http://acm.hdu.edu.cn/showproblem.php?pid=3938
Portal
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
Sample Output
36
13
1
13
36
1
3
6
2
16
13
1 #include <cstdio>
2 #include <cstring>
3 #include <cmath>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7 #define N 10005
8 #define M 50005
9 /*
10 因为询问比较多，所以需要离线
11 在线的意思就是每一个询问单独处理复杂度O（多少多少)，
12 离线是指将所有的可能的询问先一次都处理出来，
13 最后对于每个询问O(1)回答
14
15 对于每个询问有一个长度L，问有多少条路径的长度<=L
16 而且该路径的长度是T，T是从u到v上最长的边
17 只要求得有多少个点对使得点对之间的最大的边小于L即可。
18
19 先从小到大排序一遍询问的边的长度L，从小到大枚举u->v之间的边的长度
20 如果两个集合没有联通，那么联通之后路径的条数为sum[x]*sum[y]
21 因为长的L必定包含了短的L的答案，所以要累加起来
22 */
23 int fa[N],sum[N];
24
25 struct node1
26 {
27 int id,ans,l;
28 }query[N];
29
30 struct node2
31 {
32 int u,v,len;
33 }edge[M];
34
35 bool cmp1(node2 a,node2 b)
36 {
37 return a.len<b.len;
38 }
39
40 bool cmp2(node1 a,node1 b)
41 {
42 return a.id<b.id;
43 }
44
45 bool cmp3(node1 a,node1 b)
46 {
47 return a.l<b.l;
48 }
49
50 int Find(int x)
51 {
52 if(x==fa[x]) return x;
53 return fa[x]=Find(fa[x]);
54 }
55
56 int Merge(int x,int y)
57 {
58 int fx=Find(x),fy=Find(y);
59 if(fx==fy) return 0;
60 int tmp;
61 if(fx<fy){
62 fa[fy]=fx;
63 tmp=sum[fx]*sum[fy];
64 sum[fx]+=sum[fy];
65 }
66 else{
67 fa[fx]=fy;
68 tmp=sum[fx]*sum[fy];
69 sum[fy]+=sum[fx];
70 }
71 return tmp;
72 }
73
74 int main()
75 {
76 int n,m,q;
77 while(~scanf("%d%d%d",&n,&m,&q)){
78 for(int i=1;i<=n;i++){
79 fa[i]=i;
80 sum[i]=1;
81 }
82 for(int i=0;i<m;i++){
83 scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].len);
84 }
85 for(int i=0;i<q;i++){
86 scanf("%d",&query[i].l);
87 query[i].id=i;
88 query[i].ans=0;
89 }
90 sort(edge,edge+m,cmp1);
91 sort(query,query+q,cmp3);
92 int cnt=0;
93 for(int i=0;i<q;i++){
94 while(edge[cnt].len<=query[i].l&&cnt<m){
95 int x=edge[cnt].u;
96 int y=edge[cnt].v;
97 int fx=Find(x);
98 int fy=Find(y);
99 if(fx==fy) cnt++;
100 else{
101 query[i].ans+=Merge(x,y);
102 cnt++;
103 }
104 }
105 if(i>0) query[i].ans+=query[i-1].ans;
106 }
107 sort(query,query+q,cmp2);
108 for(int i=0;i<q;i++){
109 printf("%d\n",query[i].ans);
110 }
111 }
112 return 0;
113 }