HDU3938 Portal

                                                                         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 <iostream>#include <algorithm>#include <cstdio>#include <cstring>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<m&&node[j].val<=p[i].L){               cnt += Union(node[j].from,node[j].to);               j++;            }            ans[p[i].id] = cnt;        }        for(int i = 0;i < q;++i)          printf("%d\n",ans[i]);    }    return 0;}