思路：显然的mq这样的做法就T成狗...那么可以给边排个序，询问也排个序，然后扫过去，每次两个连通块合并的时候的答案就是(cnt[u]+cnt[v])*(cnt[u]+cnt[v]-1) - (cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1))嘛，然后就做完了
#include<bits/stdc++.h>
using namespace std;
const int maxn = 20000+6;
const int maxm = 100000+7;
int fa[maxn],cnt[maxn],ans[maxn];
int Find(int x){return x==fa[x]?x:fa[x]=Find(fa[x]);}
int n,m,qq;
struct Node
{
int u,v,w;
}e[maxm];
struct node
{
int w,id;
}q[maxn];
bool cmp1(node a,node b){return a.w<b.w;}
bool cmp(Node a,Node b){return a.w<b.w;}
void merge(int u,int v)
{
fa[u]=v;
cnt[v]+=cnt[u];
}
void init()
{
for(int i = 0;i<=n;i++)
{
fa[i]=i;
cnt[i]=1;
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&qq);
init();
for(int i= 1;i<=m;i++)
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
sort(e+1,e+1+m,cmp);
for(int i = 1;i<=qq;i++)
scanf("%d",&q[i].w),q[i].id=i;
sort(q+1,q+1+qq,cmp1);
int j = 1;
int res = 0;
for(int i = 1;i<=qq;i++)
{
while(j<=m && e[j].w<=q[i].w)
{
int u = Find(e[j].u);
int v = Find(e[j].v);
if(u!=v)
{
int tmp = cnt[u]+cnt[v];
res += (tmp*(tmp-1))-(cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1));
merge(u,v);
}
j++;
}
ans[q[i].id]=res;
}
for(int i = 1;i<=qq;i++)
printf("%d\n",ans[i]);
}
}
Description
cities and
bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is
minutes, how many pairs of city
are there that Jack can travel from city
to
Input
, which represents the number of test case.
For each test case, the first line consists of three integers
and
where
. The Undirected Kingdom has
cities and
bidirectional roads, and there are
queries.
Each of the following
lines consists of three integers
and
where
and
. It takes Jack
minutes to travel from city
to city
and vice versa.
Then
lines follow. Each of them is a query consisting of an integer
where
is the time limit before Jack goes berserk.
Output
lines for each test case. Each of them contains one integer as the number of pair of cities
which Jack may travel from
to
within the time limit
.
Note that
and
are counted as different pairs and
and
Sample Input
15 5 32 3 63341 5 15724 3 5 5705 4 3 12382 1 3 21726 6000 10000 13000
Sample Output
2612
