Travel
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1852 Accepted Submission(s): 641
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 aren
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?
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.
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.
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.
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.
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
Source
Recommend
hujie
通过并查集计算新加入的，每次的计算公式为原来的s+合并后的两个集合数量乘积。。。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string>
#include <queue>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define eps 1e-8
#define INF 20000000005
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d",&x,&y,&z);
#define rdLL(x) scanf("%I64d",&x)
#define MAXn 20005
#define MAXe 100005
#define QUEnum 5005
struct Node{
int sta,end,val;
}node[MAXe]; struct Que{
int queval , num;
}que[QUEnum]; int sett[MAXn]; int sett_find(int x)
{
if(sett[x]<0) return x;
return sett[x]=sett_find(sett[x]);
} bool cmp1(Node a,Node b){
return a.val<b.val;
} bool cmp2(Que a,Que b){
return a.queval<b.queval;
}
int main ()
{
int Case;
rd(Case);
while(Case--)
{
memset(sett,-1,sizeof(sett));
int sum[MAXn];
int res[QUEnum];
for(int i=0;i<MAXn;i++) sum[i]=1;
int n,e,quenum;
rd3(n,e,quenum); for(int i=0;i<e;i++)
rd3(node[i].sta,node[i].end,node[i].val); sort(node,node+e,cmp1); ///这里的排序不是+n 而是e 找了两个小时
for(int i=0;i<quenum;i++) {
rd(que[i].queval);
que[i].num=i;
} sort(que,que+quenum,cmp2);
int j=0,s=0; ///计数有多少能到达
for(int i=0 ; i<quenum ;i++) ///不能在这里添加j<e，因为大于的还需要计算
{
while( j<e && node[j].val <= que[i].queval ){
int a=node[j].sta,b=node[j].end; int p = sett_find(a);
int q = sett_find(b);
if(p!=q){
s += (sum[q]*sum[p]); ////((sum[q]*(sum[q]+1))>>1) - ((sum[p]*(sum[p]+1))>>1) +
sett[q] = p;
sum[p]=sum[q]+sum[p]; ///p作为根节点
}
j++;
}
res[ que[i].num ] = (s<<1);
}
for(int i=0;i<quenum;i++)
printf("%d\n",res[i]);
}
return 0;
}