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
题目要求a、b两点之间所有的路径中的最大边的最小值。
这题可以用并查集做，因为询问次数很多，所以要离线操作。设数组num[i],只记录i所在集合的总顶点数（一开始以为是记录边的，其实是记录顶点的），先对所有的边排序，从小到大，（其实可以看做是依次枚举最大边的最小值，因为如果这条线段的两端点不在一个集合，即两个集合的边不相连，那么这条线段就起到了桥梁的作用，也就是说任意两个集合的点相连都要通过这条边，因为之前两个集合的最大边的最小值都符合条件，那么如果这条线段也符合条件，就可以把两个集合的点都连起来），因为能量大的值一定包括能量小的路径，所以每次初始化的时候p[i].sum=p[i-1].sum;
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
int pre[10006],num[10006];
struct edge
{
int from,to,len;
}e[50006];
struct node
{
int id,num,sum;
}q[10006];
bool cmp1(edge a,edge b)
{
int temp;
if(a.len>b.len){
temp=a.from;a.from=b.from;b.from=temp;
temp=a.to;a.to=b.to;b.to=temp;
return a.len<b.len;
}
return a.len<b.len;
}
bool cmp2(node a,node b)
{
int temp;
if(a.num>b.num){
temp=a.id;a.id=b.id;b.id=temp;
return a.num<b.num;
}
return a.num<b.num;
}
bool cmp3(node a,node b)
{
int temp;
if(a.id>b.id){
temp=a.sum;a.sum=b.sum;b.sum=temp;
return a.id<b.id;
}
return a.id<b.id;
}
int find(int x)
{
int r=x,i,j=x;
while(r!=pre[r])r=pre[r];
while(j!=pre[j]){
i=pre[j];
pre[j]=r;
j=i;
}
return r;
}
int main()
{
int n,m,i,j,a,b,c,p,t1,t2,ans;
while(scanf("%d%d%d",&n,&m,&p)!=EOF)
{
for(i=0;i<=n;i++){
pre[i]=i;num[i]=1;
}
for(i=1;i<=m;i++){
scanf("%d%d%d",&e[i].from,&e[i].to,&e[i].len);
}
sort(e+1,e+1+m,cmp1);
for(i=1;i<=p;i++){
scanf("%d",&q[i].num);
q[i].id=i;
}
sort(q+1,q+p+1,cmp2);
ans=1;
q[0].sum=0;
for(i=1;i<=p;i++){
q[i].sum=q[i-1].sum;
while(ans<=m && e[ans].len<=q[i].num){
t1=find(e[ans].from);t2=find(e[ans].to);
if(t1==t2){
ans++;continue;
}
pre[t1]=t2;
q[i].sum+=num[t1]*num[t2];
num[t2]+=num[t1];
ans++;
}
}
sort(q+1,q+1+p,cmp3);
for(i=1;i<=p;i++){
printf("%d
",q[i].sum);
}
}
return 0;
}