Travel
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 46 Accepted Submission(s): 20
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
类似Krsukal。首先确认一个事实，等待时间长的点对一定包含等待时间短的。
把询问离线然后排序，把边按照权值排序，每次只考虑比上次等待时间长，比这次等待时间短的边，用并查集维护连通分量的总个数。
#include<bits/stdc++.h> using namespace std; const int maxq = 5e3+5; int Qry[maxq]; long long ans[maxq]; bool cmp(int a,int b) { return Qry[a] < Qry[b]; } int rk[maxq]; int n,m,q; const int maxm = 1e5+5,maxn = 2e4+5; struct Edge { int u,v,w; bool operator < (const Edge &rh) const { return w < rh.w; } void IN() { scanf("%d%d%d",&u,&v,&w); } }edges[maxm]; int pa[maxn],cnt[maxn]; int fdst(int x) { return x==pa[x]?x:pa[x]=fdst(pa[x]); } int main() { //freopen("in.txt","r",stdin); int T; scanf("%d",&T); while(T--){ scanf("%d%d%d",&n,&m,&q); for(int i = 0; i < m ;i++){ edges[i].IN(); } for(int i = 0; i < q; i++){ rk[i] = i; scanf("%I64d",Qry+i); } sort(rk,rk+q,cmp); sort(edges,edges+m); for(int i = 1; i <= n; i++) pa[i] = i,cnt[i] = 1; long long cur = 0; for(int i = 0,j = 0; i < q; i++){ int id = rk[i]; int cq = Qry[id]; while(j < m && edges[j].w <= cq){ int u = fdst(edges[j].u),v = fdst(edges[j].v); if(u != v){ pa[u] = v; cur += 2LL*cnt[u]*cnt[v]; cnt[v] += cnt[u]; } j++; } ans[id] = cur; } for(int i = 0; i < q; i++){ printf("%I64d ",ans[i]); } } return 0; }