HDU 5627 Clarke and MST &意义下最大生成树 贪心

Clarke and MST

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5627

Description

Clarke is a patient with multiple personality disorder. One day he turned into a learner of graph theory.
He learned some algorithms of minimum spanning tree. Then he had a good idea, he wanted to find the maximum spanning tree with bit operation AND.
A spanning tree is composed by n−1 edges. Each two points of n points can reach each other. The size of a spanning tree is generated by bit operation AND with values of n−1 edges.
Now he wants to figure out the maximum spanning tree.

Input

The first line contains an integer T(1≤T≤5), the number of test cases.
For each test case, the first line contains two integers n,m(2≤n≤300000,1≤m≤300000), denoting the number of points and the number of edge respectively.
Then m lines followed, each line contains three integers x,y,w(1≤x,y≤n,0≤w≤109), denoting an edge between x,y with value w.
The number of test case with n,m>100000 will not exceed 1.

Output

For each test case, print a line contained an integer represented the answer. If there is no any spanning tree, print 0.

Sample Input

1
4 5
1 2 5
1 3 3
1 4 2
2 3 1
3 4 7

Sample Output

1

Hint

题意

让你求&意义下的最大生成树

题解：

贪心，我们从高位到低位贪心，如果含有这一位的边能够构成一棵树的话，我们就可以直接把其他不含有这一位的边全部去掉

然后重复这个行为

这个贪心显然正确啦，至于判断能否构成一颗树，就用并查集就好啦

代码

#include<algorithm>
#include<stdio.h>
#include<iostream>
using namespace std;
const int maxn = 3e5+6;
int x[maxn],y[maxn];
int w[maxn];
int flag[maxn];
int fa[maxn];
int fi(int x)
{
    return x == fa[x]?x:fa[x]=fi(fa[x]);
}
int uni(int x,int y)
{
    int p = fi(x),q = fi(y);
    if(p != q)
        fa[p]=fa[q];
}
int main()
{
    int t;scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        int n,m;scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
            scanf("%d%d%d",&x[i],&y[i],&w[i]);
        int ans = 0;
        for(int i=30;i>=0;i--)
        {
            for(int j=1;j<=n;j++)
                fa[j]=j;
            for(int j=1;j<=m;j++)
            {
                if(((w[j]&ans)==ans)&&(w[j]>>i&1))
                    flag[j]=1;
                else
                    flag[j]=0;
            }
            for(int j=1;j<=m;j++)
                if(flag[j])
                    uni(x[j],y[j]);
            int p = fi(1);
            int f = 1;
            for(int j=1;j<=n;j++)
                if(fi(j)!=p)
                    f = 0;
            if(f)
                ans|=(1<<i);
        }
        cout<<ans<<endl;
    }
}