c/c++语言开发共享洛谷P3366 【模板】最小生成树(Boruvka算法)

题意

题目链接

sol

自己yy着写了一下boruvka算法。

算法思想很简单，就是每次贪心的用两个联通块之间最小的边去合并。

复杂度(o(n log n))，然鹅没有kruskal跑的快，但是好像在一类生成树问题上很有用

#include<bits/stdc++.h>  #define pair pair<int, int>  #define fi first #define se second #define pb push_back #define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? eof : *p1++) char buf[(1 << 22)], *p1 = buf, *p2 = buf; using namespace std; const int maxn = 5001; inline int read() {     char c = getchar(); int x = 0, f = 1;     while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}     while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();     return x * f; } int n, m; namespace dsu {     int fa[maxn], siz[maxn];     void init(int n) {         for(int i = 1; i <= n; i++) fa[i] = i, siz[i] = 1;     }     int find(int x) {         return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);     }     void unionn(int x, int y) {         int fx = find(x), fy = find(y);         if(siz[fx] > siz[fy]) swap(fx, fy);         fa[fx] = fy; siz[fy] += siz[fx];     } }; using namespace dsu; vector<pair> v[maxn]; int link[maxn], val[maxn]; void boruvka() {     init(n); int ans = 0;     bool flag;     do {         flag = 0;         memset(link, -1, sizeof(link));         memset(val, 0x3f, sizeof(val));         for(int x = 1; x <= n; x++) {             int fx = find(x);             for(auto &tmp : v[x]) {                 int to = tmp.fi, w = tmp.se, fy = find(to);                 if(fx == fy || (w > val[fx])) continue;                 link[fx] = fy; val[fx] = w;             }         }         for(int x = 1; x <= n; x++) {             int fx = find(x);             if((~link[fx]) && find(fx) != find(link[fx]))                  unionn(fx, link[fx]), ans += val[fx], flag = 1;          }        }while(flag);     int f1 = find(1);     for(int i = 2; i <= n; i++) if(find(i) != f1) return (void) puts("orz");     cout << ans; } signed main() {     n = read(); m = read();     for(int i = 1; i <= m; i++) {         int x = read(), y = read(), w = read();         v[x].push_back({y, w});         v[y].push_back({x, w});     }     boruvka(); }