Boruvka求最小生成树

作者: 亖呉?盀
来源: 51数据库
2021-07-28

$因为牛客多校我第一次见到这个算法：算法大致过程就是因为牛客多校我第一次见到这个算法：
算法大致过程就是$
对于每个连通块，每次找出一条最小的出边
然后把两个连通块合并
每次可以是连通块个数减半
然后就得到了一个O((n+m)log n)的做法

代码模板：

#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <limits.h>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <set>
// #define mid ((l + r) >> 1) 
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define count Count
#define pb push_back
#define f first
#define s second
using namespace std;
const int N = 4e5+10, mod = 1e9 + 9;
const long double eps = 1e-5;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
typedef pair<double,double> PDD;
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
template<typename T, typename... Args> void read(T &first, Args& ... args) 
{
    read(first);
    read(args...);
}
struct node {
    int to, next, w;
}e[N], mi[N];
int fa[N];
int head[N], cnt;
int n,m;
inline int find(int x)
{
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}

inline void add(int from, int to, int w)
{
    e[cnt] = {to,head[from],w};
    head[from] = cnt ++;
}

inline int Borucka()
{
    int ans = 0, edg = 0;
    while(1)
    {
        ms(mi,INF);
        for(int i = 1; i <= n; ++ i)
          for(int j = head[i]; ~j; j = e[j].next)
          {
              int v = e[j].to, w = e[j].w;
              if(find(v) != find(i) && w < mi[find(i)].w)
              {
                  mi[find(i)].w = w, mi[find(i)].to = v;
              }
          }
          int F = 0;
          for(int i = 1; i <= n; ++ i)
          {
              int v = mi[i].to, w = mi[i].w;
              if(mi[i].to != mi[0].to && find(i) != find(v)) 
                 fa[find(v)] = find(i), ans += w, edg ++, F = 1;
          }
          if(!F) break;
    }
    if(edg == n - 1) return ans;
    else return -1;
}

int main()
{
    read(n,m);
    ms(head,-1);
    for(int i = 0; i <= n + 1; ++ i) fa[i] = i;
    for(int i = 0; i < m; ++ i)
    {
        int l, r, w;
        read(l,r,w);
        add(l,r,w);
        add(r,l,w);
    }
    int ans = Borucka();
    if(ans == -1) cout << "orz";
    else cout << ans << endl;
}