CF1375G. Tree Modification

Solution

假设我们取定了根，那么只可能从深度大的点接到深度小的点，我们每次取一个高度为2的子树接到该子树的父亲，这样取一定不劣，操作次数相当于是偶数深度点（根深度为0）的个数减一。

注意到所有奇数深度的点为根的答案都相同，偶数同理。
因此整合之后，相当于是奇数深度和偶数深度点的个数的较小值减一。

时间复杂度$O(n)$。

Code

#include <bits/stdc++.h>using namespace std;template<typename T> inline bool upmin(T &x, T y) { return y < x ? x = y, 1 : 0; } template<typename T> inline bool upmax(T &x, T y) { return x < y ? x = y, 1 : 0; }#define MP(A,B) make_pair(A,B) #define PB(A) push_back(A) #define SIZE(A) ((int)A.size()) #define LEN(A) ((int)A.length()) #define FOR(i,a,b) for(int i=(a);i<(b);++i) #define fi first #define se secondtypedef long long ll; typedef unsigned long long ull; typedef long double lod; typedef pair<int, int> PR; typedef vector<int> VI; const lod eps = 1e-9; const lod pi = acos(-1); const int oo = 1 << 30; const ll loo = 1ll << 60; const int mods = 1e9 + 7; const int inv2 = (mods + 1) >> 1; const int MAXN = 500005; const int INF = 0x3f3f3f3f; //1061109567 /*--------------------------------------------------------------------*/namespace FastIO{constexpr int SIZE = (1 << 21) + 1;int num = 0, f;char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf, *oT = obuf + SIZE - 1, c;#define gc() (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)), (iS == iT ? EOF : *iS ++)) : *iS ++)inline void flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}inline void putc(char c) {*oS ++ = c;if (oS == oT) flush();}inline void getc(char &c) {for (c = gc(); c != '<' && c != '>' && c != EOF; c = gc());}inline void reads(char *st) {char c;int n = 0;getc(st[++ n]);for (c = gc(); c == '<' || c == '>' ; c = gc()) st[++ n] = c;st[n + 1] = '\0';}template<class I>inline void read(I &x) {for (f = 1, c = gc(); c < '0' || c > '9' ; c = gc()) if (c == '-') f = -1;for (x = 0; c >= '0' && c <= '9' ; c = gc()) x = (x << 3) + (x << 1) + (c & 15);x *= f;}template<class I>inline void print(I x) {if (x < 0) putc('-'), x = -x;if (!x) putc('0');while (x) que[++ num] = x % 10 + 48, x /= 10;while (num) putc(que[num --]);}struct Flusher_{~Flusher_(){flush();}} io_Flusher_; } using FastIO :: read; using FastIO :: putc; using FastIO :: reads; using FastIO :: print;vector<int> e[MAXN]; int dep[MAXN], num[2]; void dfs(int x, int father) {dep[x] = dep[father] + 1;++ num[dep[x] & 1];for (auto v : e[x]) {if (v == father) continue;dfs(v, x);} } signed main() { #ifndef ONLINE_JUDGEfreopen("a.in", "r", stdin); #endifint n;read(n);for (int i = 1, u, v; i < n ; ++ i) read(u), read(v), e[u].PB(v), e[v].PB(u); dfs(1, 0);print(min(num[0], num[1]) - 1);return 0; }

总结

以上是生活随笔为你收集整理的CF1375G. Tree Modification(贪心，黑白染色)的全部内容，希望文章能够帮你解决所遇到的问题。

如果觉得生活随笔网站内容还不错，欢迎将生活随笔推荐给好友。