Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.
Input
There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).
Output
Output the answer to each query on a separate line.
Sample Input
10 10 10
7 2 1
6 8 3
4 5 8
5 8 2
2 8 9
6 4 5
2 1 5
8 10 5
7 3 7
7 8 8
10
6
1
5
9
1
8
2
7
6
7 2 1
6 8 3
4 5 8
5 8 2
2 8 9
6 4 5
2 1 5
8 10 5
7 3 7
7 8 8
10
6
1
5
9
1
8
2
7
6
Sample Output
36
13
1
13
36
1
36
2
16
13
13
1
13
36
1
36
2
16
13
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并查集&&离线算法
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1 /* 2 求最大边的最小值，用贪心排序 3 */ 4 #include<iostream> 5 #include<string> 6 #include<cstring> 7 #include<cmath> 8 #include<cstdio> 9 #include<algorithm> 10 using namespace std; 11 int fa[50001],num[50001]; 12 int ans[50001]; 13 int n,m,q; 14 int i,j,k; 15 void init() 16 { 17 int it; 18 for(int it=1;it<=n;it++) 19 { 20 fa[it]=it; 21 num[it]=1; 22 } 23 } 24 struct node 25 { 26 int st,en,len; 27 }e[50001]; 28 struct out 29 { 30 int id; 31 int value; 32 }A[50001]; 33 bool cmp1(node a,node b) 34 { 35 return a.len<b.len; 36 } 37 bool cmp2(out a,out b) 38 { 39 return a.value<b.value; 40 } 41 int find(int x) 42 { 43 int temp=x; 44 int root; 45 while(x!=fa[x]) 46 x=fa[x]; 47 root=x; 48 x=temp; 49 while(x!=fa[x]) 50 { 51 temp=x; 52 fa[x]=root; 53 x=fa[temp]; 54 } 55 return root; 56 } 57 int Unon(int a,int b) 58 { 59 int x=find(a); 60 int y=find(b); 61 if(x==y)return 0; 62 int temp1=num[x]*num[y]; 63 num[x]+=num[y]; 64 num[y]=0;//此处很重要，主要为了去重 65 fa[y]=x; 66 return temp1; 67 } 68 int main() 69 { 70 while(scanf("%d %d %d",&n,&m,&q)!=EOF) 71 { 72 init(); 73 for(i=0;i<m;i++) 74 scanf("%d %d %d",&e[i].st,&e[i].en,&e[i].len); 75 sort(e,e+m,cmp1); 76 for(i=0;i<q;i++) 77 { 78 scanf("%d",&A[i].value); 79 A[i].id=i; 80 81 } 82 sort(A,A+q,cmp2); 83 int temp=0; 84 k=0; 85 for(i=0;i<q;i++) 86 { 87 while(k<m&&e[k].len<=A[i].value) 88 { 89 temp+=Unon(e[k].st,e[k].en); 90 k++; 91 } 92 ans[A[i].id]=temp; 93 } 94 for(i=0;i<q;i++) 95 { 96 printf("%d ",ans[i]); 97 } 98 } 99 }