hdu 5627 Clarke and MST（最大 生成树）

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

 

 

Source

BestCoder Round #72 (div.2)

 

首先贴上自己的写法，虽然不是很正宗的做法

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)  
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 300006
#define M 300006
#define inf 1e12
struct Node{
    int x,y;
    int cost;
}edge[M];
int n,m;
int fa[N];
void init(){
    for(int i=0;i<N;i++){
        fa[i]=i;
    }
}
int find(int x){
    return fa[x]==x?x:fa[x]=find(fa[x]);
}
bool cmp(Node a,Node b){
    return a.cost>b.cost;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        init();
        for(int i=0;i<m;i++){
            int a,b,c;
            scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
        }
        sort(edge,edge+m,cmp);
        int flag=1;
        int ans;
        int num=n-1;
        for(int i=0;i<m;i++){
            int root1=find(edge[i].x);
            int root2=find(edge[i].y);
            if(root1!=root2){
                if(flag){
                    ans=edge[i].cost;
                    flag=0;
                }else{
                    ans&=edge[i].cost;
                }
                fa[root1]=root2;
                num--;
            }
        }
        if(num!=0){
            printf("0\n");
        }
        else{
            printf("%d\n",ans);
        }
    }
        
    
    return 0;
}

官方题解：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N =  300000 + 10;

struct Edge{
    int from,to,dis;
}a[N],b[N];
int fa[N];
int find(int x){
    if(x==fa[x]) return x;
    return fa[x] = find(fa[x]);
}
int tmp;
bool solve(int pos, Edge *a, int n, int m){
    for(int i=1;i<=n;++i)
        fa[i] = i;
    int cnt = 0;
    tmp = 0;
    for(int i=1;i<=m;++i){
        if(((a[i].dis>>pos)&1)==0)
            continue;
        
        int fu = find(a[i].from);
        int fv = find(a[i].to);
        if(fu!=fv){
            if(cnt==0)
                tmp = a[i].dis;
            else
                tmp &= a[i].dis;
            
            fa[fu] = fv;
            cnt++;
            if(cnt==n-1)
                return true;
        }
    }
    return false;
}
int main() {
    
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        
        for(int i=1;i<=m;++i){
            scanf("%d%d%d",&a[i].from,&a[i].to,&a[i].dis);
        }
        int ans = 0;
        for(int i=30;i>=0;--i){
            if(solve(i,a,n,m)){
                ans = tmp;
                int mm = 0;
                for(int i=1;i<=m;++i){
                    if((a[i].dis>>i)&1)
                        b[++mm] = a[i];
                }
                m = mm;
                for(int i=1;i<=m;++i)
                    a[i] = b[i];
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}