杭电OJ3938离线+并查集

Posted 2020-11-03 MekakuCityActor

http://acm.hdu.edu.cn/showproblem.php?pid=3938

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1921    Accepted Submission(s): 955

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

10 10 10 7 2 1 6 8 3 4 5 8 5 8 2 2 8 9 6 4 5 2 1 5 8 10 5 7 3 7 7 8 8 10 6 1 5 9 1 8 2 7 6

 

Sample Output

36 13 1 13 36 1 36 2 16 13

题目大意：给一个图，然后会有Q次询问，询问路径上最大权值的边的权值小于 L的路径数。

题目分析：

【1】由于Q的范围是10^4，而如果一个一个查找的话，会TLE。而观察由于每次询问并未确定对路径的起点或者终点，所以每次询问的小L值可以累加得到询问时的大L的值，也就是每次询问的查找是有重叠的，所以就可以将Q次询问的L值进行从小到大的排序，小L慢慢累加得到大L，避免了不必要的重复查询

【解题的第一步-询问存起来等待离线】

【2】由于所连接的边的权值必须不大于L，所以也要将所有边进行排序

【解题第二步-->对边进行排序，等待查询是否能够连接】

【3】而由于是求路径数，两个图【一个顶点数为A,一个为B】连成一个图之后所能增加的路径数是A*B，所以可以利用并查集判断需要连接的两个图是否已经连接，如果没有则进行连接，并且路径数增加A*B，且根节点记录的图的顶点数增加A（或者B）

【解题最关键的一步，从小到大连接边，并更新L[i]的路径数，其中利用了两个连通图连接之后路径数增加A*B的原理】

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 struct edge{
 7     int to1;
 8     int to2;
 9     int lrn;
10 }EDGE[50005];
11 struct qury{
12     int id;
13     int x;
14 }qwq[10005];
15 int n,m,q;
16 int pre[10005],num[10005];;
17 int find(int x)
18 {
19 
20     int xx=x;
21     while(x!=pre[x])
22     {
23         x=pre[x];
24     }
25     while(pre[xx]!=x)
26     {
27         int t=pre[xx];
28         pre[xx]=x;
29         xx=t;
30     }
31     return x;
32 }
33 bool cmp1(struct edge orz1,struct edge orz2)
34 {
35     return orz1.lrn<orz2.lrn;
36 }
37 bool cmp2(struct qury orz3,struct qury orz4)
38 {
39     return orz3.x<orz4.x;
40 }
41 int main()
42 {
43     while(scanf("%d%d%d",&n,&m,&q)==3)
44     {
45     for(int i = 1 ; i <= n ; i++)
46     {
47         pre[i]=i;
48         num[i]=1;
49     }
50     for(int i = 0 ; i < m ; i++)
51     {
52         scanf("%d%d%d",&EDGE[i].to1,&EDGE[i].to2,&EDGE[i].lrn);
53     }
54     sort(EDGE,EDGE+m,cmp1);
55     for(int i = 0; i < q ; i++)
56     {
57         qwq[i].id=i;
58         scanf("%d",&qwq[i].x);
59     }
60     sort(qwq,qwq+q,cmp2);
61     int j=0;
62     long long sum=0;
63     long long ans[10005];
64     memset(ans,0,sizeof(ans));
65     for(int i = 0 ; i < q ; i++)
66     {
67         for(;j<m;j++)
68         {
69             if(EDGE[j].lrn>qwq[i].x)break;
70             int wqw1=find(EDGE[j].to1);
71             int wqw2=find(EDGE[j].to2);
72             if(wqw1!=wqw2)
73             {
74                 pre[wqw1]=wqw2;
75                 sum+=num[wqw1]*num[wqw2];
76                 num[wqw2]+=num[wqw1];
77             }    
78         }
79         ans[qwq[i].id]=sum;
80     
81     }
82     for(int i = 0 ;i < q ; i++)
83     {
84         printf("%lld\n",ans[i]);
85     }
86 }
87     return 0;
88  }