51nod 1640最小生成树&最大生成树

思路：首先对全图求一次最小生成树，最出来生成树的自大边便是所有情况中最小的（因为kruscal和prime每次选边都是选从小选到大），然后再从最大的边从大到小开始枚举最大生成树，求出权值和即可。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=3e5+50;
int n,m,u,v,w,r[maxn],fa[maxn],t,max_edge;
long long cost;

struct edge{
int from,to,weight;
}e[maxn];

bool cmp_min(edge a,edge b){
return a.weight<b.weight;
}

bool cmp_max(edge a,edge b){
return a.weight>b.weight;
}

void init(){
cost=0;
memset(r,0,sizeof(r));
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 a,int b){
a=find(a);
b=find(b);
if(a==b) return;
if(r[a]>r[b]) fa[b]=a;
else{
fa[a]=b;
if(r[a]==r[b]) r[a]++;
}
}

void MIN(){
init();
sort(e+1,e+1+m,cmp_min);
for(int i=1;i<=m;i++){
if(find(e[i].from)!=find(e[i].to)){
unite(e[i].from,e[i].to);
max_edge=max(max_edge,e[i].weight);
}
}
}

long long MAX(){
init();
sort(e+1,e+1+m,cmp_max);
for(int i=1;i<=m;i++){
if(e[i].weight>max_edge) continue;
if(find(e[i].to)!=find(e[i].from)){
unite(e[i].to,e[i].from);
cost+=e[i].weight;
}
}
return cost;
}

int main(){
//freopen("out.txt","w",stdout);
while(scanf("%d%d",&n,&m)!=EOF){
memset(e,0,sizeof(e));
max_edge=0;
for(int i=1;i<=m;i++)
scanf("%d%d%d",&e[i].from,&e[i].to,&e[i].weight);
MIN();
printf("%lld\n",MAX());
}
return 0;
}