标签：pid 中间 size lse div 技术分享 基础 ble int
第1行：2个数N,M中间用空格分隔，N为点的数量，M为边的数量。（2 <= N <= 1000, 1 <= M <= 50000) 第2 - M + 1行：每行3个数S E W，分别表示M条边的2个顶点及权值。(1 <= S, E <= N，1 <= W <= 10000)
输出最小生成树的所有边的权值之和。
9 14 1 2 4 2 3 8 3 4 7 4 5 9 5 6 10 6 7 2 7 8 1 8 9 7 2 8 11 3 9 2 7 9 6 3 6 4 4 6 14 1 8 8
37
就是最简单的求最小生成树，用并查集
/* data:2018.05.10 author:gsw link:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1212 account:2684524947@qq.com */ #define ll long long #define IO ios::sync_with_stdio(false); #include<iostream> #include<algorithm> #include<math.h> #include<stdio.h> #include<string.h> using namespace std; class Edge { public: int a,b,c; }; Edge edge[50005]; int pre[1005];int n,m; bool camp(Edge aa,Edge bb) { return aa.c<bb.c; } int Find(int x) { int r=x; while(r!=pre[r]) r=pre[r]; int i=x,j; while(pre[i]!=r) { j=pre[i]; pre[i]=r; i=j; } return r; } void init() { memset(edge,0,sizeof(edge)); for(int i=1;i<=n;i++) pre[i]=i; } int mix(int x,int y) { //cout<<x<<" y="<<y<<endl; int fx=Find(x),fy=Find(y); //cout<<fx<<" fy="<<fy<<endl; if(fx!=fy) { pre[fy]=fx; return 1; } return 0; } ll kusal() { sort(edge,edge+m,camp); ll ans=0;int js=0; for(int i=0;i<m;i++) { if(mix(edge[i].a,edge[i].b)) { //cout<<"ans="<<ans<<endl; ans+=edge[i].c; js++; } if(js==(n-1))break; } return ans; } int main() { scanf("%d%d",&n,&m); init(); int a,b,c; for(int i=0;i<m;i++) { scanf("%d%d%d",&a,&b,&c); edge[i].a=a;edge[i].b=b;edge[i].c=c; } cout<<kusal()<<endl; return 0; }
标签：pid 中间 size lse div 技术分享 基础 ble int
原文地址：https://www.cnblogs.com/fantastic123/p/9021895.html
