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=;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=;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=;
         int ans;
         int num=n-;
         for(int i=;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=;
                 }else{
                     ans&=edge[i].cost;
                 }
                 fa[root1]=root2;
                 num--;
             }
         }
         if(num!=){
             printf("0\n");
         }
         else{
             printf("%d\n",ans);
         }
     }

     return ;
 }

官方题解：

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

 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=;i<=n;++i)
         fa[i] = i;
     int cnt = ;
     tmp = ;
     for(int i=;i<=m;++i){
         if(((a[i].dis>>pos)&)==)
             continue;

         int fu = find(a[i].from);
         int fv = find(a[i].to);
         if(fu!=fv){
             if(cnt==)
                 tmp = a[i].dis;
             else
                 tmp &= a[i].dis;

             fa[fu] = fv;
             cnt++;
             if(cnt==n-)
                 return true;
         }
     }
     return false;
 }
 int main() {

     int t,n,m;
     scanf("%d",&t);
     while(t--){
         scanf("%d%d",&n,&m);

         for(int i=;i<=m;++i){
             scanf("%d%d%d",&a[i].from,&a[i].to,&a[i].dis);
         }
         int ans = ;
         for(int i=;i>=;--i){
             if(solve(i,a,n,m)){
                 ans = tmp;
                 int mm = ;
                 for(int i=;i<=m;++i){
                     if((a[i].dis>>i)&)
                         b[++mm] = a[i];
                 }
                 m = mm;
                 for(int i=;i<=m;++i)
                     a[i] = b[i];
             }
         }
         cout<<ans<<endl;
     }
     return ;
 }