HDU 5627 (生成树)

Clarke and MST

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 282    Accepted Submission(s): 157

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)

题意是需要找到一棵生成树使得树上的&和最大。

拆位，从最大位开始依次判断这个位上为1的那些边能不能组成一棵ST，如果可以就更新边集，复杂度O(32*n)。

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <time.h>
#include <vector>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 311111
#define maxm 611111

struct node {
	int u, v;
	long long w;
};
vector <node> edge[2];
int n, m;


int fa[maxn];
int find (int x) {
	return fa[x] == x ? fa[x] : fa[x] = find (fa[x]);
}
bool ok (int id) { //判断能不能构成一个联通分量
	for (int i = 1; i <= n; i++)
		fa[i] = i;
	int Max = edge[id].size ();
	for (int i = 0; i < Max; i++) {
		int u = edge[id][i].u, v = edge[id][i].v;
		int p1 = find (u), p2 = find (v);
		if (p1 != p2)
			fa[p1] = p2 ;
	}
	int gg = find (1);
	for (int i = 2; i <= n; i++) {
		if (find (i) != gg)
			return 0;
	}
	return 1;
}

bool is_1 (long long num, int id) { //判断num的id位是不是1
	id--;
	num >>= id;
	return (num&1);
}

bool work (int id) { //判断id位能否为1
	int Max = edge[0].size ();
	for (int i = 0; i < Max; i++) {
		if (is_1 (edge[0][i].w, id)) {
			edge[1].push_back (edge[0][i]);
		}
	}
	if (ok (1)) {
		edge[0].clear ();
		int Max = edge[1].size ();
		for (int i = 0; i < Max; i++) {
			edge[0].push_back (edge[1][i]);
		}
		return 1;
	}
	return 0;
}

int main () {
	int t;
	scanf ("%d", &t);
	while (t--) {
		scanf ("%d%d", &n, &m);
		edge[0].clear (); edge[1].clear ();
		for (int i = 1; i <= m; i++) {
			int u, v; long long w;
			scanf ("%d%d%lld", &u, &v, &w);
			edge[0].push_back ((node){u, v, w});
		}
		if (!ok (0)) {
			printf ("0\n");
			continue;
		}
		long long ans = 0; 
		for (int i = 32; i >= 1; i--) { 
			edge[1].clear ();
			if (work (i)) { 
				ans <<= 1;
				ans++;
			}
			else 
				ans <<= 1;
		}
		printf ("%lld\n", ans);
	}
	return 0;
}