Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 977    Accepted Submission(s): 490

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

这题挺不错的，学到了离线算法。题目要求两点之间最大边的最小值小于某值的个数。其实就是求点对。

集合划分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
using namespace std;
int n,m,q,set[50010],num[50010],ans[50010];
struct node
{
      int x,y,w;
}e[50010];
struct nod
{
      int x,id;
}aa[50010];
bool cmp(node a,node b)
{
      return a.w<b.w;
}
bool cmp1(nod a,nod b)
{
      return a.x<b.x;
}
int find(int x)
{
      if(x!=set[x])
            set[x]=find(set[x]);
      return set[x];
}
int Union(int a,int b)
{
      int fx,fy;
      fx=find(a),fy=find(b);
      if(fx==fy) return 0;
      int t=num[fx]*num[fy];
      num[fx]+=num[fy];
      num[fy]=0;
      set[fy]=fx;
      return t;
}
int main()
{
      while(scanf("%d%d%d",&n,&m,&q)!=EOF)
      {
           memset(e,0,sizeof(e));
           memset(ans,0,sizeof(ans));
           for(int i=1;i<=n;i++)
                  set[i]=i,num[i]=1;
           for(int i=1;i<=m;i++)
           {
                 scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].w);
           }
           sort(e+1,e+1+m,cmp);
           for(int i=1;i<=q;i++)
           {
                 scanf("%d",&aa[i].x);
                 aa[i].id=i;
           }
           sort(aa+1,aa+1+q,cmp1);
           int k=1,cnt=0;
           for(int i=1;i<=q;i++)
           {
                 while(k<=m&&e[k].w<=aa[i].x)
                 {
                       cnt+=Union(e[k].x,e[k].y);
                       k++;
                 }
                 ans[aa[i].id]=cnt;
           }
           for(int i=1;i<=q;i++)
                  printf("%d\n",ans[i]);
      }
      return 0;
}