首页 > ACM题库 > HDU-杭电 > HDU 4750-Count The Pairs-最小生成树-[解题报告]HOJ
2015
09-17

HDU 4750-Count The Pairs-最小生成树-[解题报告]HOJ

Count The Pairs

问题描述 :

Parade Show

  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?

输入:

  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.

输出:

  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.

样例输入:

2 1
0 1 2
3
1
2
3
3 3
0 1 2
0 2 4
1 2 5
5
0 
2
3
4
5

样例输出:

2
2
0
6
6
4
4
0

关键是利用“所有从s到t的路径上的最长边的最小值”这个条件,我们所要的只是这个最小值:

对于(s,t),这个最小值一定是s到t的最短路上的最长边,则所有可能被取到的边是最小生成树上的边

因为最小生成树利用的就是贪心,树上的边都是图上每两点间的最短路会经过的边

实现:

在kruskal的过程中,添加一条边,权值为w,两个连通分量u和v变为连通,有num[u]*num[v]*2对pair会取到这个w,因为(s,t)和(t,s)不同所以*2

统计,对于一个f,对所有t<=f的guests更新

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <algorithm>
using namespace std;
#define N 10010
#define M 500500
struct Edge
{
    int u,v,w;
}edge[M];
bool cmp(Edge u,Edge v)
{
    return u.w<v.w;
}
struct Q
{
    int t,idx,val;
    Q(int t=0,int idx=0,int val=0)
    {
        this->t=t;this->idx=idx;this->val=val;
    }
}q[N*14];
bool cmp2(Q u,Q v)
{
    return u.t<v.t;
}
int n,m,p,fa[N],num[N],ans[N*10];
int _find(int u)
{
    return fa[u]==u?u:fa[u]=_find(fa[u]);
}
void update(int f,int val)
{
    int left=0,right=p,mid;
    while(left<right)
    {
        mid=left+right>>1;
        if(q[mid].t<=f) left=mid+1;
        else right=mid;
    }
    if(q[left].val>f) return;
    q[0].val+=val;
    q[left].val-=val;
}
int main ()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=0;i<m;++i)
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
        scanf("%d",&p);
        for(int i=0;i<p;++i)
        {
            scanf("%d",&q[i].t);
            q[i].idx=i;q[i].val=0;
        }
        sort(q,q+p,cmp2);
        sort(edge,edge+m,cmp);

        for(int i=0;i<n;++i) fa[i]=i,num[i]=1;
        int cc=0;
        for(int i=0;i<m;++i)
        {
            int u=_find(edge[i].u);
            int v=_find(edge[i].v);
            if(u==v) continue;
            update(edge[i].w,num[u]*num[v]*2);
            fa[u]=v;
            num[v]+=num[u];

            if(++cc>=n-1) break;
        }
        for(int i=0,k=0;i<p;++i)
        {
            k+=q[i].val;
            q[i].val=k;
        }
        for(int i=0;i<p;++i) ans[q[i].idx]=q[i].val;
        for(int i=0;i<p;++i) printf("%d\n",ans[i]);
    }
    return 0;
}

参考:http://blog.csdn.net/jackyguo1992/article/details/11895471