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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, print a line contained an integer represented the answer. If there is no any spanning tree, print 0.

[align=left]Sample Input[/align]

1

4 5

1 2 5

1 3 3

1 4 2

2 3 1

3 4 7

[align=left]Sample Output[/align]

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;
}