【HDU - 5627】Clarke and MST（最大生成树，与运算性质，最小生成树MST变形）

题干：

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−1n−1 edges. Each two points of nn points can reach each other. The size of a spanning tree is generated by bit operation AND with values of n−1n−1 edges.
Now he wants to figure out the maximum spanning tree.

Input

The first line contains an integer T(1≤T≤5)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)n,m(2≤n≤300000,1≤m≤300000), denoting the number of points and the number of edge respectively.
Then mm lines followed, each line contains three integers x,y,w(1≤x,y≤n,0≤w≤109)x,y,w(1≤x,y≤n,0≤w≤109), denoting an edge between x,yx,y with value ww.
The number of test case with n,m>100000n,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

Sample Output

题目大意：

给一棵n个点m条边的无向图，问你最大与运算生成树多少权值是多少，如果不连通输出0。

解题报告：

贪心不难发现其实就是求最大生成树。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 6e5 + 5;
struct Edge {int u,v;ll w;
} e[MAX];
int n,m;
bool cmp(Edge a,Edge b) {return a.w > b.w;
}
int f[MAX];
int getf(int v) {return f[v] == v ? v : f[v] = getf(f[v]);
}
void merge(int u,int v) {int t1 = getf(u),t2 = getf(v);f[t2] = t1;
}
int main()
{int t;cin>>t;while(t--) {scanf("%d%d",&n,&m);for(int i = 1; i<=n; i++) f[i] = i;for(int a,b,i = 1; i<=m; i++) {ll w;scanf("%d%d%lld",&a,&b,&w);e[i].u = a;e[i].v = b;e[i].w = w;}sort(e+1,e+m+1,cmp);ll ans = 1LL<<62;ans--;ll no = ans;
//      cout <<ans <<endl;
//      while(ans) {
//          printf("%lld",ans%2);
//          ans>>=1;
//      }int cnt = 0;for(int i = 1; i<=m; i++) {if(getf(e[i].u) != getf(e[i].v)) {ans = ans & e[i].w;merge(e[i].u,e[i].v);cnt++;}}if(cnt != n-1) printf("0\n");else printf("%lld\n",ans);} return 0 ;
}

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【HDU - 5627】Clarke and MST（最大生成树，与运算性质，最小生成树MST变形）

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