【杭电OJ3938】【离线+并查集】

http://acm.hdu.edu.cn/showproblem.php?pid=3938

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1921    Accepted Submission(s): 955

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

题目大意：给一个图，然后会有Q次询问，询问路径上最大权值的边的权值小于 L的路径数。

题目分析：

【1】由于Q的范围是10^4，而如果一个一个查找的话，会TLE。而观察由于每次询问并未确定对路径的起点或者终点，所以每次询问的小L值可以累加得到询问时的大L的值，也就是每次询问的查找是有重叠的，所以就可以将Q次询问的L值进行从小到大的排序，小L慢慢累加得到大L，避免了不必要的重复查询

【解题的第一步-询问存起来等待离线】

【2】由于所连接的边的权值必须不大于L，所以也要将所有边进行排序

【解题第二步-->对边进行排序，等待查询是否能够连接】

【3】而由于是求路径数，两个图【一个顶点数为A,一个为B】连成一个图之后所能增加的路径数是A*B，所以可以利用并查集判断需要连接的两个图是否已经连接，如果没有则进行连接，并且路径数增加A*B，且根节点记录的图的顶点数增加A（或者B）

【解题最关键的一步，从小到大连接边，并更新L[i]的路径数，其中利用了两个连通图连接之后路径数增加A*B的原理】

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 struct edge{
     int to1;
     int to2;
     int lrn;
 }EDGE[];
 struct qury{
     int id;
     int x;
 }qwq[];
 int n,m,q;
 int pre[],num[];;
 int find(int x)
 {

     int xx=x;
     while(x!=pre[x])
     {
         x=pre[x];
     }
     while(pre[xx]!=x)
     {
         int t=pre[xx];
         pre[xx]=x;
         xx=t;
     }
     return x;
 }
 bool cmp1(struct edge orz1,struct edge orz2)
 {
     return orz1.lrn<orz2.lrn;
 }
 bool cmp2(struct qury orz3,struct qury orz4)
 {
     return orz3.x<orz4.x;
 }
 int main()
 {
     while(scanf("%d%d%d",&n,&m,&q)==)
     {
     for(int i =  ; i <= n ; i++)
     {
         pre[i]=i;
         num[i]=;
     }
     for(int i =  ; i < m ; i++)
     {
         scanf("%d%d%d",&EDGE[i].to1,&EDGE[i].to2,&EDGE[i].lrn);
     }
     sort(EDGE,EDGE+m,cmp1);
     for(int i = ; i < q ; i++)
     {
         qwq[i].id=i;
         scanf("%d",&qwq[i].x);
     }
     sort(qwq,qwq+q,cmp2);
     int j=;
     long long sum=;
     long long ans[];
     memset(ans,,sizeof(ans));
     for(int i =  ; i < q ; i++)
     {
         for(;j<m;j++)
         {
             if(EDGE[j].lrn>qwq[i].x)break;
             int wqw1=find(EDGE[j].to1);
             int wqw2=find(EDGE[j].to2);
             if(wqw1!=wqw2)
             {
                 pre[wqw1]=wqw2;
                 sum+=num[wqw1]*num[wqw2];
                 num[wqw2]+=num[wqw1];
             }
         }
         ans[qwq[i].id]=sum;

     }
     for(int i =  ;i < q ; i++)
     {
         printf("%lld\n",ans[i]);
     }
 }
     return ;
  }