离线的并查集

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): 820    Accepted Submission(s): 425

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

题意：给出一个连通图，以及每条路径的长度，定义从u走到v的所有边的最长边作为消耗的能量，q次询问，每次给出一个L能量，问存在多少这样的<u,v>路径对；

分析：很明显是并查集，但是每询问一次就做一次并查集肯定会超时，所以应采用离线算法，先把所有答案全求出来，然后一块输出；首先把边权从小到大排序，然后把询问也从小到大排序，对于每次询问，只加入小于询问的边到一个集合中，用h数组记录每个集合中的元素，sum记录当加完这条边后此时的u，v点对数目；

程序：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"map"
#include"string"
#include"queue"
#include"stdlib.h"
#include"math.h"
#include"algorithm"
#include"vector"
#define M 100009
#define eps 1e-10
#define inf 1000000000
#define mod 1000000000
#define INF 1000000000
using namespace std;
struct node
{
    int u,v,w;
}e[M*5],q[M];
int f[M],h[M];
__int64 sum,ans[M];
int cmp(node a,node b)
{
    return a.w<b.w;
}
int finde(int x)
{
    if(x!=f[x])
        f[x]=finde(f[x]);
    return f[x];
}
void make(int a,int b)
{
    int x=finde(a);
    int y=finde(b);
    if(x==y)return;
    else if(x>y)
    {
        f[x]=y;
        int tt=h[y];
        h[y]+=h[x];
        sum+=h[y]*(h[y]-1)/2-h[x]*(h[x]-1)/2-tt*(tt-1)/2;
    }
    else
    {
        f[y]=x;
        int tt=h[x];
        h[x]+=h[y];
        sum+=h[x]*(h[x]-1)/2-h[y]*(h[y]-1)/2-tt*(tt-1)/2;
    }
}
int main()
{
    int n,m,k,i;
    while(scanf("%d%d%d",&n,&m,&k)!=-1)
    {
        for(i=0;i<m;i++)
            scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
        sort(e,e+m,cmp);
        for(i=0;i<k;i++)
        {
            scanf("%d",&q[i].w);
            q[i].u=i;
        }
        sort(q,q+k,cmp);
        for(i=1;i<=n;i++)
        {
            f[i]=i;h[i]=1;
        }
        sum=0;
        int j=0;
        for(i=0;i<k;i++)
        {
            while(j<m&&e[j].w<=q[i].w)
            {
                make(e[j].u,e[j].v);
                j++;
            }
            ans[q[i].u]=sum;
        }
        for(i=0;i<k;i++)
            printf("%I64d
",ans[i]);
    }
    return 0;
}