Portal
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 931 Accepted Submission(s): 466
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
36
2
16
13
Source
2011 Multi-University Training Contest 10 - Host by HRBEU
题目意思:
给定一张无向图，要你求出满足小于或等于给定边长的最多边长的种类数目。当然关键是有n次询问。
思路：
对于一颗有n条路径的无向图，可以知道，当与另一个m条路径的无向图合并为一个全新的无向图时： 此时的路径为 n*m;
当然这题的重点不在这里，此题关键是有q次询问，对于每一次询问，若我们都去重新求解一次，将会很花时间，无疑TLE倒哭。%>_<%
于是采用离线处理（这里是开了解题报告才想起来，这么巧妙的地方，果然是too yong 哇！ ）：
离线的思路为： 先将询问的权值从小到大排序，由于大的权值包含了小的权值，于是整个过程，居然只需要一次就搞定。 俺是乡下人，第一次见识到这点，吓尿了！╮(╯▽╰)╭
代码：
1 #define LOCAL
2 #include<cstdio>
3 #include<cstring>
4 #include<iostream>
5 #include<algorithm>
6 #include<cstdlib>
7 using namespace std;
8 const int maxn=10005;
9 /*for point*/
10 struct node{
11 int a,b,c;
12 bool operator <(const node &bn)const {
13 return c<bn.c;
14 }
15 }sac[maxn*5];
16 /*for query*/
17 struct query{
18 int id,val;
19 bool operator <(const query &bn)const {
20 return val<bn.val;
21 }
22 }qq[maxn];
23
24 int father[maxn],rank[maxn];
25 int key[maxn];
26
27 void init(int n){
28 for(int i=0;i<=n;i++){
29 father[i]=i;
30 rank[i]=1;
31 }
32 }
33
34 int fin(int x){
35 while(x!=father[x])
36 x=father[x];
37 return x;
38 }
39
40 int unin(int x,int y)
41 {
42 x=fin(x);
43 y=fin(y);
44 int ans=0;
45 if(x!=y){
46 ans=rank[x]*rank[y];
47 if(rank[x]<rank[y]){
48 rank[y]+=rank[x];
49 father[x]=y;
50 }
51 else{
52 rank[x]+=rank[y];
53 father[y]=x;
54 }
55 }
56 return ans;
57 }
58
59 int main()
60 {
61 #ifdef LOCAL
62 freopen("test.in","r",stdin);
63 freopen("test1.out","w",stdout);
64 #endif
65 int n,m,q;
66 while(scanf("%d%d%d",&n,&m,&q)!=EOF)
67 {
68 init(n);
69 for(int i=0;i<m;i++){
70 scanf("%d%d%d",&sac[i].a,&sac[i].b,&sac[i].c);
71 }
72
73 for(int i=0;i<q;i++){
74 scanf("%d",&qq[i].val);
75 qq[i].id=i;
76 }
77 sort(sac,sac+m);
78 sort(qq,qq+q);
79 int i,j,res=0;
80 for(j=0,i=0;i<q;i++){
81 while(j<m&&sac[j].c<=qq[i].val){
82 res+=unin(sac[j].a,sac[j].b);
83 ++j;
84 }
85 key[qq[i].id]=res;
86 }
87 for(i=0;i<q;i++){
88 printf("%d\n",key[i]);
89 }
90 }
91 // system("comp");
92
93 return 0;
94 }