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=10005;
/*for point*/
struct node{
    int a,b,c;
    bool operator <(const node &bn)const {
       return c<bn.c;
    }
}sac[maxn*5];
/*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=0;i<=n;i++){
        father[i]=i;
        rank[i]=1;
    }
}

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=0;
    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=0;i<m;i++){
        scanf("%d%d%d",&sac[i].a,&sac[i].b,&sac[i].c);
     }

     for(int i=0;i<q;i++){
       scanf("%d",&qq[i].val);
        qq[i].id=i;
     }
     sort(sac,sac+m);
     sort(qq,qq+q);
     int i,j,res=0;
     for(j=0,i=0;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=0;i<q;i++){
            printf("%d\n",key[i]);
     }
  }
  //  system("comp");

   return 0;
}