hdu5627 Clarke and MST (并查集)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 311    Accepted Submission(s): 173

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

14 51 2 51 3 31 4 22 3 13 4 7

 

Sample Output

1

 

题意：给你n个点m条边以及m条边的权值，让你构造一棵生成树，使得生成树的权值and和最大。

思路：我们可以贪心地枚举二进制中的每一位，从30到0，然后判断有该位的值为1的所有权值中能不能形成一棵生成树，如果有，那么这些权值的边为可选边，看能不能组成下一位。

#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<string>#include<algorithm>using namespace std;typedef long long ll;#define inf 99999999#define pi acos(-1.0)#define maxn 300050struct node{    int x,y,w;}e[maxn];int ok[maxn],pre[maxn],ran[maxn];void makeset(int x){    pre[x]=x;    ran[x]=0;}int findset(int x){    int i,j=x,r=x;    while(r!=pre[r])r=pre[r];    while(j!=pre[j]){        i=pre[j];        pre[j]=r;        j=i;    }    return r;}int main(){    int n,m,i,j,T,t;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&m);        for(i=1;i<=m;i++){            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);        }        for(i=1;i<=m;i++)ok[i]=1;        int sum=0;        for(t=30;t>=0;t--){            for(i=1;i<=n;i++){                makeset(i);            }            int ans=n;            for(i=1;i<=m;i++){                if(ok[i] && (e[i].w&(1<<t) ) ){                    int x=findset(e[i].x);                    int y=findset(e[i].y);                    if(x==y)continue;                    ans--;                    if(ran[x]>ran[y]){                        pre[y]=x;                    }                    else{                        pre[x]=y;                        if(ran[x]==ran[y])ran[y]++;                    }                }            }            if(ans==1){                sum|=(1<<t);                for(i=1;i<=m;i++){                    if(ok[i] && (e[i].w&(1<<t) )){                        ok[i]=1;                    }                    else ok[i]=0;                }            }        }        printf("%d\n",sum);    }    return 0;}