hdu 4750 (最小生成树)

Count The Pairs

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 67    Accepted Submission(s): 32

Problem Description

  With the 60th anniversary celebration of Nanjing University of Science and Technology coming soon, the university sets n tourist spots to welcome guests. Of course, Redwood forests in our university and its Orychophragmus violaceus must be recommended as top ten tourist spots, probably the best of all. Some undirected roads are made to connect pairs of tourist spots. For example, from Redwood forests (suppose it’s a) to fountain plaza (suppose it’s b), there may exist an undirected road with its length c. By the way, there is m roads totally here. Accidently, these roads’ length is an integer, and all of them are different. Some of these spots can reach directly or indirectly to some other spots. For guests, they are travelling from tourist spot s to tourist spot t, they can achieve some value f. According to the statistics calculated and recorded by us in last years, We found a strange way to calculate the value f:
  From s to t, there may exist lots of different paths, guests will try every one of them. One particular path is consisted of some undirected roads. When they are travelling in this path, they will try to remember the value of longest road in this path. In the end, guests will remember too many longest roads’ value, so he cannot catch them all. But, one thing which guests will keep it in mind is that the minimal number of all these longest values. And value f is exactly the same with the minimal number.
  Tom200 will recommend pairs (s, t) (start spot, end spot points pair) to guests. P guests will come to visit our university, and every one of them has a requirement for value f, satisfying f>=t. Tom200 needs your help. For each requirement, how many pairs (s, t) you can offer?

 

Input

  Multiple cases, end with EOF.
  First line:n m
  n tourist spots ( 1<n<=10000), spots’ index starts from 0.
  m undirected roads ( 1<m<=500000).

  Next m lines, 3 integers, a b c
  From tourist spot a to tourist spot b, its length is c. 0<a, b<n, c(0<c<1000000000), all c are different.

  Next one line, 1 integer, p (0<p<=100000)
  It means p guests coming.

  Next p line, each line one integer, t(0<=t)
  The value t you need to consider to satisfy f>=t.

 

Output

  For each guest's requirement value t, output the number of pairs satisfying f>=t.
  Notice, (1,2), (2,1) are different pairs.

 

Sample Input

2 10 1 231233 30 1 20 2 41 2 550 2345

 

Sample Output

22066440

 

Source

2013 ACM/ICPC Asia Regional Nanjing Online

 

Recommend

liuyiding

 

这道题开始想用最小瓶颈生成树做，预处理出任意两点的f值，这样就可以找到答案了，不过复杂度是O(n^2)的，但看给的是10s，于是想试试，结果交上去mle了，各种减内存都无果。
后来跟学长讨论了思路，他提出了一种对这题更有针对性的算法，依然是要利用最小生成树的思路，每次选权值最小的边加入图中，记这个边的两个端点分别所在的两个个连通分量中的点个数为a，b，则以这条边为最长边的点对有2*a*b个，这个证明很好想，这里就略过了。最后在查询时需要离线算法。时间复杂度O(m)。

#include<cstdio>#include<cmath>#include<cstring>#include<vector>#include<algorithm>#include<map>#include<iostream>using namespace std;const int maxn = 10000 + 10;const int maxm = 500000 + 5;struct Edge{    int x,y,dis;}e[maxm];bool cmp(Edge a,Edge b){    return a.dis < b.dis;}int fa[maxn],Rank[maxn];int Find(int x){return x==fa[x]?x:fa[x]=Find(fa[x]);}map<int,int> M;map<int,int>::reverse_iterator it;struct ans{    int id,t,num;}q[maxn*10];bool cmp2(ans a,ans b){    return a.t > b.t;}bool cmp3(ans a,ans b){    return a.id < b.id;}int main(){    int n,m;    while(scanf("%d%d",&n,&m) != EOF){        int a,b,c;        for(int i = 0;i < m;i++){            scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].dis);        }        sort(e,e+m,cmp);        for(int i = 0;i < n;i++){            fa[i] = i;            Rank[i] = 1;        }        M.clear();        for(int i = 0;i < m;i++){            int a = e[i].x;            int b = e[i].y;            int X = Find(a);            int Y = Find(b);            if(X != Y){                int tema = Rank[X];                int temb = Rank[Y];                M[e[i].dis] += tema*temb*2;                fa[X] = Y;                Rank[Y] += Rank[X];            }        }        int p;        scanf("%d",&p);        for(int i = 0;i < p;i++){            scanf("%d",&q[i].t);            q[i].id = i;        }        sort(q,q+p,cmp2);        int tem = 0;        it = M.rbegin();        for(int i = 0;i < p;){            if(it->first >= q[i].t){                tem += it->second;                it++;            }            else{                q[i++].num = tem;            }            while(it == M.rend() && i < p) q[i++].num = tem;        }        sort(q,q+p,cmp3);        for(int i = 0;i < p;i++){            printf("%d\n",q[i].num);        }    }    return 0;}