Travel
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 2882 Accepted Submission(s): 984
Problem Description
Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are
n
cities and m
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
x
minutes, how many pairs of city (a,b)
are there that Jack can travel from city a
to b
without going berserk?
Input
The first line contains one integer
T,T≤5,
which represents the number of test case.
For each test case, the first line consists of three integers n,m and q where n≤20000,m≤100000,q≤5000. The Undirected Kingdom has n cities and m bidirectional roads, and there are q queries.
Each of the following m lines consists of three integers a,b and d where a,b∈{1,...,n} and d≤100000. It takes Jack d minutes to travel from city a to city b and vice versa.
Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.
For each test case, the first line consists of three integers n,m and q where n≤20000,m≤100000,q≤5000. The Undirected Kingdom has n cities and m bidirectional roads, and there are q queries.
Each of the following m lines consists of three integers a,b and d where a,b∈{1,...,n} and d≤100000. It takes Jack d minutes to travel from city a to city b and vice versa.
Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.
Output
You should print q
lines for each test case. Each of them contains one integer as the number of pair of cities
(a,b)
which Jack may travel from a
to b
within the time limit x.
Note that (a,b) and (b,a) are counted as different pairs and a and b must be different cities.
Note that (a,b) and (b,a) are counted as different pairs and a and b must be different cities.
Sample Input
1 5 5 3 2 3 6334 1 5 15724 3 5 5705 4 3 12382 1 3 21726 6000 10000 13000
Sample Output
2 6 12
大体题意:
告诉你n个城市和m个边,和你能忍耐的时间,问你有几个城市对,(i,j),使得从i到j 的时间小于等于你所忍耐的时间,你没经过一个城市,你可以选择休息,休息的话,你所经历的时间全部清零。
思路:
结构体存边,给边按照权值排序,因为优先遍历权值小的嘛!
将询问也存下来,先按照询问权值排序,求出答案后再按id排回去即可!
当枚举到一个边时,发现两个端点所处在两个不同的连通块,答案就加上两个联通快内部个数的乘积的二倍!
然后合并联通块。
详细见代码:
告诉你n个城市和m个边,和你能忍耐的时间,问你有几个城市对,(i,j),使得从i到j 的时间小于等于你所忍耐的时间,你没经过一个城市,你可以选择休息,休息的话,你所经历的时间全部清零。
思路:
结构体存边,给边按照权值排序,因为优先遍历权值小的嘛!
将询问也存下来,先按照询问权值排序,求出答案后再按id排回去即可!
当枚举到一个边时,发现两个端点所处在两个不同的连通块,答案就加上两个联通快内部个数的乘积的二倍!
然后合并联通块。
详细见代码:
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 100000 + 10;
typedef long long ll;
int fa[maxn];
ll num[maxn];
struct Node{
int u,v,w;
bool operator < (const Node & rhs) const {
return w < rhs.w;
}
}p[maxn];
struct Query{
int id,v;
ll ans;
bool operator < (const Query& rhs){
return v < rhs.v;
}
}pq[maxn];
int find(int x){
return fa[x] == x ? x : (fa[x] = find(fa[x]));
}
bool cmp_id(const Query& lhs,const Query& rhs){
return lhs.id < rhs.id;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
int n,m,q;
scanf("%d %d %d",&n,&m,&q);
for (int i = 0; i <= n; ++i){
fa[i] = i;
num[i] = 1;
}
for (int i = 0; i < m; ++i){
scanf("%d %d %d",&p[i].u,&p[i].v,&p[i].w);
}
sort(p,p+m);
for (int i = 0; i < q; ++i){
scanf("%d",&pq[i].v);
pq[i].id = i;
}
sort(pq,pq+q);
int j = 0;
ll ans = 0ll;
for (int i = 0; i < q; ++i){
while(j < m && p[j].w <= pq[i].v){
int x = find(p[j].u);
int y = find(p[j].v);
if (x != y){
ans += (ll)num[x] * (ll)num[y] * 2ll;
num[x] += num[y];
fa[y] = x;
}
++j;
}
pq[i].ans = ans;
}
sort(pq,pq+q,cmp_id);
for (int i = 0; i < q; ++i){
printf("%I64d\n",pq[i].ans);
}
}
return 0;
}