【杭电OJ3938】【离线+并查集】
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 }
