zoukankan      html  css  js  c++  java
  • 最短路径———迪杰斯特拉算法

    北大:3767

    I Wanna Go Home
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 2523   Accepted: 1049

    Description

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<stdlib.h>
     4 #define maxint 999999
     5 int c[610][610],dist[610],p[610];
     6 void dj(int n)
     7 {
     8     int i,j,temp,u,s[610];
     9     for(i=1;i<=n;i++)
    10     {
    11         dist[i]=c[1][i];
    12         s[i]=0;
    13     }
    14     s[1]=1;
    15     dist[1]=0;
    16     for(i=2;i<=n;i++)
    17     {
    18         temp=maxint;
    19         for(j=1;j<=n;j++)
    20             if(!s[j]&&temp>dist[j])
    21             {
    22                 temp=dist[j];
    23                 u=j;
    24             }
    25             s[u]=1;
    26             for(j=1;j<=n;j++)
    27                 if(!s[j]&&dist[j]>dist[u]+c[u][j])
    28                     dist[j]=dist[u]+c[u][j];
    29     }
    30     if(dist[2]==maxint)  printf("-1\n");
    31     else
    32     printf("%d\n",dist[2]);
    33 }
    34 int main()
    35 {
    36     int i,j,x,y,z,n,m;
    37     while(scanf("%d",&n),n)
    38     {
    39         for(i=1;i<=n;i++)
    40             for(j=1;j<=n;j++)
    41                 c[i][j]=maxint;
    42         scanf("%d",&m);
    43         for(i=0;i<m;i++)
    44         {
    45             scanf("%d%d%d",&x,&y,&z);
    46             c[x][y]=z;
    47             c[y][x]=z;
    48         }
    49         for(i=1;i<=n;i++)
    50             scanf("%d",&p[i]);
    51         for(i=1,j=1;i<=n;i++)
    52             for(j=1;j<=n;j++)
    53             {   
    54                 if(p[i]==1&&p[j]==2)
    55                     c[j][i]=maxint;
    56                 else if(p[i]==2&&p[j]==1)
    57                        c[i][j]=maxint;
    58             }
    59     dj(n);
    60     }
    61     return 0;
    62 }
    63     

     

     

    The country is facing a terrible civil war----cities in the country are divided into two parts supporting different leaders. As a merchant, Mr. M does not pay attention to politics but he actually knows the severe situation, and your task is to help him reach home as soon as possible.

    "For the sake of safety,", said Mr.M, "your route should contain at most 1 road which connects two cities of different camp."

    Would you please tell Mr. M at least how long will it take to reach his sweet home?

    Input

    The input contains multiple test cases.

    The first line of each case is an integer N (2<=N<=600), representing the number of cities in the country.

    The second line contains one integer M (0<=M<=10000), which is the number of roads.

    The following M lines are the information of the roads. Each line contains three integers A, B and T, which means the road between city A and city B will cost time T. T is in the range of [1,500].

    Next part contains N integers, which are either 1 or 2. The i-th integer shows the supporting leader of city i.

    To simplify the problem, we assume that Mr. M starts from city 1 and his target is city 2. City 1 always supports leader 1 while city 2 is at the same side of leader 2.

    Note that all roads are bidirectional and there is at most 1 road between two cities.

    Input is ended with a case of N=0.

    Output

    For each test case, output one integer representing the minimum time to reach home.

    If it is impossible to reach home according to Mr. M's demands, output -1 instead.

    Sample Input

    2
    1
    1 2 100
    1 2
    3
    3
    1 2 100
    1 3 40
    2 3 50
    1 2 1
    5
    5
    3 1 200
    5 3 150
    2 5 160
    4 3 170
    4 2 170
    1 2 2 2 1
    0
    

    Sample Output

    100
    90
    540
    
  • 相关阅读:
    Kafka 配置
    Zookeeper的Watcher机制
    Ubuntu18.04下希捷移动硬盘Seagate Backup Plus读写慢
    Spring Boot 使用Jar打包发布, 并使用 Embedded Jetty/Tomcat 容器
    再谈C#委托与事件
    C#委托和事件例析
    PHP:session无法使用
    C++:实现类似MFC的IsKindOf功能
    C++:复制构造函数
    C++:运算符重载
  • 原文地址:https://www.cnblogs.com/xiaofanke/p/2650161.html
Copyright © 2011-2022 走看看