hdu 3948 Portal (kusral+离线)

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 931    Accepted Submission(s): 466

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

 

题目意思:

            给定一张无向图，要你求出满足小于或等于给定边长的最多边长的种类数目。当然关键是有n次询问。

     思路：

               对于一颗有n条路径的无向图，可以知道，当与另一个m条路径的无向图合并为一个全新的无向图时： 此时的路径为 n*m;

               当然这题的重点不在这里，此题关键是有q次询问，对于每一次询问，若我们都去重新求解一次，将会很花时间，无疑TLE倒哭。%>_<%

               于是采用离线处理（这里是开了解题报告才想起来，这么巧妙的地方，果然是too yong 哇！ ）：

      离线的思路为：   先将询问的权值从小到大排序，由于大的权值包含了小的权值，于是整个过程，居然只需要一次就搞定。  俺是乡下人，第一次见识到这点，吓尿了！╮(╯▽╰)╭

  代码：

 #define LOCAL
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #include<cstdlib>
 using namespace std;
 const int maxn=;
 /*for point*/
 struct node{
     int a,b,c;
     bool operator <(const node &bn)const {
        return c<bn.c;
     }
 }sac[maxn*];
 /*for query*/
 struct query{
    int id,val;
    bool operator <(const query &bn)const {
        return val<bn.val;
    }
 }qq[maxn];

 int father[maxn],rank[maxn];
 int key[maxn];

 void init(int n){
     for(int i=;i<=n;i++){
         father[i]=i;
         rank[i]=;
     }
 }

 int fin(int x){
   while(x!=father[x])
          x=father[x];
   return x;
 }

 int unin(int x,int y)
 {
    x=fin(x);
    y=fin(y);
    int ans=;
     if(x!=y){
        ans=rank[x]*rank[y];
      if(rank[x]<rank[y]){
          rank[y]+=rank[x];
          father[x]=y;
      }
     else{
           rank[x]+=rank[y];
          father[y]=x;
     }
     }
    return ans;
 }

 int main()
 {
     #ifdef LOCAL
        freopen("test.in","r",stdin);
        freopen("test1.out","w",stdout);
     #endif
   int n,m,q;
   while(scanf("%d%d%d",&n,&m,&q)!=EOF)
   {
         init(n);
        for(int i=;i<m;i++){
         scanf("%d%d%d",&sac[i].a,&sac[i].b,&sac[i].c);
      }

      for(int i=;i<q;i++){
        scanf("%d",&qq[i].val);
         qq[i].id=i;
      }
      sort(sac,sac+m);
      sort(qq,qq+q);
      int i,j,res=;
      for(j=,i=;i<q;i++){
          while(j<m&&sac[j].c<=qq[i].val){
             res+=unin(sac[j].a,sac[j].b);
             ++j;
         }
         key[qq[i].id]=res;
      }
      for(i=;i<q;i++){
             printf("%d\n",key[i]);
      }
   }
   //  system("comp");

    return ;
 }