hdu 5441 Travel（带权并查集）

题目：​​http://acm.hdu.edu.cn/showproblem.php?pid=5441​​

Travel

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1586    Accepted Submission(s): 565

Problem Description

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

Input

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

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.
br />Note that (a,b) and (b,a) are counted as different pairs and a and b

 

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

分析：

两个连通块连通后，增加的对数：(t1+t2)(t1+t2-1)-t1(t1-1)-t2(t2-1) = 2*t1*t2.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=2e4+10,N=1e5+10,M=5010;
struct node{
    int l,r,w;
}edge[N];
struct anode{
    int id,wid,val;
}ans[M];
int cnt,father[maxn],sum[maxn];
void addedge(int a,int b,int c){
    edge[cnt].l=a;
    edge[cnt].r=b;
    edge[cnt++].w=c;
}
int find(int x){
    if(x==father[x]) return x;
    return father[x]=find(father[x]);
}
bool cmp1(node a,node b){
    return a.w<b.w;
}
bool cmp2(anode a,anode b){
    return a.wid<b.wid;
}
bool cmp3(anode a,anode b){
    return a.id<b.id;
}
int main()
{
    //freopen("cin.txt","r",stdin);
    int t,n,m,q;
    cin>>t;
    while(t--){
        scanf($quot;%d%d%d",&n,&m,&q);
        for(int i=1;i<=n;i++){
            sum[i]=1;
            father[i]=i;
        }
        cnt=0;
        memset(ans,0,sizeof(ans));
        int a,b,c;
        for(int i=0;i<m;i++){
            scanf("%d%d%d",&a,&b,&c);
            addedge(a,b,c);
        }
        sort(edge,edge+cnt,cmp1);
        for(int i=0;i<q;i++){
            scanf("%d",&ans[i].wid);
            ans[i].id=i;
            ans[i].val=0;
        }
        sort(ans,ans+q,cmp2);
        int k=0,res=0;
        for(int i=0;i<q;i++){
            ans[i].val=ans[i-1].val;
            for(;k<cnt;k++){
                if(edge[k].w>ans[i].wid)break;
                int fx=find(edge[k].l),fy=find(edge[k].r);
                if(fx==fy) continue;
                father[fx]=fy;
                ans[i].val+=2*sum[fy]*sum[fx];
                sum[fy]+=sum[fx];
            }
        }
        sort(ans,ans+q,cmp3);
        for(int i=0;i<q;i++){
            printf("%d\n",ans[i].val);
        }
    }
    return 0;
}