HDU 5627Clarke and MST

Clarke and MST

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 315    Accepted Submission(s): 176

Problem 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

从大到小按位枚举。因为是&  所以把边集中这个位为1的都拿出来，看这些边能不能构成生成树。具体 看代码。（这里用的并查集判断是不是树，也可以用bfs或dfs判断）

/* ***********************************************
Author        :guanjun
Created Time  :2016/2/16 19:02:06
File Name     :bc72c.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 300000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
int fa[maxn],w[maxn],p[maxn],q[maxn],n,m;
int findfa(int x){
    if(x==fa[x])return x;
    return fa[x]=findfa(fa[x]);
}
void Union(int a,int b){
    int x=findfa(a);
    int y=findfa(b);
    if(x>y)fa[x]=y;
    else if(y>x)fa[y]=x;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t,x,y,z;
    cin>>t;
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&x,&y,&z);
            w[i]=z;
            p[i]=x;
            q[i]=y;
        }
        int ans=0;
        for(int i=30;i>=0;i--){
            x=(1<<i);
            for(int ii=1;ii<=n;ii++)fa[ii]=ii;
            for(int j=0;j<m;j++){
                if((w[j]&x)&&((w[j]&ans)==ans)){
                    //cout<<p[j]<<" "<<q[j]<<endl;
                    Union(p[j],q[j]);
                }
            }
            int f=fa[1];
            int mark=1;
            for(int ii=1;ii<=n;ii++){
                if(fa[ii]!=f){
                    mark=0;break;
                }
            }
            if(mark)ans+=x;
        }
        printf("%d\n",ans);
    }
    return 0;
}