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ARC116E Spread of Information

感觉并没有那么难,赛事也基本想到了 ,实现比较困难。

Solution

二分是显然的。

然后重点是如何判断是否可以构造出来,赛事我用的是树分块的构造方法,比较low,是$O(n^2)$(可能跑不满),主要是没想清楚怎么判断两个无关的树上叶子节点怎么匹配,发现只需要叶子回溯到父亲的时候将自己儿子中是关键节点的最高节点返回回去即可。

注意特判1

Code

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#include <bits/stdc++.h>
#define LL long long
#define PII pair<int, int>
#define PLL pair<LL, LL>
using namespace std;
template <class T>
inline void read(T &res)
{
res = 0;
bool flag = 0;
char c = getchar();
while (c < '0' || '9' < c) { if (c == '-') flag = 1; c = getchar(); }
while ('0' <= c && c <= '9') res = (res << 3) + (res << 1) + (c ^ 48), c = getchar();
if (flag) res = -res;
}
template <class T, class... ARC>
inline void read(T &res, ARC &...com) { read(res), read(com...); }
template <class T>
void write(T res)
{
if (res < 0)
putchar('-'), res = -res;
if (res > 9)
write(res / 10);
putchar(res % 10 + '0');
}
template <>
inline void write(char c) { putchar(c); }
template <>
inline void write(const char *s) { while (*s) putchar(*s++); }
template <class T, class... ARC>
inline void write(T res, ARC... com) { write(res), write(com...); }
const int N = 2e5 + 5;
int n, m;
int idx;
int e[N << 1], ne[N << 1], h[N];
inline void add(int x, int y)
{
idx ++;
e[idx] = y, ne[idx] = h[x], h[x] = idx;
}
int cnt, dfn[N], len, type[N];
void dfs(int x, int y)
{
dfn[x] = 0, type[x] = 1;
int res = 0, maxn = 0;
for(int i = h[x], j; ~i;i = ne[i])
{
j = e[i];
if(j == y) continue;
dfs(j, x);
res = min(res, dfn[j]);
maxn = max(maxn, dfn[j] + type[j]);
}
if(res + maxn + 1 <= 0 && res < 0) dfn[x] = res + 1, type[x] = 0;
else{
if(maxn < len) dfn[x] = maxn;
else dfn[x] = -len, cnt ++, type[x] = 0;
}
}
inline bool check(int mid)
{
cnt = 0, len = mid;
dfs(1, 1);
if(dfn[1] > 0 || type[1]) cnt ++;
return cnt <= m;
}
int main()
{
memset(h, -1, sizeof(h));
read(n, m);
for(int i = 1, o, u;i < n;i ++) read(o, u), add(o, u), add(u, o);
int l = 0, r = n, k;
while(l <= r)
{
int mid = l + r >> 1;
if(check(mid)) r = mid - 1, k = mid;
else l = mid + 1;
}
write(k);
return 0;
}