HDU 5441

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1300    Accepted Submission(s): 469

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 mbidirectional 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 band 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

15 5 32 3 63341 5 157243 5 57054 3 123821 3 2172660001000013000

 

Sample Output

2612

 

//打表  由于有多少个联通块不知道 所有利用并查集记录根的个数

#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#include <map>#include <queue>using namespace std;const int maxn=100000+10;const int inf=(1<<30);int pre[maxn];struct Node{    int to,from,v;}p[maxn];bool cmp(Node a,Node b){    return a.v<b.v;}int fin(int x){    int r=x;    while(pre[r]!=r)        r=pre[r];    int k=x;    while(pre[k]!=r)    {        int j=pre[k];        pre[k]=r;        k=j;    }    return r;}int dp[maxn];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m,q;        memset(pre,0,sizeof(pre));        memset(dp,0,sizeof(dp));        map <int,int> ma;        scanf("%d%d%d",&n,&m,&q);        for(int i=1;i<=n;i++)            pre[i]=i;        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&p[i].to,&p[i].from,&p[i].v);        }        sort(p,p+m,cmp);        int cnt=0;        for(int i=1;i<maxn;i++)//打表        {            dp[i]=dp[i-1];  //每次的            if(cnt==m)continue;            if(i>=p[cnt].v)            {                int fx=fin(p[cnt].to);                int fy=fin(p[cnt].from);                if(fx!=fy)    //当两个根不同时分三种情况                {                    if(ma[fx]==0&&ma[fy]==0)  //如果两个点都是第一次出现 结果加上2                    {                        pre[fx]=fy;                        ma[fy]+=2;                        dp[i]+=2;                    }                    else if(ma[fx]==0||ma[fy]==0)  //一个未出现过                    {                        if(ma[fx]>ma[fy])                        {                            dp[i]+=2*ma[fx];                            pre[fy]=fx;                            ma[fx]++;                        }                        else                        {                            dp[i]+=2*ma[fy];                            pre[fx]=fy;                            ma[fy]++;                        }                    }                    else  //两个都出现过 两个联通块合并                    {                        dp[i]+=2*ma[fx]*ma[fy];                        pre[fx]=fy;                        ma[fy]+=ma[fx];                        ma[fx]=0;                    }                }                cnt++;            }        }        for(int i=0;i<q;i++)        {            int res;            scanf("%d",&res);            printf("%d\n",dp[res]);        }    }    return 0;}