HDU 5627Clarke and MST
阅读原文时间:2023年07月09日阅读:2

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#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 << ; const double eps=1e-; 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=;i=;i--){
x=(<<i);
for(int ii=;ii<=n;ii++)fa[ii]=ii;
for(int j=;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[];
int mark=;
for(int ii=;ii<=n;ii++){
if(fa[ii]!=f){
mark=;break;
}
}
if(mark)ans+=x;
}
printf("%d\n",ans);
}
return ;
}

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