Travel
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1313 Accepted Submission(s): 472
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 mbidirectional 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 mbidirectional 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
Source
题目大意：无向图。给你m条边，u - v 每条边有一个权值w。表示从城市u - v、v - u 的时间是w。然后q次询问。每次询问一个x值。表示问你当时间容忍为x时，可以走多少个城市对。每次到一个城市后可以休息，时间容忍会重新变为x。
解题思路：首先按边权把边排序。用并查集统计加入第k条边时可以走的城市对个数，记录在pair_cnt数组中。同时需要维护每个集合中的点的个数。最后在排完序后的边中二分找到第一个大于x的边的编号ck。然后在pair_cnt数组中找到编号ck-1即为答案。
#include<bits/stdc++.h>
using namespace std;
const int maxn=(1e4)*3;
const int maxm=1e5+200;
const int INF=0x3f3f3f3f;
struct Edge{
int u,v,w;
Edge(){}
Edge(int _u,int _v,int _w){
u=_u; v=_v; w=_w;
}
}edges[maxm];
int cnt[maxn],pair_cnt[maxm];
int fa[maxn];
void init(int n){
for(int i=0;i<=n;i++){
fa[i]=i;
cnt[i]=1;
}
}
int Find(int x){
if(fa[x]!=x){
return fa[x]=Find(fa[x]);
}
return x;
}
int Union(int u,int v){
int fu=Find(u);
int fv=Find(v);
int ret;
if(fu<fv){
ret=cnt[fu]*cnt[fv]*2;
fa[fv]=fu;
cnt[fu]+=cnt[fv];
return ret;
}else if(fu>fv){
ret=cnt[fu]*cnt[fv]*2;
fa[fu]=fv;
cnt[fv]+=cnt[fu];
return ret;
}else{
return 0;
}
}
bool cmp(Edge a,Edge b){
return a.w<b.w;
}
int BinSearch(int l,int r,int key){
while(l<r){
int md=(l+r)/2;
if(key<edges[md].w){
r=md;
}else if(key>edges[md].w){
l=md+1;
}else{
return md+1;
}
}
return r;
}
int main(){
int t,n,m,q;
scanf("%d",&t);
while(t--){
scanf("%d%d%d",&n,&m,&q);
init(n);
for(int i=1;i<=m;i++){
scanf("%d%d%d",&edges[i].u,&edges[i].v,&edges[i].w);
}
sort(edges+1,edges+1+m,cmp);
edges[0].w=-1;edges[m+1].w=INF;
for(int i=1;i<=m;i++){
pair_cnt[i]=pair_cnt[i-1]+Union(edges[i].u,edges[i].v);
}
int tmp;
for(int i=0;i<q;i++){
scanf("%d",&tmp);
printf("%d
",pair_cnt[BinSearch(0,m+1,tmp)-1]);
}
}
return 0;
}