洛谷P3258 [JLOI2014]松鼠的新家

(Large extbf{Description: } large{一棵树，给出一个游历顺序，每次经过一点一次，此点点权加1，最后一个点不用再加，输出所有点的点权。})

(Large extbf{Solution: } large{很明显的树上点差分，注意一下细节即可。\觉得这道题属实水，谈一下带给我的启发吧。记得有这么一句话 ext{Think twice, code once.}\一开始码完有个细节没注意，然后看了一下标签，线段树？？当时就怀疑自己想假了，但是后来仔细想想没问题，改了下细节就A了。\总之，以后码前先仔细想清楚自己算法的正确性，别半路醒悟；而且事先想好一些细节处理，代码一气呵成。})

(Large extbf{Code: })

#include <cstdio>
#include <algorithm>
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
using namespace std;
const int N = 3e5 + 5;
int n, cnt, head[N], top[N], f[N], dep[N], dir[N], son[N], size[N], ans[N];

struct Edge {
	int to, next;	
}e[N << 1];

inline int read() {
	char ch = gc();
	int ans = 0;
	while (ch > '9' || ch < '0') ch = gc();
	while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
	return ans;
}

inline void add(int x, int y) {
	e[++cnt].to = y;
	e[cnt].next = head[x];
	head[x] = cnt;	
}

inline void dfs1(int x, int fa) {
	int Max = 0;
	f[x] = fa;
	size[x] = 1;
	dep[x] = dep[fa] + 1;
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		dfs1(u, x);
		size[x] += size[u];
		if (size[u] > Max) Max = size[u], son[x] = u;
	}
}

inline void dfs2(int x, int tp) {
	top[x] = tp;
	if (!son[x]) return;
	dfs2(son[x], tp);
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == son[x] || u == f[x]) continue;
		dfs2(u, u);
	}
}

inline int LCA(int x, int y) {
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) swap(x, y);
		x = f[top[x]];
	}	
	return dep[x] < dep[y] ? x : y;
}

inline void modify(int x, int y) {
	int lca = LCA(x, y);
	++ans[y], ++ans[x], --ans[lca], --ans[f[lca]];
}

inline void dfs(int x, int fa) {
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		dfs(u, x);
		ans[x] += ans[u];
	}	
}

int main() {
	n = read();
	int x, y;
	rep(i, 1, n) dir[i] = read(); 
	rep(i, 2, n) x = read(), y = read(), add(x, y), add(y, x);
	dfs1(1, 0); 
	dfs2(1, 1);
	rep(i, 2, n) modify(dir[i - 1], dir[i]);
	dfs(1, 0);
	rep(i, 1, n) {
		if (i == dir[1]) printf("%d
", ans[i]);
		else printf("%d
", ans[i] - 1);
	}
	return 0;
}