网站开发e r图,网页设计代码的意思,网站改不了分类,网站右下角图片广告代码传送门 文章目录题意#xff1a;思路#xff1a;题意#xff1a; 思路#xff1a;
首先考虑最小值#xff0c;如果从一个叶子结点出发到任意叶子的距离都为偶数#xff0c;那么只需要一个值就可以满足条件。如果有奇数的#xff0c;考虑111 ^ 222 ^ 303030#xff0c;…传送门
文章目录题意思路题意 思路
首先考虑最小值如果从一个叶子结点出发到任意叶子的距离都为偶数那么只需要一个值就可以满足条件。如果有奇数的考虑111 ^ 222 ^ 303030我们可以在非连接叶节点的边上交错填1,21,21,2在叶节点的边上填333这样就可以保证满足条件。所以答案为111或333。 考虑最大值由于我们填的数可以无限大 所以考虑对于每个边都先分配一个不同的值假设有kkk个叶子连接在一个点上那么我们需要减去k−1k-1k−1因为这些点的边都需要填一样的值。 这个过程可以直接求也可以dpdpdp来求。 复杂度O(N)O(N)O(N)。
直接求
// Problem: D. Edge Weight Assignment
// Contest: Codeforces - Codeforces Round #633 (Div. 2)
// URL: https://codeforces.com/contest/1339/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize(Ofast,no-stack-protector,unroll-loops,fast-math)
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tunenative)
//#pragma GCC optimize(2)
#includecstdio
#includeiostream
#includestring
#includecstring
#includemap
#includecmath
#includecctype
#includevector
#includeset
#includequeue
#includealgorithm
#includesstream
#includectime
#includecstdlib
#define X first
#define Y second
#define L (u1)
#define R (u1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].ltr[u].r1)
#define Len(u) (tr[u].r-tr[u].l1)
#define random(a,b) ((a)rand()%((b)-(a)1))
#define db puts(---)
using namespace std;//void rd_cre() { freopen(d://dp//data.txt,w,stdout); srand(time(NULL)); }
//void rd_ac() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//AC.txt,w,stdout); }
//void rd_wa() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//WA.txt,w,stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pairint,int PII;const int N1000010,mod1e97,INF0x3f3f3f3f;
const double eps1e-6;int n;
int d[N],cd[N],a[N];
int f[N];
vectorintv[N];
bool flag;void dfs1(int u,int fa,int d) {for(auto x:v[u]) {if(xfa) continue;dfs1(x,u,d1);}if(v[u].size()1) {if(d%21) flag1;}
}void dfs2(int u,int fa) {int add0;for(auto x:v[u]) {if(xfa) continue;dfs2(x,u);if(v[x].size()1) add1;else f[u]f[x]1;}f[u]add;
}int solve1() {for(int i1;in;i) if(d[i]1) { dfs1(i,0,0); break; }if(flag) return 3;else return 1;
}int solve2() {int ansn-1;for(int i1;in;i) {if(d[i]1) continue; int cnt0;for(auto x:v[i]) {if(v[x].size()1) cnt;}ans-max(0,cnt-1);}return ans;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);scanf(%d,n);for(int i1;in-1;i) {int a,b; scanf(%d%d,a,b);v[a].pb(b); v[b].pb(a);d[a]; d[b];}printf(%d %d\n,solve1(),solve2());return 0;
}
/**/
树形dpdpdp
// Problem: D. Edge Weight Assignment
// Contest: Codeforces - Codeforces Round #633 (Div. 2)
// URL: https://codeforces.com/contest/1339/problem/D
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize(Ofast,no-stack-protector,unroll-loops,fast-math)
//#pragma GCC target(sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tunenative)
//#pragma GCC optimize(2)
#includecstdio
#includeiostream
#includestring
#includecstring
#includemap
#includecmath
#includecctype
#includevector
#includeset
#includequeue
#includealgorithm
#includesstream
#includectime
#includecstdlib
#define X first
#define Y second
#define L (u1)
#define R (u1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].ltr[u].r1)
#define Len(u) (tr[u].r-tr[u].l1)
#define random(a,b) ((a)rand()%((b)-(a)1))
#define db puts(---)
using namespace std;//void rd_cre() { freopen(d://dp//data.txt,w,stdout); srand(time(NULL)); }
//void rd_ac() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//AC.txt,w,stdout); }
//void rd_wa() { freopen(d://dp//data.txt,r,stdin); freopen(d://dp//WA.txt,w,stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pairint,int PII;const int N1000010,mod1e97,INF0x3f3f3f3f;
const double eps1e-6;int n;
int d[N],cd[N],a[N];
int f[N];
vectorintv[N];
bool flag;void dfs1(int u,int fa,int d) {for(auto x:v[u]) {if(xfa) continue;dfs1(x,u,d1);}if(v[u].size()1) {if(d%21) flag1;}
}void dfs2(int u,int fa) {int add0;for(auto x:v[u]) {if(xfa) continue;dfs2(x,u);if(v[x].size()1) add1;else f[u]f[x]1;}f[u]add;
}int solve1() {for(int i1;in;i) if(d[i]1) { dfs1(i,0,0); break; }if(flag) return 3;else return 1;
}int solve2() {for(int i1;in;i) {if(d[i]2) {dfs2(i,0);return f[i];}}return 0;
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0);scanf(%d,n);for(int i1;in-1;i) {int a,b; scanf(%d%d,a,b);v[a].pb(b); v[b].pb(a);d[a]; d[b];}printf(%d %d\n,solve1(),solve2());return 0;
}
/**/
