[Ybt 1349] 最优布线问题 ——最小生成树[Kruskal]

Kruskal算法

基本思想: 并查集+贪心
操作
首先从小到大排序【贪心】 
看最小的边，如果这边的两个结点不是同一个根，【用并查集操作】，则可以合并为一棵新树

Ybt 1349

#include<bits/stdc++.h>using namespace std;struct RoadType{int GoTop,ToTop,Weight;bool operator < (const RoadType &x) const{return Weight<x.Weight;} }Road[10005];int Father[1005];void Init(int Many)//并查集初始化 {for (int i=1; i<=Many; i++) Father[i]=i;}int Find(int Top)//并查集中寻找父节点 {Father[Top]=Father[Top]==Top?Top:Find(Father[Top]);return Father[Top];}void Union(int A, int B)//并查集中合并操作 {Father[Find(A)]=Find(B);}int main(){int Many,Un,RMany=0,Choose=0,C_Top=0,Sum=0;cin >> Many;for (int i=1; i<=Many; i++){for (int j=1; j<=i; j++) cin >> Un;for (int j=i+1; j<=Many; j++){cin >> Road[++RMany].Weight;Road[RMany].GoTop=i;Road[RMany].ToTop=j;}}sort(Road+1,Road+1+RMany);//贪心最基本操作: 排序 Init(Many);while (Choose < Many-1 && C_Top <= RMany)//一个n节点的图，最小生成树，只需要选择n-1条边，自己可以画图想一想 {C_Top++;if (Find(Road[C_Top].GoTop) != Find(Road[C_Top].ToTop))//如果不是同一根【即不是同一棵树】，合并 {Union(Road[C_Top].GoTop,Road[C_Top].ToTop);Sum+=Road[C_Top].Weight; Choose++;}}cout << Sum;}