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  • hdu 5627 Clarke and MST(最大 生成树)

    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 n1 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 n1 edges.  Now he wants to figure out the maximum spanning tree.
     
    Input
    The first line contains an integer T(1T5), the number of test cases.  For each test case, the first line contains two integers n,m(2n300000,1m300000), denoting the number of points and the number of edge respectively. Then m lines followed, each line contains three integers x,y,w(1x,yn,0w109), 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
     


    首先贴上自己的写法,虽然不是很正宗的做法

     1 #pragma comment(linker, "/STACK:1024000000,1024000000")
     2 #include<iostream>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<cmath>
     6 #include<math.h>
     7 #include<algorithm>
     8 #include<queue>
     9 #include<set>
    10 #include<bitset>
    11 #include<map>
    12 #include<vector>
    13 #include<stdlib.h>
    14 #include <stack>
    15 using namespace std;
    16 #define PI acos(-1.0)
    17 #define max(a,b) (a) > (b) ? (a) : (b)  
    18 #define min(a,b) (a) < (b) ? (a) : (b)
    19 #define ll long long
    20 #define eps 1e-10
    21 #define MOD 1000000007
    22 #define N 300006
    23 #define M 300006
    24 #define inf 1e12
    25 struct Node{
    26     int x,y;
    27     int cost;
    28 }edge[M];
    29 int n,m;
    30 int fa[N];
    31 void init(){
    32     for(int i=0;i<N;i++){
    33         fa[i]=i;
    34     }
    35 }
    36 int find(int x){
    37     return fa[x]==x?x:fa[x]=find(fa[x]);
    38 }
    39 bool cmp(Node a,Node b){
    40     return a.cost>b.cost;
    41 }
    42 int main()
    43 {
    44     int t;
    45     scanf("%d",&t);
    46     while(t--){
    47         scanf("%d%d",&n,&m);
    48         init();
    49         for(int i=0;i<m;i++){
    50             int a,b,c;
    51             scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].cost);
    52         }
    53         sort(edge,edge+m,cmp);
    54         int flag=1;
    55         int ans;
    56         int num=n-1;
    57         for(int i=0;i<m;i++){
    58             int root1=find(edge[i].x);
    59             int root2=find(edge[i].y);
    60             if(root1!=root2){
    61                 if(flag){
    62                     ans=edge[i].cost;
    63                     flag=0;
    64                 }else{
    65                     ans&=edge[i].cost;
    66                 }
    67                 fa[root1]=root2;
    68                 num--;
    69             }
    70         }
    71         if(num!=0){
    72             printf("0
    ");
    73         }
    74         else{
    75             printf("%d
    ",ans);
    76         }
    77     }
    78         
    79     
    80     return 0;
    81 }
    View Code


    官方题解:

     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <string.h>
     4 #include <algorithm>
     5 using namespace std;
     6 const int N =  300000 + 10;
     7 
     8 struct Edge{
     9     int from,to,dis;
    10 }a[N],b[N];
    11 int fa[N];
    12 int find(int x){
    13     if(x==fa[x]) return x;
    14     return fa[x] = find(fa[x]);
    15 }
    16 int tmp;
    17 bool solve(int pos, Edge *a, int n, int m){
    18     for(int i=1;i<=n;++i)
    19         fa[i] = i;
    20     int cnt = 0;
    21     tmp = 0;
    22     for(int i=1;i<=m;++i){
    23         if(((a[i].dis>>pos)&1)==0)
    24             continue;
    25         
    26         int fu = find(a[i].from);
    27         int fv = find(a[i].to);
    28         if(fu!=fv){
    29             if(cnt==0)
    30                 tmp = a[i].dis;
    31             else
    32                 tmp &= a[i].dis;
    33             
    34             fa[fu] = fv;
    35             cnt++;
    36             if(cnt==n-1)
    37                 return true;
    38         }
    39     }
    40     return false;
    41 }
    42 int main() {
    43     
    44     int t,n,m;
    45     scanf("%d",&t);
    46     while(t--){
    47         scanf("%d%d",&n,&m);
    48         
    49         for(int i=1;i<=m;++i){
    50             scanf("%d%d%d",&a[i].from,&a[i].to,&a[i].dis);
    51         }
    52         int ans = 0;
    53         for(int i=30;i>=0;--i){
    54             if(solve(i,a,n,m)){
    55                 ans = tmp;
    56                 int mm = 0;
    57                 for(int i=1;i<=m;++i){
    58                     if((a[i].dis>>i)&1)
    59                         b[++mm] = a[i];
    60                 }
    61                 m = mm;
    62                 for(int i=1;i<=m;++i)
    63                     a[i] = b[i];
    64             }
    65         }
    66         cout<<ans<<endl;
    67     }
    68     return 0;
    69 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/5188332.html
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