hdu 3938 Portal 计算两点路上最大距离中的最小值<=L的路径个数离线并查集

最新推荐文章于 2018-08-12 19:37:00 发布

原创最新推荐文章于 2018-08-12 19:37:00 发布 · 893 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#integer #struct #each #input #path #output

acm_图论同时被 2 个专栏收录

71 篇文章

订阅专栏

acm_并查集

16 篇文章

订阅专栏

本文介绍了一种用于计算图中特定能量限制下最长边路径数量的算法。该算法通过按边权重排序并使用并查集来高效地解决这一问题。输入包含节点数、边数及查询次数等信息，输出则为不同能量水平下可形成的路径数量。

Problem Description

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.

Input

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).

Output

Output the answer to each query on a separate line.

Sample Input

Sample Output

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
//按照边大小排序,依次合并两端点,则以此边为路上最大长度
//的路径共有 num[x]*num[y];
//将要查询的L从小到大排序，依次处理即可
const int maxn=51000;
int fath[maxn];
int num[maxn];
int n;
void init()
{
for(int i=1;i<=n;i++) fath[i]=i,num[i]=1;
}
int find(int x)
{
return fath[x]==x?x:fath[x]=find(fath[x]);
}
int uion(int x,int y)
{
x=find(x),y=find(y);
if(x==y) return 0;
int cnt=num[x]*num[y];
if(num[x]<num[y])
{
fath[x]=y;
num[y]+=num[x];
}
else
{
fath[y]=x;
num[x]+=num[y];
}
return cnt;
}
struct Edge
{
int u,v,w;
};
Edge edge[maxn];
bool cmpEdge(Edge h,Edge k)
{
return h.w<k.w;
}
struct Node
{
int val;
int index;
};
Node ind[maxn];
bool cmpNode(Node h,Node k)
{
return h.val<k.val;
}
int ans[maxn];
int main()
{
int m,q;
while(scanf("%d%d%d",&n,&m,&q)==3)
{
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
}
sort(edge,edge+m,cmpEdge);
for(int i=0;i<q;i++)
{
scanf("%d",&ind[i].val);
ind[i].index=i;
}
sort(ind,ind+q,cmpNode);
init();
int sum=0;
int l=0;
for(int i=0;i<m;i++)
{
int u=edge[i].u,v=edge[i].v,w=edge[i].w;
int tmp=uion(u,v);
sum+=tmp;
while(l<q&&w>ind[l].val) ans[ind[l++].index]=sum-tmp;
}
for(int i=l;i<q;i++) ans[ind[i].index]=ans[ind[i-1].index];
for(int i=0;i<q;i++) printf("%d\n",ans[i]);
}
return 0;
}