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  • CF982C Cut 'em all! DFS 树 * 二十一

     Cut 'em all!
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    You're given a tree with nn 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 nn (1n1051≤n≤105) denoting the size of the tree.

    The next n1n−1 lines contain two integers uu, vv (1u,vn1≤u,v≤n) each, describing the vertices connected by the ii-th edge.

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

    Output

    Output a single integer kk — the maximum number of edges that can be removed to leave all connected components with even size, or 1−1 if it is impossible to remove edges in order to satisfy this property.

    Examples
    input
    Copy
    4
    2 4
    4 1
    3 1
    output
    Copy
    1
    input
    Copy
    3
    1 2
    1 3
    output
    Copy
    -1
    input
    Copy
    10
    7 1
    8 4
    8 10
    4 7
    6 5
    9 3
    3 5
    2 10
    2 5
    output
    Copy
    4
    input
    Copy
    2
    1 2
    output
    Copy
    0
    Note

    In the first example you can remove the edge between vertices 11 and 44. 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 1−1.

    #include <map>
    #include <set>
    #include <stack>
    #include <cmath>
    #include <queue>
    #include <cstdio>
    #include <vector>
    #include <string>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #define debug(a) cout << #a << " " << a << endl
    using namespace std;
    const int maxn = 1e5 + 10;
    const int mod = 1e9 + 7;
    typedef long long ll;
    int hd[maxn], ne[maxn*2], to[maxn*2], num, n, siz[maxn], ans;
    void add( int x, int y ) {
        to[++num]=y, ne[num]=hd[x], hd[x]=num;
    }
     
    void dfs( int x, int fa ) {
        siz[x]=1;
        for( int i = hd[x]; i; i = ne[i] ) {
            if(to[i]!=fa){
                dfs(to[i],x);
                siz[x]+=siz[to[i]];
            }
        }
        if(!(siz[x]&1)) 
            siz[x]=0,ans++;
    }
     
    int main(){
        std::ios::sync_with_stdio(false);
        scanf("%d",&n);
        int uu, vv;
        for( int i = 1; i < n; i ++ ) {
            scanf("%d%d",&uu,&vv);
            add(uu,vv), add(vv,uu);
        }
        if( n & 1 ) { 
            puts("-1"); 
            return 0;
        }
         
        dfs( 1, -1 ), ans--;
         
        printf("%d
    ", ans );
        return 0;
    }
    彼时当年少,莫负好时光。
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  • 原文地址:https://www.cnblogs.com/l609929321/p/9249990.html
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