2015区域赛长春网络赛 Travel

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1528    Accepted Submission(s): 541

Problem Description

Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are  n  cities and  m  bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is  x  minutes, how many pairs of city  (a,b)  are there that Jack can travel from city  a  to  b  without going berserk?

 

Input

The first line contains one integer  T,T≤5 , which represents the number of test case.

For each test case, the first line consists of three integers  n,m  and  q  where  n≤20000,m≤100000,q≤5000 . The Undirected Kingdom has  n  cities and  m bidirectional roads, and there are  q  queries.

Each of the following  m  lines consists of three integers  a,b  and  d  where  a,b∈{1,...,n}  and  d≤100000 . It takes Jack  d  minutes to travel from city  a  to city  b and vice versa.

Then  q  lines follow. Each of them is a query consisting of an integer  x  where  x  is the time limit before Jack goes berserk.

 

Output

You should print  q  lines for each test case. Each of them contains one integer as the number of pair of cities  (a,b)  which Jack may travel from  a  to  b  within the time limit  x .

Note that  (a,b)  and  (b,a)  are counted as different pairs and  a  and  b  must be different cities.

 

Sample Input

   
   
   
   

    
    
    
    1
5 5 3
2 3 6334
1 5 15724
3 5 5705
4 3 12382
1 3 21726
6000
10000
13000
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    2
6
12
   
   
   
   

 

比赛的时候没有看出来是并查集，结果没有做出来。。。

用并查集做：

先对边进行排序储存，从小到大；再对询问从小到大进行排序储存。

然后每次一个询问，进行边的合并和联通分量的合并，计算出每次合并增加的点对数。。。其实就是

c(n1+n2)-(c(n1,2)+c(n2,2))*2=n1*n2*2;

代码如下：

#include<iostream>
#include<fstream>
#include<set>
#include<stack>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;

typedef long long LL;
const int MAX=0xfffffff;

const int mx=22000;
struct Node{
    int id;
    int value;
    int num;
    bool operator<(const Node&a)const{
        return value<a.value;
    }
}node[mx];
struct Edge{
    int u,v,w;
    bool operator<(const Edge &a)const{
        return w<a.w;
    }
}edge[mx*5];

int fa[mx],num[mx];
int _find(int x)
{
    return fa[x]==x?x:fa[x]=_find(fa[x]);
}
void _union(int x,int y)
{
    x=_find(x),y=_find(y);
    if(y<x) swap(x,y);
    fa[y]=x;
    num[x]+=num[y];
}
bool cmp2(Node a,Node b)
{
    return a.id<b.id;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m,q;
        scanf("%d %d %d",&n,&m,&q);

        for(int i=1;i<=n;i++)
        fa[i]=i,num[i]=1;

        for(int i=0;i<m;i++)
        {
            scanf("%d %d %d",&edge[i].u,&edge[i].v,&edge[i].w);
        }
        sort(edge,edge+m);
        for(int i=0;i<q;i++)
        {
            scanf("%d",&node[i].value);
            node[i].id=i;
            node[i].num=0;
        }
        sort(node,node+q);
        int j=0;
        int ans=0;
        for(int i=0;i<q;i++)
        {
            while(j<m&&edge[j].w<=node[i].value)
            {
                int u=_find(edge[j].u);
                int v=_find(edge[j].v);
                j++;
                if(u==v)  continue;
                ans+=2*num[u]*num[v];
                _union(u,v);
            }
            node[i].num=ans;
        }
        sort(node,node+q,cmp2);
        for(int i=0;i<q;i++)
            printf("%d\n",node[i].num);
    }
    return 0;
}