多校 HDU - 6614 AND Minimum Spanning Tree （二进制）

传送门

AND Minimum Spanning Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 378    Accepted Submission(s): 190

Problem Description

You are given a complete graph with N vertices, numbered from 1 to N. 
The weight of the edge between vertex x and vertex y (1<=x, y<=N, x!=y) is simply the bitwise AND of x and y. Now you are to find minimum spanning tree of this graph.

 

Input

The first line of the input contains an integer T (1<= T <=10), the number of test cases. Then T test cases follow. Each test case consists of one line containing an integer N (2<=N<=200000).

 

 

Output

For each test case, you must output exactly 2 lines. You must print the weight of the minimum spanning tree in the 1st line. In the 2nd line, you must print N-1 space-separated integers f2, f3, … , fN, implying there is an edge between i and fi in your tree(2<=i<=N). If there are multiple solutions you must output the lexicographically smallest one. A tree T1 is lexicographically smaller than tree T2, if and only if the sequence f obtained by T1 is lexicographically smaller than the sequence obtained by T2.

 

Sample Input

2 3 2

 

Sample Output

1 1 1 0 1

Hint

In the 1st test case, w(1, 2) = w(2, 1) = 0, w(1, 3) = w(3, 1) = 1, w(2, 3) = w(3, 2) = 2. There is only one minimum spanning tree in this graph, i.e. {(1, 2), (1, 3)}. For the 2nd test case, there is also unique minimum spanning tree.

 

Source

2019 Multi-University Training Contest 4

 

 

题意：

有N个结点，两两之间的权值为两个点取与

比如说结点2和结点3之间的权值 W(2,3) = 10&11 = 10

问整个图的最小生成树

第一行输出最小生成树的权值和

第二行分别输出2到N节点连接的点（如果一个节点连接了2个点的话，需要输出小的那个）

思路：

举个例子：
10011（二进制）最好和哪个结点连接呢
无疑是 00100（二进制）
感受一下？
其实本质就是找到第一个非1的位置，其他的所有位置都置为0就OK。

这里有个特殊情况
如果是111？（二进制）
那么最优的连接结点就是1000。
但是如果N恰好就是7，那么就没有1000（十进制是8）来和111（十进制是7）组队了。
111只能去和1连接（被迫）
所以第一行答案要么是1要么是0

判断一下N是不是二的幂次减1就行
第二行的话可以按照上面的规律去找，即找到第一个非1的位置为1，其他的都为0

#include"bits/stdc++.h"
using namespace std;
typedef long long LL;
int a[200011];
bool judge(int x){
    int n = x&(x-1);
    return n == 0;
}
/*
找到二进制中第一个不是0的
比如说   1001
返回的是 0010
1111返回的是10000
*/
int getMin(int x){
    int ans = 1; 
    while(true){
        int k = x&1;
        x >>= 1;
        if(k == 0){
            break;
        }
        ans <<= 1;
    }
    return ans;
} 
int main()
{
     int _;
    for(scanf("%d",&_);_--;){
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(int)*(n+1));
        judge(n+1)?puts("1"):puts("0");
        for(int i = 2 ; i <= n ; i++){
            int Min = getMin(i);
            Min > n? printf("1"):printf("%d",Min);//判断一下找到的数是不是大于n
            i == n ? printf("\n"):printf(" ");
        }
    } 
}
/*
2
3
2
*/