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C. Cut 'em all!time limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

You're given a tree with?$n$vertices.

Your task is to determine the maximum possible number of edges that can be removed?

in such a way that all the remaining connected components will have even size.

Input

The first line contains an integer?$n$?($1\le n\le {10}^5$) denoting the size of the tree.

The next?$n?1$?lines contain two integers?$u$,?$v$?($1\le u,v\le n$) each, describing the vertices connected by the?$i$-th edge.

It's guaranteed that the given edges form a tree.

Output

Output a single integer?$k$?— the maximum number of edges that can be removed to leave all connected components with even size, or?$?1$

if it is impossible to remove edges in order to satisfy this property.

ExamplesinputCopy4 2 4 4 1 3 1 outputCopy1inputCopy3 1 2 1 3 outputCopy-1inputCopy10 7 1 8 4 8 10 4 7 6 5 9 3 3 5 2 10 2 5 outputCopy4inputCopy2 1 2 outputCopy0 Note

In the first example you can remove the edge between vertices?$1$and?$4$.?

The graph after that will have two connected components with two vertices in each.

In the second example you can't remove edges in such a way that all components have even number of vertices,?

so the answer is?$?1$

.

題目描述：

? ? 給你一個有n個結(jié)點的樹，其中有n-1條邊，詢問你最多你可以移動多少條邊，使得每個強連通分量的個數(shù)均為偶數(shù)。

題目分析：

? ? 因為我們知道，如果所給的結(jié)點的個數(shù)是奇數(shù)，那么，因為奇數(shù)必定是由一個偶數(shù)和一個奇數(shù)組成，故總會有一個強連通分量的個數(shù)為奇數(shù)，因此當(dāng)n為奇數(shù)使，答案為-1。

? ? 那么當(dāng)n等于偶數(shù)的時，我們只需要進行dfs遍歷整張圖，統(tǒng)計一下某個結(jié)點以及其兒子結(jié)點的數(shù)量，如果遞歸到最底時統(tǒng)計出的數(shù)量為偶數(shù)，則答案+1即可。

#include<bits/stdc++.h> using namespace std; const int maxn=2e5+5; struct edge {int to,next; }q[maxn]; int head[maxn]; int size[maxn]; int cnt=0; int ans=0; void init() {memset(head,-1,sizeof(head));cnt=0; } void add_edge(int from,int to) {q[cnt].next=head[from];q[cnt].to=to;head[from]=cnt++; } void dfs(int x,int fa) {size[x]=1;for(int i=head[x];i!=-1;i=q[i].next){if(q[i].to==fa) continue;dfs(q[i].to,x);size[x]+=size[q[i].to];}if(size[x]%2==0){ans++;size[x]=0;} } int main() {int n,u,v;scanf("%d",&n);init();for(int i=0;i<n-1;i++){scanf("%d%d",&u,&v);add_edge(u,v);add_edge(v,u);}if(n&1){printf("-1\n");return 0;}dfs(1,-1);printf("%d\n",ans-1);return 0; }

#include<bits/stdc++.h> using namespace std; const int maxn=1e5+5; vector<int>G[maxn]; int son[maxn]; int n,ans; void dfs(int x,int fa) {son[x]=1;for(int i=0;i<G[x].size();i++){int k=G[x][i];if(k!=fa){dfs(k,x);son[x]+=son[k];}}if(son[x]%2==0){ans++;son[x]=0;} } int main() {ans=0;memset(son,0,sizeof(son));scanf("%d",&n);for(int i=0;i<n-1;i++){int a,b;scanf("%d%d",&a,&b);G[a].push_back(b);G[b].push_back(a);}if(n%2==1){printf("-1\n");return 0;}dfs(1,-1);printf("%d\n",ans-1);return 0; }

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