HDU3938 Portal - c++编程基础

Portal

题目链接：Click Here~

题目分析：

ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of portals will cost min{T} energies. T in a path between point V and point U is the length of the longest edge in the path. There may be lots of paths between two points. Now ZLGG owned L energies and he want to know how many kind of path he could make.

There are multiple test cases. The first line of input contains three integer N, M and Q (1 < N ≤ 10,000, 0 < M ≤ 50,000, 0 < Q ≤ 10,000). N is the number of points, M is the number of edges and Q is the number of queries. Each of the next M lines contains three integers a, b, and c (1 ≤ a, b ≤ N, 0 ≤ c ≤ 10^8) describing an edge connecting the point a and b with cost c. Each of the following Q lines contain a single integer L (0 ≤ L ≤ 10^8).
题目有点难看懂，一开始看不懂什么意思。后来才知道，告诉你两个顶点u和v.然后，这两个点之间的花费是这两个点中最长距离路的大小。然后，每次会给你一个能力值，叫你求出在能量范围内的路径方法数。
算法分析：并查集+离线。这道题，不是我自己想出来的。是看了别人的解体报告才知道做法的。虽然以前有用过离线的做法，但是我没有想到联系到这道题上。为什么可以用到离线呢？因为我们知道小能量的范围一定可以满足大能量的要求。所以，我们可以从小到大的求出，在离线输出。

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std; const int maxn = 1e4 + 5; int f[maxn],rank[maxn],ans[maxn]; void Init() { for(int i = 0;i < maxn;++i) f[i] = i,rank[i] = 1; } struct Node{ int from,to,val; bool operator<(const Node& a)const{ return val < a.val; } }node[5*maxn]; struct Point{ int id,L; bool operator<(const Point& a)const{ return L < a.L; } }p[maxn]; int Find(int x) { if(x == f[x]) return f[x]; return f[x] = Find(f[x]); } int Union(int u,int v) { int a = Find(u),b = Find(v); if(a == b)return 0; int tmp = rank[a]*rank[b]; rank[a] += rank[b]; f[b] = a; return tmp; } int main() { int n,m,q; while(~scanf("%d%d%d",&n,&m,&q)) { for(int i = 0;i < m;++i) scanf("%d%d%d",&node[i].from,&node[i].to,&node[i].val); for(int i = 0;i < q;++i){ scanf("%d",&p[i].L); p[i].id = i; } sort(node,node+m); sort(p,p+q); Init(); int j = 0,cnt = 0; for(int i = 0;i < q;++i){ while(j