HDU 3938 Portal(并查集构建MST+离线处理)*

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2063    Accepted Submission(s): 1008

 

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

 

 

Source

2011 Multi-University Training Contest 10 - Host by HRBEU

 

 

Recommend

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#pragma comment(linker, "/STACK:102400000,102400000")
#include
using namespace std;

#define debug puts("YES");
#define rep(x,y,z) for(int (x)=(y);(x)<(z);(x)++)

#define lrt int l,int r,int rt
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define ll long long

const int  maxn =1e5+5;
const int mod=1e9+7;
/*
题目大意：就是说两个点之间如果要建路的话，
权重是多个路径中的最长边的最小值。

要注意到生成树可以保证所有点连通，
那么最小生成树可以维护这样的集合，
边一定选取的都是最小的，
就是说用最小的代价边来维护连通区域。

这道题目明显应该离线查询，
对查询的权值从小到大排序，
然后对每个查询值不断的通过Krustal并查集方法扩展，
两个连通块的合并对答案的贡献是两个连通块大小的乘积。

至此，答案可以迭代出来。

*/

int n,m,q;
int ans;
///并查集结构
int pre[maxn],sum[maxn];
int Find(int x){return pre[x]==x?x:pre[x]=Find(pre[x]);}
void Union(int x,int y)
{
    int fx=Find(x),fy=Find(y);
    if(fx==fy) return ;///同一节点的就不比了
    ans+=sum[fx]*sum[fy];
    if(sum[fx]>sum[fy])
    {
        pre[fy]=fx;
        sum[fx]+=sum[fy];
    }
    else
    {
        pre[fx]=fy;
        sum[fy]+=sum[fx];
    }
    return ;
}
///边节点和查询节点
struct node{int x,y,z;};///公用
node edge[maxn],query[maxn];
bool cmp1(node p,node q)
{
    return p.z