Travel
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2618 Accepted Submission(s): 922
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.
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.
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
思路：并查集；
我们用离线查询，然后将边按照升序排列，询问按照升序排列，然后我们可以用并查集维护联通的点，也就是本来不在同一个集合中的点只要有小于d的边就把他两合并，那么小于
d长度的点的组合，就是所有合法的集合求Cn2*2；
由于后面的d是比前面的大的所以前面我们所求得对后面的是有贡献的，我们就不需要从新循环找了
1 #include<stdio.h> 2 #include<algorithm> 3 #include<iostream> 4 #include<stdlib.h> 5 #include<math.h> 6 using namespace std; 7 typedef long long LL; 8 int bin[30000]; 9 LL du[30000]; 10 typedef struct pp 11 { 12 int x; 13 int y; 14 int cost; 15 } ss; 16 typedef struct ak 17 { 18 int x; 19 int id; 20 } dd; 21 ss aa[200005]; 22 dd kk[6000]; 23 LL an[6000]; 24 bool cmp(pp p,pp q) 25 { 26 return p.cost<q.cost; 27 } 28 bool cmp1(dd p,dd q) 29 { 30 return p.x<q.x; 31 } 32 int main(void) 33 { 34 int i,j,k; 35 int n,m; 36 int ask; 37 scanf("%d",&k); 38 while(k--) 39 { 40 scanf("%d %d",&n,&m); 41 scanf("%d",&ask); 42 for(i=0; i<m; i++) 43 { 44 scanf("%d %d %d",&aa[i].x,&aa[i].y,&aa[i].cost); 45 } 46 for(i=0; i<ask; i++) 47 { 48 scanf("%d",&kk[i].x); 49 kk[i].id=i; 50 } 51 aa[m].cost=1e9; 52 sort(aa,aa+m,cmp); 53 sort(kk,kk+ask,cmp1); 54 for(i=0; i<=20000; i++) 55 { 56 du[i]=1; 57 bin[i]=i; 58 } 59 LL cnt=0; 60 int t=0; 61 for(i=0; i<=m; i++) 62 { 63 int x=aa[i].x; 64 int y=aa[i].y; 65 while(aa[i].cost>kk[t].x&&t<ask) 66 { 67 an[kk[t].id]=cnt; 68 t++; 69 } 70 if(t==ask) 71 break; 72 int xx,yy; 73 for(xx=x; xx!=bin[xx];) 74 { 75 xx=bin[xx]; 76 } 77 for(yy=y; yy!=bin[yy];) 78 { 79 yy=bin[yy]; 80 } 81 if(xx!=yy) 82 { 83 cnt=cnt-(LL)du[xx]*(LL)(du[xx]-1)/2; 84 cnt=cnt-(LL)du[yy]*(LL)(du[yy]-1)/2; 85 if(du[xx]>du[yy]) 86 { 87 du[xx]+=du[yy]; 88 bin[yy]=xx; 89 cnt+=(LL)du[xx]*(LL)(du[xx]-1)/2; 90 } 91 else 92 { 93 du[yy]+=du[xx]; 94 bin[xx]=yy; 95 cnt+=(LL)du[yy]*(LL)(du[yy]-1)/2; 96 } 97 } 98 } 99 for(i=0; i<ask; i++) 100 { 101 printf("%lld ",an[i]*2); 102 } 103 } 104 return 0; 105 }