最小生成树kruskal模版

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prim算法适合稠密图，其时间复杂度为O(n^2)，其时间复杂度与边的数目无关，

而kruskal算法的时间复杂度为O(eloge)跟边的数目有关，适合稀疏图。

kruskal : 按权排,每次取权最小且不构成回路的加入,直到取完n-1条边 (用并查集解决不构成回路)

/*hdu 1233 最小生成树kruskal*/#include <cstdio>  #include <cstring>  #include <cstdlib>  #include <string>  #include <iostream>  #include <algorithm>  #include <sstream>  #include <math.h>  #include <queue>  #include <stack>  #include <vector>  #include <deque>  #include <set>  #include <map>  #include <time.h>;#define cler(arr, val)    memset(arr, val, sizeof(arr))  #define IN     freopen ("in.txt" , "r" , stdin);  #define OUT  freopen ("out.txt" , "w" , stdout);  using namespace std;typedef long long  ll;const int MAXN = 100010;//点数的最大值  const int MAXM = 20006;//边数的最大值  const int INF = 0x3f3f3f3f;const int mod = 10000007;int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }const int maxn = 100 + 5;int fa[maxn];int n;struct edges {int x, y, d;}e[maxn*(maxn-1)/2];int cmp(edges a1, edges a2) {return a1.d < a2.d;}void init() {for (int i = 1; i <= n; i++) fa[i] = i;}int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}void unite(int x, int y){int tx = find(x), ty = find(y);if (tx != ty)fa[tx] = ty;}int kruskal(int num){init();int ans = 0;int sum = 0;//加进去的边数for (int i = 0; i < num; i++) //num条边{if (find(e[i].x) != find(e[i].y)){ans += e[i].d;unite(e[i].x, e[i].y);sum++;}if (n - 1 == sum)return ans;}}int main(){while (scanf("%d", &n) && n){if (n == 1){puts("0");continue;}int bian = n*(n - 1) / 2;for (int i = 0; i < bian; i++)scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].d);sort(e, e + bian, cmp);printf("%d\n", kruskal(bian));}return 0;}

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