hdu 3938 Portal（离线并查集）

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 747    Accepted Submission(s): 373

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m条边，每条边都有一个花费，每两点a到b可形成多条路径，定义走过a b路径的花费为a到b的 各条路径中花费最大的边的最小值，给出q个询问，每个询问给出拥有多少权值，对于每个询问，要输出能走过的路径条数。

思路：类似kruskal算法的思想，用并查集离线处理。将互达的点看成一个集合，集合的点的个数为rank[i]。按边的花费从小到大给边排序，然后从小到大枚举每条边，若这条边的两个端点不在一个同集合中，设分别为集合a和集合b，那么这条边必为集合a的点到集合b的点的路径中花费最大的边的最小值，那么答案就可以加上rank[a]*rank[b]。

AC代码：

#include 
#include 
#include 
#include 
#include 
#define L(r) (r<<1)
#define R(r) (r<<1|1)
using namespace std;

const int maxn = 10005;
const int maxm = 50005;
const int INF = 1000000000;

int n, m, q;
int fa[maxn], rank[maxn];
struct Edge
{
    int u, v, w;
} et[maxm];
struct node
{
    int L, id;
} query[maxn];
bool cmp1(Edge a, Edge b)
{
    return a.w < b.w;
}
bool cmp2(node a, node b)
{
    return a.L < b.L;
}
void init()
{
    for(int i = 1; i <= n; i++)
    {
        rank[i] = 1;
        fa[i] = i;
    }
}
int find(int x)
{
    return fa[x] == x ? x : fa[x] = find(fa[x]);
}
void Union(int a, int b)
{
    if(rank[a] >= rank[b])
    {
        fa[b] = a;
        rank[a] += rank[b];
    }
    else
    {
        fa[a] = b;
        rank[b] += rank[a];
    }
}

int main()
{
    while(~scanf("%d%d%d", &n, &m, &q))
    {
        init();
        for(int i = 0; i < m; i++)
            scanf("%d%d%d", &et[i].u, &et[i].v, &et[i].w);
        for(int i = 0; i < q; i++)
        {
            scanf("%d", &query[i].L);
            query[i].id = i;
        }
        sort(et, et + m, cmp1);
        sort(query, query + q, cmp2);
        int cnt = 0;
        int ans[maxm] = {0};
        for(int i = 0; i < q; i++)
        {
            if(i > 0) ans[query[i].id] = ans[query[i - 1].id];
            while(et[cnt].w <= query[i].L  && cnt < m)
            {
                int ra = find(et[cnt].u);
                int rb = find(et[cnt].v);
                if(ra != rb)
                {
                    ans[query[i].id] += rank[ra] * rank[rb];
                    Union(ra, rb);
                }
                cnt++;
            }
        }
        for(int i = 0; i < q; i++)
            printf("%d\n", ans[i]);
    }
    return 0;
}