zoukankan      html  css  js  c++  java
  • hdu 2612 多终点BFS

    Find a way

    Problem Description
    Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
    Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest. 
    Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.
     
    Input
    The input contains multiple test cases.
    Each test case include, first two integers n, m. (2<=n,m<=200). 
    Next n lines, each line included m character.
    ‘Y’ express yifenfei initial position.
    ‘M’    express Merceki initial position.
    ‘#’ forbid road;
    ‘.’ Road.
    ‘@’ KCF
     
    Output
    For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.
     
    Sample Input
    4 4 Y.#@ .... .#.. @..M 4 4 Y.#@ .... .#.. @#.M 5 5 Y..@. .#... .#... @..M. #...#
     
    Sample Output
    66 88 66
     

    思路:两次BFS存下对每个kfc的最短距离,之后两两相加取min

    代码:

     1 #include "cstdio"
     2 #include "stdlib.h"
     3 #include "iostream"
     4 #include "algorithm"
     5 #include "string"
     6 #include "cstring"
     7 #include "queue"
     8 #include "cmath"
     9 #include "vector"
    10 #include "map"
    11 #include "set"
    12 #define mj
    13 #define db double
    14 #define ll long long
    15 using namespace std;
    16 const int N=1e8+2;
    17 const int mod=1e9+7;
    18 //const ll inf=1e16+10;
    19 #define inf 0x3f3f3f
    20 typedef pair<int,int> P;
    21 int n,m;
    22 char  s[300][300];
    23 int d[300][300],k[205][205];
    24 int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};
    25 int t[2][205*205];
    26 int bfs(int sx,int sy,int id)
    27 {
    28     queue<P> q;
    29     for(int i=0;i<205;i++){
    30         for(int j=0;j<205;j++){
    31             d[i][j]=N;
    32         }
    33     }
    34     q.push(P(sx,sy));
    35     d[sx][sy]=0;
    36     while(q.size()){
    37         P p;
    38         p=q.front(),q.pop();
    39         for(int i=0;i<4;i++){
    40             int nx=p.first+dx[i],ny=p.second+dy[i];
    41             if(0<=nx&&nx<n&&0<=ny&&ny<m&&s[nx][ny]!='#'&&d[nx][ny]==N){
    42                 d[nx][ny]=d[p.first][p.second]+1;
    43                 if(s[nx][ny]=='@') t[id][k[nx][ny]]=d[nx][ny];//到该点的距离
    44                 q.push(P(nx,ny));
    45             }
    46         }
    47     }
    48     return 0;
    49 }
    50 int main()
    51 {
    52     int xx[3],yy[3];
    53     while(scanf("%d%d",&n,&m)==2){
    54         memset(t,inf, sizeof(t));
    55         int c=0,cnt=0,ma=N;
    56         for(int i=0;i<n;i++){
    57             scanf("%s",s[i]);
    58             for(int j=0;j<m;j++){
    59                 if(s[i][j]=='Y') xx[c]=i,yy[c++]=j;
    60                 else if(s[i][j]=='M') xx[c]=i,yy[c++]=j;
    61                 else if(s[i][j]=='@'){
    62                     k[i][j]=cnt++;
    63                 }
    64             }
    65         }
    66         bfs(xx[0],yy[0],0),bfs(xx[1],yy[1],1);
    67         for(int i=0;i<cnt;i++){
    68             ma=min(t[0][i]+t[1][i],ma);
    69 //            printf("%d %d
    ",t[0][i],t[1][i]);
    70         }
    71         printf("%d
    ",11*ma);
    72     }
    73     return 0;
    74 }
  • 相关阅读:
    美团配送系统架构演进实践
    系统学习NLP(二十一)--SWEM
    转:大众点评信息流基于文本生成的创意优化实践
    从Encoder到Decoder实现Seq2Seq模型
    从YOLOv1到YOLOv3,目标检测的进化之路
    23岁融了一千万,被创新工场投资,创业就是解决问题。专访丨陈海沙
    通俗理解word2vec
    关于眼下分词的想法
    Angular 2的12个经典面试问题汇总(文末附带Angular測试)
    ubuntu14.04-64位机配置android开发环境,ADT,sdk,eclipsea
  • 原文地址:https://www.cnblogs.com/mj-liylho/p/7190577.html
Copyright © 2011-2022 走看看