AND Minimum Spanning Tree（二进制）HDU - 6614

 

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.

InputThe 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).OutputFor 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

题意：

有N个结点，两结点之间边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）

分析：看到这道题目的时候当时一脸懵逼，最小生成树的权值之和好求，但是输出相连接结点的话。。。还是规律。。

二进制的10011最优的连接结点无疑是00100（二进制），，本质就是找到第一个非1的位置，其他位置都为0就OK.

但是看一个特例：111的最有连接结点是1000，但是如果N最大为7（十进制），那就没有8（二进制1000)来进行配对了，因此111只能被迫与1进行配对。

总结下来：第一行最小生成树的权值之和只能是1或者0，然后特例只要判断n+1是否为2的幂指数就可以。

AC代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 2e5+10;
#define LL long long
#define INF 0x3f3f3f3f
int T;
int a[maxn];
bool judge(int x){
    int k=x&(x-1);
    if(!k)return true;
    return false;
}
int ft(int x){
    int ans=1;
    while(1){
        int k=x&1;
        x>>=1;
        if(!k) break;
        ans<<=1;
    }
    return ans;
}

int main(){
    scanf("%d",&T);
    while(T--){
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        if(judge(n+1)) printf("1\n");
        else printf("0\n");
        for(int i=2;i<=n;i++){
            int k=ft(i);
            if(k>n) printf("1");
            else printf("%d",k);
            if(i==n) printf("\n");
            else printf(" ");
        }
    }

    return 0;
}