HDU3938Portal（并查集离线应用）求路的条数

Portal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1017    Accepted Submission(s): 514

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

 

Source

2011 Multi-University Training Contest 10 - Host by HRBEU 

题意：有n个点m条边，现在问给出一个能量值L,有多少条路，两个点之间虽有多条路可走但只算一条，这条路的边的最大权值要小于等于给出的L值。

#include
#include
using namespace std;

const int N = 10005;

struct EDG
{
    int u,v,c;
};
struct ANS
{
    int i,c;
};

EDG edg[N*5];
ANS ans[N];
int fath[N],node[N],n;

int cmp1(const EDG &a,const EDG &b)
{
    return a.c0)
    {
        init();
        for(int i=1;i<=m;i++)
            scanf("%d%d%d",&edg[i].u,&edg[i].v,&edg[i].c);
        for(int i=1;i<=q;i++)
        {
            scanf("%d",&ans[i].c); ans[i].i=i;
        }
        sort(edg+1,edg+1+m,cmp1);
        sort(ans+1,ans+1+q,cmp2);
        int sum,u,v,tq,tm;
        sum=0; tq=tm=1;
        while(tm<=m)
        {
            if(edg[tm].c<=ans[tq].c)
            {
                u=findfath(edg[tm].u);
                v=findfath(edg[tm].v);
                if(u!=v)
                {
                    sum+=node[u]*node[v];
                    fath[u]=v;
                    node[v]+=node[u];
                }
                tm++;
            }
            else while(tq<=q&&edg[tm].c>ans[tq].c)
            {
                aa[ans[tq].i]=sum; tq++;
            }
            if(tq>q)
                break;

        }
        while(tq<=q)
         {
             aa[ans[tq].i]=sum; tq++;
         }
        for(int i=1;i<=q;i++)
            printf("%d\n",aa[i]);
    }
}