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文章目錄

• 題意：
• 思路：

題意：

思路：

首先考慮最小值，如果從一個葉子結點出發到任意葉子的距離都為偶數，那么只需要一個值就可以滿足條件。如果有奇數的，考慮$1$ ^ $2$ ^ $3=0$，我們可以在非連接葉節點的邊上交錯填$1,2$，在葉節點的邊上填$3$，這樣就可以保證滿足條件。所以答案為$1$或$3$。
考慮最大值，由于我們填的數可以無限大， 所以考慮對于每個邊都先分配一個不同的值，假設有$k$個葉子連接在一個點上，那么我們需要減去$k?1$，因為這些點的邊都需要填一樣的值。
這個過程可以直接求，也可以$dp$來求。
復雜度$O(N)$。

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,tune=native") //#pragma GCC optimize(2) #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<map> #include<cmath> #include<cctype> #include<vector> #include<set> #include<queue> #include<algorithm> #include<sstream> #include<ctime> #include<cstdlib> #define X first #define Y second #define L (u<<1) #define R (u<<1|1) #define pb push_back #define mk make_pair #define Mid (tr[u].l+tr[u].r>>1) #define Len(u) (tr[u].r-tr[u].l+1) #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 pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f; const double eps=1e-6;int n; int d[N],cd[N],a[N]; int f[N]; vector<int>v[N]; bool flag;void dfs1(int u,int fa,int d) {for(auto x:v[u]) {if(x==fa) continue;dfs1(x,u,d+1);}if(v[u].size()==1) {if(d%2==1) flag=1;} }void dfs2(int u,int fa) {int add=0;for(auto x:v[u]) {if(x==fa) continue;dfs2(x,u);if(v[x].size()==1) add=1;else f[u]+=f[x]+1;}f[u]+=add; }int solve1() {for(int i=1;i<=n;i++) if(d[i]==1) { dfs1(i,0,0); break; }if(flag) return 3;else return 1; }int solve2() {int ans=n-1;for(int i=1;i<=n;i++) {if(d[i]==1) continue; int cnt=0;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 i=1;i<=n-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; } /**/

樹形$dp$

// 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,tune=native") //#pragma GCC optimize(2) #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<map> #include<cmath> #include<cctype> #include<vector> #include<set> #include<queue> #include<algorithm> #include<sstream> #include<ctime> #include<cstdlib> #define X first #define Y second #define L (u<<1) #define R (u<<1|1) #define pb push_back #define mk make_pair #define Mid (tr[u].l+tr[u].r>>1) #define Len(u) (tr[u].r-tr[u].l+1) #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 pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f; const double eps=1e-6;int n; int d[N],cd[N],a[N]; int f[N]; vector<int>v[N]; bool flag;void dfs1(int u,int fa,int d) {for(auto x:v[u]) {if(x==fa) continue;dfs1(x,u,d+1);}if(v[u].size()==1) {if(d%2==1) flag=1;} }void dfs2(int u,int fa) {int add=0;for(auto x:v[u]) {if(x==fa) continue;dfs2(x,u);if(v[x].size()==1) add=1;else f[u]+=f[x]+1;}f[u]+=add; }int solve1() {for(int i=1;i<=n;i++) if(d[i]==1) { dfs1(i,0,0); break; }if(flag) return 3;else return 1; }int solve2() {for(int i=1;i<=n;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 i=1;i<=n-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; } /**/

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