HDU 5441 2015年长春icpc网络赛 并查集

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 4204    Accepted Submission(s): 1408

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 aren 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

 

Source

2015 ACM/ICPC Asia Regional Changchun Online

题意：

给出一些点、边以及边的权重，有q 次询问，问你：任意连通的两点间所有边的权重都小于 x ，这样的点有多少对

题解：

离线处理

把所有信息保存下来，对于边，权值按照从小到大排序，询问的 x 也按照从小到大排序

此时遍历每一条边，当小于 x 的时候，我们就加入并查集，计算并查集里面点对的增量

对于每一棵并查集树：

若有 n 个点，那么有 n*(n-1) 种点对

对于两棵需要合并的树，它们分别有 a 、 b 个点

那么需要先得到点对的增量是 (a+b)*(a+b-1) - a*(a-1) -b*(b-1) ,化简得到 2*a*b

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define maxn 100007
#define LL long long

struct edge{
    int from,to,cost;
    bool operator < (const edge& x) const {
        return cost < x.cost;
    }
}e[maxn];
int a[maxn],ta[maxn],fa[maxn],sum[maxn],ans[maxn];

int find_it(int x){
    int tempx=x,t;
    while(tempx!=fa[tempx]) tempx=fa[tempx];
    while(x!=fa[x])  t=fa[x],fa[x]=tempx,x=t;
    return tempx;
}

int unite(int num1,int num2){
    int tx=find_it(num1);
    int ty=find_it(num2),t=0;
    if(tx!=ty){
        fa[tx]=ty;
        t=2*sum[tx]*sum[ty];
        sum[ty]+=sum[tx];
    }
    return t;
}

int main()
{
    int T;
    int n,m,q;
    //freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d",&n,&m,&q);
        for(int i=0;i<m;i++)
            scanf("%d%d%d",&e[i].from,&e[i].to,&e[i].cost);
        sort(e,e+m);
        for(int i=0;i<q;i++)
            scanf("%d",&a[i]),ta[i]=a[i];
        sort(a,a+q);

        memset(ans,0,sizeof(ans));
        for(int i=0;i<=n;i++)  fa[i]=i,sum[i]=1;

        int pos=0,t=0;
        for(int i=0;i<m;i++){
            while(pos<q&&e[i].cost>a[pos]) ans[a[pos++]]=t;
            if(pos<q) t+=unite(e[i].from,e[i].to);
        }
        while(pos<q) ans[a[pos++]]=t;

        for(int i=0;i<q;i++)
            printf("%d\n",ans[ta[i]]);
    }
    return 0;
}