HDU 3938 Portal（离线+Kruskal+并查集）

题目：
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, 易做图 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

分析与总结：
做这题学到了什么是“离线算法”的概念。所谓“离线”，就是把所有的数据都输入之后再计算，“在线”就是边输入边计算。
用在这题中，是因为输入中的“询问部分”，有Q 个问，每个L可以有多少种不同路径。由于大的L必定会包含到小的L，所以把所有问题都输入，再从大到小排序，再计算，可以减少很多计算量。
这题还需要用到的是并查集中的“权值”，用rank数组表示，也就是某个棵树k有rank【k】个结点。同一个树之间的点都是连通的，任何点都可以通往其它的任意点，那么当两颗树合并成一棵树时，将会增加rank[a]*rank[b]条路径。

代码：
[cpp]
#include<cstdio>
#include<algorithm>
using namespace std;
#define N 10005
int f[N], rank[N], ans[N], n, m, Q;

struct Edge{
    int u, v, val;
    friend bool operator < (const Edge&a,const Edge&b){
        return a.val < b.val;
    }
}arr[N*5];

struct Query{
    int id, L;
    friend bool operator<(const Query&a,const Query&b){
        return a.L<b.L;
    }
}q[N];

void init(){
    for(int i=0; i<=n; ++i)
        f[i]=i, rank[i]=1;
}
int find(int x){
    int i, j=x;
    while(j!=f[j]) j=f[j];
    while(x!=j){ i=f[x]; f[x]=j; x=i; }
    return j;
}
int Union(int x,int y){
    int a=find(x), b=find(y);
    if(a==b) return 0;
    int t=rank[a]*rank[b];
    rank[a] += rank[b];
    f[b] = a;
    return t;
}

int main(){
    while(~scanf("%d%d%d",&n,&m,&Q)){

        for(int i=0; i<m; ++i)
            scanf("%d%d%d",&arr[i].u,&arr[i].v,&arr[i].val);
        for(int i=0; i<Q; ++i)
            scanf("%d",&q[i].L), q[i].id=i;

        sort(arr,arr+m);
        sort(q,q+Q);

        int cnt=0, j=0;
        init();
        for(int i=0; i<Q; ++i){
            while(j<m && arr[j].val<=q[i].L){
                cnt += Union(arr[j].u, arr[j].v);
                ++j;
            }
            ans[q[i].id] = cnt;
        }

        for(int i=0; i<Q; ++i)
            printf("%d\n", ans[i]);
    }
    return 0;
}

补充：软件开发 , C++ ,