hdu 5441 (并查集)

題意：給你n個點，m條邊構成無向圖。q個詢問，每次一個值，求有多少條路，路中的邊權都小于這個值

a->b 和 b->a算兩種

思路：把權值從小到大排序，詢問從小到大排序，如果相連則用并查集相連形成聯通塊

x個點可以形成：x * （x - 1）

如果新增的路使兩個聯通塊和并則數量 增長了：

（num[1]+num[2]）×（num[1]+num[2]-1) - num[1] × (num[1]-1) - num[2] ×(num[2]-1)

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
int T,n,m,k;
int num[20005],par[20005],p[20005];struct node
{int u,v,w;bool operator<(const node&a)const{return w < a.w;}
} pnode[100005];struct term
{int id,we;bool operator<(const term&a)const{return we<a.we;}
} te[20005];int fin(int x)
{return x == par[x]?  x : par[x] = fin(par[x]);
}void merg(int x,int y)
{int x1 = fin(x);int x2 = fin(y);if(x1 < x2){par[x2]= x1;num[x1] += num[x2];}else{par[x1] = x2;num[x2] += num[x1];}
}int main()
{scanf("%d",&T);while(T--){scanf("%d%d%d",&n,&m,&k);for(int i = 0; i <= n; i++){par[i] = i;num[i] = 1;}for(int i = 0; i < m; i++)scanf("%d%d%d",&pnode[i].u,&pnode[i].v,&pnode[i].w);sort(pnode,pnode+m);for(int i = 0; i < k; i++){te[i].id = i;scanf("%d",&te[i].we);}sort(te,te+k);int tt = 0;ll ans = 0;for(int i = 0; i < k; i++){while(tt < m &&  pnode[tt].w <= te[i].we ){int u = fin(pnode[tt].u);int v = fin(pnode[tt].v);tt++;if(u == v)continue;ans += (num[u]+num[v])*(num[u]+num[v]-1)-num[u]*(num[u]-1) - num[v]*(num[v]-1);merg(u,v);}p[te[i].id] = ans;}for(int i = 0;i <k;i++)printf("%d\n",p[i]);}return 0;
}