cf628 div2

https://codeforces.com/contest/13

 

#include <bits/stdc++.h>

using namespace std;
#define int long long
const int maxn = 1e5 + 5;
int n;
int u, v;
vector<int> g[maxn];
int ans[maxn];

signed main() {
    //freopen("in", "r", stdin);
    ios::sync_with_stdio(0);
    cin >> n;

    for (int i = 1; i < n; i++) {
        cin >> u >> v;
        g[u].push_back(i);
        g[v].push_back(i);
        ans[i] = -1;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (g[i].size() >= 3) {
            for (int j = 0; j < 3; j++) {
                ans[g[i][j]] = j;
            }
            cnt = 3;
            break;
        }
    }

    for (int i = 1; i < n; i++) {
        if (ans[i] == -1)
            ans[i] = cnt++;
    }
    for (int i = 1; i < n; i++)
        cout << ans[i] << endl;
    return 0;
}

这是一道图论的构造题，有时候不必纠结样例一样不，比如说第二个样例的输出结果

 给了n 个点，n - 1条边，每条边的两个端点u,v 找u,v对应的最大的最小

n 个顶点的树是n--1条边 

如果是链的话，这几个数字怎么排都可以

如果不是的话，节点的度= 3的时候，排0,1,2

>3的话，就变成任意链了，自己画一下就知道了

 

 

 

#include <bits/stdc++.h>
using namespace std;
#define int long long
int u,v,x;
signed main(){
    //freopen("in","r",stdin);
    ios::sync_with_stdio(0);//用这个不能用"
"
    cin >> u >> v;
   // cout << u << v << endl;
    if(v < u) cout << "-1";
    else if(v == u) {
        if(u) cout << 1 << endl;
        cout << u;//当=0时,最后一组杨样例给的是0
    }
    else{
        x = (v - u) / 2;
        if((v - u) & 1) cout << "-1";//这里是(v-u)，不是x
        else{
            if((x & u) == 0) {
                cout << 2 << endl;
                cout << u + x << " " << x;
            }
            else {
                cout << 3 << endl;
                cout << u << " " << x << " " << x;
            }
        }
    }
    return 0;
}