HDOJ 5627 Clarke and MST （位运算最大生成树kruskul）

Clarke and MST

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 266    Accepted Submission(s): 152

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 ofn−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：给你一个图，有边权，构建一棵生成树，使得边权经过and(&)运算后结果最大，求这棵生成树最后的结果，不存在输出0

思路：用kruskul方法，排序时按照边权含有1的数量排序，优先添加1多的边，然后过程中计算就好了



ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)using namespace std;int gcd(int a,int b){return b?gcd(b,a%b):a;}LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}//headstruct s{    int a,b;    int num;    int v;}p[MAXN];ll ans;int pri[MAXN];bool cmp(s aa,s bb){    return aa.num>bb.num;}int find(int x){    int r=x;    while(r!=pri[r])    r=pri[r];    int i=x,j;    while(i!=r)    {        j=pri[i];        pri[i]=r;        i=j;    }    return r;}void connect(int xx,int yy,int k){    int nx=find(xx);    int ny=find(yy);    if(nx!=ny)    {        pri[nx]=pri[ny];        ans=ans&k;        //printf("a=%d b=%d\n",xx,yy);        //printf("kkk:%I64d\n",ans);    }}int main(){    int t,i;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        for(i=1;i<=n;i++)        pri[i]=i;        for(i=0;i<m;i++)        {            int x,y,z;            scanf("%d%d%d",&x,&y,&z);            int cnt=0;            int zz=z;            while(z)            {                z&=(z-1);                cnt++;            }            p[i].a=x;p[i].b=y;p[i].num=cnt;p[i].v=zz;        }        sort(p,p+m,cmp);        ans=p[0].num;        pri[p[0].a]=p[0].b;        for(i=1;i<m;i++)        {            //printf("...%d\n",p[i].num);            connect(p[i].a,p[i].b,p[i].v);        }        int ccnt=0;        for(i=1;i<=n;i++)        if(pri[i]==i)        ccnt++;        if(ccnt>1)        printf("0\n");        else        printf("%I64d\n",ans);    }    return 0;}