zoukankan      html  css  js  c++  java
  • POJ 2607 Fire Station

    Fire Station

    Time Limit: 5000ms
    Memory Limit: 65536KB
    This problem will be judged on PKU. Original ID: 2607
    64-bit integer IO format: %lld      Java class name: Main
     
    A city is served by a number of fire stations. Some residents have complained that the distance from their houses to the nearest station is too far, so a new station is to be built. You are to choose the location of the fire station so as to reduce the distance to the nearest station from the houses of the disgruntled residents. 
    The city has up to 500 intersections, connected by road segments of various lengths. No more than 20 road segments intersect at a given intersection. The location of houses and firestations alike are considered to be at intersections (the travel distance from the intersection to the actual building can be discounted). Furthermore, we assume that there is at least one house associated with every intersection. There may be more than one firestation per intersection. 
     

    Input

    The first line of input contains two positive integers: f,the number of existing fire stations (f <= 100) and i, the number of intersections (i <= 500). The intersections are numbered from 1 to i consecutively. f lines follow; each contains the intersection number at which an existing fire station is found. A number of lines follow, each containing three positive integers: the number of an intersection, the number of a different intersection, and the length of the road segment connecting the intersections. All road segments are two-way (at least as far as fire engines are concerned), and there will exist a route between any pair of intersections.
     

    Output

    You are to output a single integer: the lowest intersection number at which a new fire station should be built so as to minimize the maximum distance from any intersection to the nearest fire station.
     

    Sample Input

    1 6
    2
    1 2 10
    2 3 10
    3 4 10
    4 5 10
    5 6 10
    6 1 10
    

    Sample Output

    5
    

    Source

     
    解题:枚举最短路,坑爹,居然文件结束输入路径
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <climits>
     7 #include <vector>
     8 #include <queue>
     9 #include <cstdlib>
    10 #include <string>
    11 #include <set>
    12 #include <stack>
    13 #define LL long long
    14 #define INF 0x3f3f3f3f
    15 #define pii pair<int,int>
    16 using namespace std;
    17 const int maxn = 2050;
    18 struct arc{
    19     int to,cost,next;
    20     arc(int x = 0,int y = 0,int z = -1){
    21         to = x;
    22         cost = y;
    23         next = z;
    24     }
    25 };
    26 arc e[maxn*maxn];
    27 int head[maxn],d[maxn],dd[maxn];
    28 int tot,n,m;
    29 bool done[maxn];
    30 void add(int u,int v,int w){
    31     e[tot] = arc(v,w,head[u]);
    32     head[u] = tot++;
    33     e[tot] = arc(u,w,head[v]);
    34     head[v] = tot++;
    35 }
    36 void dijkstra(int s,int *d){
    37     for(int i = 1; i <= n; ++i) done[i] = false;
    38     d[s] = 0;
    39     priority_queue< pii,vector< pii >,greater< pii > >q;
    40     q.push(make_pair(d[s],s));
    41     while(!q.empty()){
    42         int u = q.top().second;
    43         q.pop();
    44         if(done[u]) continue;
    45         done[u] = true;
    46         for(int i = head[u]; ~i; i = e[i].next){
    47             if(d[e[i].to] > d[u] + e[i].cost){
    48                 d[e[i].to] = d[u] + e[i].cost;
    49                 q.push(make_pair(d[e[i].to],e[i].to));
    50             }
    51         }
    52     }
    53 }
    54 int main(){
    55     while(~scanf("%d %d",&m,&n)){
    56         int fire[maxn],tmp,u,v,w;
    57         bool isfire[maxn] = {false};
    58         memset(head,-1,sizeof(head));
    59         for(int i = tot = 0; i < m; ++i){
    60             scanf("%d",&tmp);
    61             fire[i] = tmp;
    62             isfire[tmp] = true;
    63         }
    64         while(~scanf("%d %d %d",&u,&v,&w)) add(u,v,w);
    65         for(int i = 0; i <= n; ++i) d[i] = INF;
    66         for(int i = 0; i < m; ++i) dijkstra(fire[i],d);
    67         int maxv = INF,index = 1;
    68         for(int i = 1; i <= n; ++i){
    69             if(isfire[i]) continue;
    70             memcpy(dd,d,sizeof(d));
    71             dijkstra(i,dd);
    72             int mmxx = 0;
    73             for(int k = 1; k <= n; ++k)
    74                 mmxx = max(mmxx,dd[k]);
    75             if(mmxx < maxv){
    76                 index = i;
    77                 maxv = mmxx;
    78             }
    79         }
    80         printf("%d
    ",index);
    81     }
    82     return 0;
    83 }
    View Code
  • 相关阅读:
    瓜子IM智能客服系统的数据架构设计(整理自现场演讲)
    正确理解IM长连接的心跳及重连机制,并动手实现(有完整IM源码)
    从游击队到正规军(二):马蜂窝旅游网的IM客户端架构演进和实践总结
    拿起键盘就是干:跟我一起徒手开发一套分布式IM系统
    12306抢票带来的启示:看我如何用Go实现百万QPS的秒杀系统
    适合新手:从零开发一个IM服务端(基于Netty,有完整源码)
    关于按钮多次点击问题
    UIButton内部子控件自定义布局-“UIEdgeInsets”
    iOS中的数据库应用
    swift中的正则表达式
  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4051890.html
Copyright © 2011-2022 走看看