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
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 ",key[i]); 89 } 90 } 91 // system("comp"); 92 93 return 0; 94 }