3938 HDU Portal（离线型并查集）

Problem Description ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.  
Input There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).  
Output Output the answer to each query on a separate line.  
Sample Input

10 10 10
7 2 1
6 8 3
4 5 8
5 8 2
2 8 9
6 4 5
2 1 5
8 10 5
7 3 7
7 8 8
10
6
1
5
9
1
8
2
7
6

 
Sample Output

36
13
1
13
36
1
36
2
16
13

/*
离线并查集，边按小到大排序，询问也按小到大排序；
对于询问x,答案就是询问x-1的值加上询问x-1的L1和询问x的L2之间的边合并带来的值
对于一条属于L1和L2之间的边，如果端点u，v在一个集合中，忽略，否则新开的路是u
所在集合的大小乘上v所在集合的大小
*/
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,m,q,ans[10010];
int hav[50010],father[50010];
struct node
{
    int u,v,w;
}edge[50010];

int cmp1(node a,node b)
{
    return a.w<b.w;
}

struct que
{
    int num,v;
}qu[10010];

int cmp2(que a,que b)
{
    return a.v<b.v;
}

void init()
{
    int i;
    for(i=0;i<m;i++)
        scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].w);
    for(i=1;i<=n;i++)
    {
        father[i]=i;
        hav[i]=1;
    }
    for(i=1;i<=q;i++)
    {
        scanf("%d",&qu[i].v);
        qu[i].num=i;
    }
    sort(edge,edge+m,cmp1);
    sort(qu+1,qu+q+1,cmp2);
}

int find(int i)
{
    int x,y,j;
    x=i;
    while(x!=father[x]) x=father[x];
    y=i;
    while(y!=father[y])
    {
        j=father[y];
        father[y]=x;
        y=j;
    }
    return x;
}

void unin(int u,int v)
{
    father[u]=v;
    hav[v]+=hav[u];
}
void solve()
{
    int i,j,u,v;
    j=0;
    ans[0]=0;
    for(i=1;i<=q;i++)
    {
        ans[qu[i].num]=ans[qu[i-1].num];
        while(edge[j].w<=qu[i].v&&j<m)
        {
            u=find(edge[j].u);
            v=find(edge[j].v);
            if(u!=v)
            {
                ans[qu[i].num]+=hav[u]*hav[v];
                unin(u,v);
            }
            j++;
        }
    }
    for(i=1;i<=q;i++)
    {
        printf("%d\n",ans[i]);
    }
}
int main()
{
    while(scanf("%d %d %d",&n,&m,&q)!=EOF)
    {
        init();
        solve();
    }
    return 0;
}