zoukankan      html  css  js  c++  java
  • hdu 5441

    Travel

    Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 4685    Accepted Submission(s): 1535


    Problem Description
    Jack likes to travel around the world, but he doesn’t like to wait. Now, he is traveling in the Undirected Kingdom. There are n cities and m bidirectional roads connecting the cities. Jack hates waiting too long on the bus, but he can rest at every city. Jack can only stand staying on the bus for a limited time and will go berserk after that. Assuming you know the time it takes to go from one city to another and that the time Jack can stand staying on a bus is x minutes, how many pairs of city (a,b) are there that Jack can travel from city a to b without going berserk?
     
    Input
    The first line contains one integer T,T5, which represents the number of test case.

    For each test case, the first line consists of three integers n,m and q where n20000,m100000,q5000. The Undirected Kingdom has n cities and mbidirectional roads, and there are q queries.

    Each of the following m lines consists of three integers a,b and d where a,b{1,...,n} and d100000. It takes Jack d minutes to travel from city a to city b and vice versa.

    Then q lines follow. Each of them is a query consisting of an integer x where x is the time limit before Jack goes berserk.

     
    Output
    You should print q lines for each test case. Each of them contains one integer as the number of pair of cities (a,b) which Jack may travel from a to b within the time limit x.

    Note that (a,b) and (b,a) are counted as different pairs and a and b must be different cities.
     
    Sample Input
    1 5 5 3 2 3 6334 1 5 15724 3 5 5705 4 3 12382 1 3 21726 6000 10000 13000
     
    Sample Output
    2 6 12
     
    Source
     
    Recommend
    hujie   |   We have carefully selected several similar problems for you:  6361 6360 6359 6358 6357 
     
     
     1 #include<cstdio>
     2 #include<algorithm>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<iostream>
     6 #include<vector>
     7 using namespace std;
     8 typedef long long ll;
     9 #define N 100009
    10 #define ls l,m,rt<<1
    11 #define rs m+1,r,rt<<1|1
    12 #define mem(a,b) memset(a,b,sizeof(a))
    13 int  t,n,m,q;
    14 struct Node{
    15     int u,v,w;
    16 }nod[N];
    17 struct Nod{
    18     int id,w;
    19 }no[N];
    20 int pre[20009];
    21 int cnt[20009];
    22 bool cmp(Node a,Node b){
    23     return a.w<b.w;
    24 }
    25 bool cmp1(Nod a,Nod b){
    26     return a.w<b.w;
    27 }
    28 void init()
    29 {
    30     for(int i=0;i<=n;i++){
    31         pre[i]=i;
    32         cnt[i]=1;
    33     }
    34 }
    35 int find(int x)
    36 {
    37     return pre[x]=x==pre[x]?x:find(pre[x]);
    38 }
    39 void unite(int u,int v)
    40 {  /*
    41     pre[v]=u;
    42     cnt[u]+=cnt[v];
    43     */
    44     pre[u]=v;
    45     cnt[v]+=cnt[u];
    46     //上面两种写法都是对的。
    47 }
    48 int ret[N];
    49 int  main()
    50 {
    51     scanf("%d",&t);
    52     while(t--)
    53     {
    54         scanf("%d%d%d",&n,&m,&q);
    55         init();
    56         for(int i=0;i<m;i++){
    57             scanf("%d%d%d",&nod[i].u,&nod[i].v,&nod[i].w);
    58             }
    59             sort(nod,nod+m,cmp);
    60             for(int i=0;i<q;i++){
    61                 scanf("%d",&no[i].w);
    62                 no[i].id=i;
    63             }
    64             sort(no,no+q,cmp1);
    65             int j=0,ans=0;
    66             int u,v;
    67             //预处理
    68             for(int i=0;i<q;i++){
    69                 while(j<m&&nod[j].w<=no[i].w){
    70                     u=find(nod[j].u);
    71                     v=find(nod[j].v);
    72                     if(u!=v){
    73                         int tmp=cnt[v]+cnt[u];
    74                         //任意一对点都行
    75                         ans+=(tmp*(tmp-1))-(cnt[u]*(cnt[u]-1))-(cnt[v]*(cnt[v]-1));
    76                         //减去已经加过的
    77                         unite(u,v);
    78                     }
    79                     j++;
    80                 }
    81                 ret[no[i].id]=ans;
    82             }            
    83             for(int i=0;i<q;i++){
    84                 printf("%d
    ",ret[i]);
    85             }        
    86     }
    87     return 0;
    88 }
  • 相关阅读:
    lucene中创建索引库
    商城后台上架商品列表查询的书写全过程
    Linux命令英文全称
    商品品牌分页、过滤、排序查询的完成流程
    axios使用步骤详解(附代码)
    使用CORS处理跨域请求
    npm 是干什么的?
    Mybatis通用Mapper介绍和使用
    FastDFS的理解和分析
    CDN服务的含义
  • 原文地址:https://www.cnblogs.com/tingtin/p/9441751.html
Copyright © 2011-2022 走看看