原题传送门
跟赛道修建很像的一道题目
不同的是,现在要全选
依然二分答案,然后
对于某点的所有儿子里面传上来的链,二分出能和自己组成一条链的往上传,剩下的是通过两两贪心组合判断可行性
注意两个点,若有偶数条链,加一条长度为0的链;根节点直接判断可行性
Code:
#include <bits/stdc++.h>
#define maxn 100010
using namespace std;
struct Edge{
int to, next;
}edge[maxn << 1];
int num, head[maxn], tot, flag, base, n, a[maxn], dp[maxn];
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;
}
void addedge(int x, int y){ edge[++num] = (Edge){y, head[x]}, head[x] = num; }
bool solve(int p){
int j = tot;
for (int i = 1; i < j; ++i, --j){
if (i == p) ++i;
if (j == p) --j;
if (a[i] + a[j] < base) return 0;
}
return 1;
}
void dfs(int u, int pre){
if (!flag) return;
dp[u] = 1;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != pre) dfs(v, u);
}
tot = 0;
for (int i = head[u]; i; i = edge[i].next){
int v = edge[i].to;
if (v != pre) a[++tot] = dp[v];
}
if (tot % 2 == 0 && u != 1) a[++tot] = 0;
sort(a + 1, a + 1 + tot);
int l = 1, r = tot, p = 0;
if (tot & 1)
while (l <= r){
int mid = (l + r) >> 1;
if (solve(mid)) p = mid, l = mid + 1; else r = mid - 1;
}
if (!solve(p)) flag = 0;
else dp[u] = a[p] + 1;
}
bool check(int mid){
flag = 1, base = mid;
dfs(1, 0);
return flag;
}
int main(){
n = read();
for (int i = 1; i < n; ++i){
int x = read(), y = read();
addedge(x, y), addedge(y, x);
}
int l = 1, r = n, ans;
while (l <= r){
int mid = (l + r) >> 1;
if (check(mid)) ans = mid, l = mid + 1; else r = mid - 1;
}
printf("%d\n", ans);
return 0;
}