HDU 5723 Abandoned country (16多校第一场)

题意

给你一个图,让你求最小生成树,并且在最小生成树中求每条边被走的期望,
n个点m条边$( 1 \leq n \leq 10000, 1\leq m \leq 100000)$

分析

  • 其实这题很水,只是我在补题的时候出现了一个致命bug,先是u = find(u); v = find(v);判断是不是一个块,如果是再加边,可是这里我至今addedge(u,v),u和v是祖先值已经改变了

代码

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/*************************************************************************
> File Name: aa.cpp
> Author: liangxianfeng
> Mail: liangxianfeng96@qq.com
> Created Time: 2016年07月22日 星期五 21时49分25秒
************************************************************************/


#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<cmath>
using namespace std;
#define ll long long
#define rep(i,n) for(int i =0; i < n; ++i)
#define CLR(x) memset(x,0,sizeof x)
#define MEM(a,b) memset(a,b,sizeof a)
#define pr(x) //cout << #x << " = " << x << " ";
#define prln(x) //cout << #x << " = " << x << endl;
const int maxn = 1e6+100;
const int INF = 0x3f3f3f3f;
struct Edge{
int u, v, c;
bool operator <(const Edge& rhs)const{
return c < rhs.c;
}
}edge[maxn];
int head[maxn], nxt[maxn], to[maxn], cost[maxn];
int edgenum ;
ll n, m;
int p[maxn];
void init(){
memset(head, -1, sizeof head);
for(int i = 0; i <= n; ++i){
p[i] = i;
}
edgenum = 0;
}
void addedge(int u, int v, int c){
nxt[edgenum] = head[u];
to[edgenum] = v;
cost[edgenum] = c;
head[u] = edgenum++;
}
int find(int x){
return p[x] = (x==p[x]?x:find(p[x]));
}
int siz[maxn];
bool vis[maxn];
void dfs(int u){
vis[u] = true;
siz[u] = 1;
for(int i = head[u]; ~i; i = nxt[i]){
int v = to[i];
if(vis[v]) continue;
dfs(v);
siz[u] += siz[v];
}
}
bool is[maxn];
void solve(){
sort(edge, edge + m);
memset(vis, 0, sizeof vis);
memset(is, 0, sizeof is);
ll ans1 = 0;
double ans2 = 0;
for(int i = 0; i < m; ++i){
int u = edge[i].u;
int v= edge[i].v;
pr(u);prln(v);
u = find(u); v = find(v);
if(u == v) continue;
p[u] = v;
is[i] = true;
//pr(u);pr(v);prln(edge[i].c);
ans1 += edge[i].c;
addedge(edge[i].u, edge[i].v, edge[i].c);
addedge(edge[i].v, edge[i].u, edge[i].c);
}
dfs(1);
double ans = 0;
for(int i = 0; i < m; ++i)
{
if(!is[i]) continue;
int u = edge[i].u;
int v= edge[i].v;
int mi = min(siz[u], siz[v]);
pr(u);pr(v);pr(siz[u]);prln(siz[v]);
prln(mi);
ans += 1.0*(ll)(n-mi)*(ll)(mi)*(ll)edge[i].c/(1.0*n*(n-1));
}
ans2 = ans*2;
printf("%lld %.2f\n", ans1, ans2);

}
int main(){
#ifdef LOCAL
freopen("/home/zeroxf/桌面/in.txt","r",stdin);
//freopen("/home/zeroxf/桌面/out.txt","w",stdout);
#endif
int t;
scanf("%d", &t);
while(t--){
scanf("%lld%lld", &n, &m);
init();
int u, v, c;
for(int i = 0; i < m; ++i){
scanf("%d%d%d", &u, &v, &c);
edge[i].u = u;
edge[i].v = v;
edge[i].c = c;
pr(u);pr(v);prln(c);
}
solve();
}
return 0;
}