HDU 3938

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1796    Accepted Submission(s): 884

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 10 107 2 16 8 34 5 85 8 22 8 96 4 52 1 58 10 57 3 77 8 810615918276

 

Sample Output

36131133613621613

 

Source

2011 Multi-University Training Contest 10 - Host by HRBEU

 

Recommend

We have carefully selected several similar problems for you:  3935 3937 3931 3932 3933 

题意：从u到v两点之间建立路径花费精力是其中路段cost最多的那个，Q次询问给你总精力为L，求出有多少种类的路径被建成

将询问按照从小到大排序，然后将每条边按照权值从小到大排序，对于当前加入树边（u，v，w），对应ans【w】=sum【u】*sum【v】，这里sum【u】表示的就是点u所在联通块点的个数。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;#define N 110000struct node {    int u,v;    int w;}edge[N*5];struct Node {    int pos;    int val;}a[N];int father[N];int num[N], ans[N];int n, m, q;bool cmp1(node a, node b){    return a.w < b.w;}bool cmp2(Node a, Node b){    return a.val < b.val;}int findfather(int x){    int a = x;    while(x != father[x])        x = father[x];    while(a != father[a]) {        int t = a;        a = father[a];        father[t] = x;    }    return x;}int Union(int a, int b){    int faA = findfather(a);    int faB = findfather(b);    if(faA == faB)  //如果合并后属于同一子集说明已经被联通两点已经被计算在内        return 0;    int t = num[faA] * num[faB];    father[faB] = faA;    num[faA] += num[faB];    num[faB] = 0;//已经合并在一个集合里防止重复        return t;}int main(){    while(scanf("%d%d%d", &n, &m, &q) != EOF) {        for(int i = 0; i <= n; i++) {            father[i] = i;            num[i] = 1;        }        for(int i = 0; i < m; i++)            scanf("%d%d%d", &edge[i].u, &edge[i].v, &edge[i].w);        sort(edge, edge + m, cmp1); //边长从小到大排序        for(int i = 0; i < q; i++) {            scanf("%d", &a[i].val);            a[i].pos = i;        }        sort(a, a+q, cmp2);        int t = 0;        int j = 0;        for(int i = 0; i < q; i++) {            while(j < m && edge[j].w <= a[i].val) {                t += Union(edge[j].u, edge[j].v);                j++;            }            ans[a[i].pos] = t;        }        for(int i = 0; i < q; i++)            printf("%d\n", ans[i]);    }    return 0;}