	for (register int i = 20; i >= 0; i--)
		if (d[lca[u][i]] >= d[v]) u = lca[u][i];
	if (u == v) return u;
	for (register int i = 20; i >= 0; i--)
		if (lca[u][i] != lca[v][i]) u = lca[u][i], v = lca[v][i];
	return lca[u][0];
}
int main() {
	n = read();
    for (register int i = 1; i <= n; i++) {
        int x = read();
        while (x) ++in[x], add(i, x), x = read();
    }
	TopSort();
	for (register int i = n; i; i--) {
		int u = h[i], x = t[head[u]].ver;
		for (register int j = t[head[u]].nxt; j; j = t[j].nxt) {
			int v = t[j].ver;
			x = LCA(x, v);
		}
		lca[u][0] = x, d[u] = d[x] + 1;
		for (register int j = 1; j <= 20; j++) lca[u][j] = lca[lca[u][j - 1]][j - 1];
	}
	for (register int i = 1; i <= n; i++) ++size[h[i]], size[lca[h[i]][0]] += size[h[i]];
    for (register int i = 1; i <= n; i++) printf("%d\n", size[i] - 1);
	return 0;
}