POJ1861&ZOJ1542--Network【最小生成树】

来源：互联网发布：数据库limit的用法编辑：程序博客网时间：2024/06/11 20:50

链接：http://poj.org/problem?id=1861

最小生成树裸题，输出生成树的最长边、节点个数、节点坐标。另外OJ上样例输出时错的，4个点的最小生成树怎么可能4条边。。

主要是熟悉手写kruskal

#include<cstring>#include<string>#include<fstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cctype>#include<algorithm>#include<queue>#include<map>#include<set>#include<vector>#include<stack>#include<ctime>#include<cstdlib>#include<functional>#include<cmath>using namespace std;#define PI acos(-1.0)#define MAXN 20010#define eps 1e-7#define INF 0x7FFFFFFF#define seed 131#define ll long long#define ull unsigned ll#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct node{    int u,v,dis;}edge[MAXN];int father[1010],point[1010][2];int n,m,ans,sum;bool cmp(node x,node y){    return x.dis<y.dis;}int find(int x){    int t = x;    while(t!=father[t]){        t = father[t];    }    int k = x;    while(k!=t){        int temp = father[k];        father[k] = t;        k = temp;    }    return t;}void kruskal(){    int i,j=0;    for(i=1;i<=n;i++)   father[i] = i;    for(i=0;i<m;i++){        int a = find(edge[i].u);        int b = find(edge[i].v);        if(a!=b){            father[a] = b;            point[j][0] = edge[i].u;            point[j][1] = edge[i].v;            sum+=edge[i].dis;            ans = edge[i].dis;            j++;            if(j>=n-1)  break;        }    }}int main(){    int i,j;    while(scanf("%d%d",&n,&m)!=EOF){        sum = 0;        for(i=0;i<m;i++){            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].dis);        }        sort(edge,edge+m,cmp);        kruskal();        printf("%d\n%d\n",ans,n-1);        for(i=0;i<n-1;i++){            printf("%d %d\n",point[i][0],point[i][1]);        }    }    return 0;}

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