hdu 3938 并查集

思路：这题的出题人是不是语文不行啊，题目意思说不清楚。

知道是求存在路径的点对数后，用并查集每次记录集合中点的数目，很容易就解决了。

#include<set>

#include<cmath>

#include<queue>

#include<cstdio>

#include<vector>

#include<string>

#include<cstdlib>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pb push_back

#define mp make_pair

#define Maxn 200100

#define Maxm 800002

#define LL __int64

#define Abs(x) ((x)>0?(x):(-x))

#define lson(x) (x<<1)

#define rson(x) (x<<1|1)

#define inf 1000000000

#define lowbit(x) (x&(-x))

#define Mod 1000000007

using namespace std;

int fa[Maxn],cnt;

LL num[Maxn],ans[Maxn];

struct Edge{

    int u,v,val;

    int operator< (const Edge &temp) const

    {

        return val<temp.val;

    }

}p[Maxm];

struct Query{

    int val,id;

    int operator< (const Query &temp) const{

        return val<temp.val;

    }

}q[Maxn];

void init()

{

    for(int i=0;i<Maxn;i++){

        fa[i]=i;

        ans[i]=0;

        num[i]=1;

    }

}

int find(int x)

{

    if(x!=fa[x])

        fa[x]=find(fa[x]);

    return fa[x];

}

LL merg(int a,int b)

{

    int x=find(a);

    int y=find(b);

    if(x==y) return 0;

    LL ans=num[x]*num[y];

    num[x]+=num[y];

    fa[y]=x;

    ++cnt;

    return ans;

}

int main()

{

    int n,m,t,i,j,u,v;

    while(scanf("%d%d%d",&n,&m,&t)!=EOF){

        init();

        for(i=1;i<=m;i++){

            scanf("%d%d%d",&p[i].u,&p[i].v,&p[i].val);

        }

        for(i=1;i<=t;i++)

            scanf("%d",&q[i].val),q[i].id=i;

        sort(p+1,p+1+m);

        sort(q+1,q+1+t);

        p[m+1].val=inf;

        p[m+1].u=1,p[m+1].v=1;

        j=1;

        LL sum=0;

        cnt=0;

        int pos=1;

        for(i=1;i<=m+1;i++){

            while(j<=t&&q[j].val<p[i].val){

                ans[q[j].id]=sum;

                j++;

            }

            sum+=merg(p[i].u,p[i].v);

        }

        while(j<=t&&q[j].val<p[i].val)

        {

            ans[q[j].id]=sum;

                j++;

        }

        for(i=1;i<=t;i++){

            printf("%I64d\n",ans[i]);

        }

    }

    return 0;

}