HDU 5627Clarke 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

edges.
Now he wants to figure out the maximum spanning tree.

Input

The first line contains an integer

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

4 5

1 2 5

1 3 3

1 4 2

2 3 1

3 4 7

Sample Output

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;
}