题解【Codeforces580C】Kefa and park

题面

在对树进行遍历时记录一下这个点的父亲点权是不是 (1)、当前连续点权为 (1) 的节点个数、根到当前结点的路径上连续点权为 (1) 的最多节点个数，然后直接对叶子节点统计答案即可。

#include <bits/stdc++.h>
#define DEBUG fprintf(stderr, "Passing [%s] line %d
", __FUNCTION__, __LINE__)
#define File(x) freopen(x".in","r",stdin); freopen(x".out","w",stdout)

using namespace std;

typedef long long LL;
typedef pair <int, int> PII;
typedef pair <int, PII> PIII;

inline int gi()
{
	int f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

inline LL gl()
{
	LL f = 1, x = 0; char c = getchar();
	while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return f * x;
}

const int INF = 0x3f3f3f3f, N = 100003, M = N << 1;

int n, m;
int tot, head[N], ver[M], nxt[M];
int qz[N];
int lx[N];

inline void add(int u, int v)
{
	ver[++tot] = v, nxt[tot] = head[u], head[u] = tot;
}

void dfs(int u, int f, int now/*当前连续点权为 1 的节点个数*/, bool lst/*当前节点的父亲的点权是不是 1*/, int mxnow/*根到当前节点的路径上点权为 1 的最多节点个数*/)
{
	lx[u] = mxnow; //记录最多的节点个数
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == f) continue;
		//分情况讨论递归
		if (qz[v] == 1)
			dfs(v, u, lst ? now + 1 : 1, true, max(mxnow, lst ? now + 1 : 1));
		else 
			dfs(v, u, 0, false, mxnow);
	}
}

bool fl[N];

inline void dfss(int u, int f)
{
	for (int i = head[u]; i; i = nxt[i])
	{
		int v = ver[i];
		if (v == f) continue;
		fl[u] = true;
		dfss(v, u);
	}
}

int main()
{
	//File("");
	n = gi(), m = gi();
	for (int i = 1; i <= n; i+=1) qz[i] = gi();
	for (int i = 1; i < n; i+=1)
	{
		int u = gi(), v = gi();
		add(u, v), add(v, u);
	}
	dfss(1, 0);
	dfs(1, 0, qz[1], (qz[1] == 1), qz[1]);
	int ans = 0;
	for (int i = 1; i <= n; i+=1)
	{ 
		if (!fl[i] && lx[i] <= m) ++ans; //对符合要求的叶子节点计数
	}
	printf("%d
", ans);
	return 0;
}