ces ephone sat custom neo solution his 最短 地理

BOI‘98 DAY 2 TASK 1

CONFERENCE CALL

PROBLEM

A telecom company would like to offer a three-party conference call service. This service enables three customers to participate in a telephone conversation simultaneously. A customer accesses the interconnection network of the telephone system via a switch. The network consists of switches linked by bidirectional lines. Over the existing network, the three participants accessing the network via three different switches, must be connected with the minimum cost. Note that it is possible to connect any two switches. You will be given all the links between the switches of the network along with their associated costs. You are requested to find the connections which will minimize the total cost of interconnecting the three switches to which the participants in a conference call are connected.

INPUT

The input is a text file named conf.inp. The first line contains two numbers : The number of switches, designated by ( N <= 100) and the number of switch-to-switch links in the network, designated by E, separated by a blank. Each of the following E input lines contains three integers, designated by i, j and c_i,j

where i and j are the numbers of two different switches and c_i,jis the cost of using the link between i and j in a conference call (1 <= c_i,j<= 100), each separated by one blank. The last line of input contains three integers, which designate the three switches that participate in the conference call, each separated by one blank. The switches are identified with integers 1 through N consecutively.

OUTPUT

The output must be a text file named conf.out. The first line must contain the total (minimal) cost of establishing a conference call among the given three switches and the number R of switch-to-switch connections needed to establish this conference call, separated by a blank. Each of the following R lines must contain two integers, which denote that the link between the switch numbered by the first integer and the switch numbered by the second integer is included in the solution -ordering of between the two switches is not important. The ordering among the last R lines of the output is also not important.

EXAMPLE

分析

這道題可以形象地理解成。班裏有45個同學，其中有三個同學想要選某一個同學家作為見面地點，要求這個見面地點到三個同學家的距離和最小。

容易想到運用單源最短路算法，分別以三個客戶端作為源點，運行最短路算法。把距離保存在三個數組當中。最後循環每一個點，查哪一個點到三個點的距離和最短。保存這個點t，這個最小費用s。

最後從t開始反向查找會經過的每一條邊。

程序

  1 #include<bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 int n, m, EdgeCount, u, v, w, a[3], f[3][105], Head[105], pre[3][105], t, k, s;
  6 struct node
  7 {
  8     int Next, Aim, Weight;
  9 }Edge[20005];
 10 struct h
 11 {
 12     int len,point;
 13     bool operator < (const struct h &x) const
 14     {
 15         return x.len<len;
 16     }
 17     bool operator > (const struct h &x) const
 18     {
 19         return x.len>len;
 20     }
 21 };
 22 struct p
 23 {
 24     int x,y;
 25     bool operator > (const struct p &P) const
 26     {
 27         return (P.x<x)||((P.x==x)&&(P.y<y));
 28     }
 29     bool operator < (const struct p &P) const
 30     {
 31         return (P.x>x)||((P.x==x)&&(P.y>y));
 32     }
 33 }b[20005];
 34 
 35 inline void Insert(int u, int v, int w)
 36 {
 37     Edge[++EdgeCount]=(node){Head[u],v,w};
 38     Head[u]=EdgeCount;
 39 }
 40 
 41 void Dijkstra(int num,int st)
 42 {
 43     priority_queue<h> Q;
 44     f[num][st]=0;
 45     while(!Q.empty()) Q.pop();
 46     Q.push((struct h){0,st});
 47     for (int j=1;j<=n;j++)
 48     {
 49         struct h x=Q.top();
 50         Q.pop();
 51         for (int i=Head[x.point];i;i=Edge[i].Next)
 52          if (f[num][x.point]+Edge[i].Weight<f[num][Edge[i].Aim])
 53         {
 54             f[num][Edge[i].Aim]=f[num][x.point]+Edge[i].Weight;
 55             Q.push((struct h){f[num][Edge[i].Aim],Edge[i].Aim});
 56             pre[num][Edge[i].Aim]=x.point;
 57         }
 58     }
 59 }
 60 
 61 int main()
 62 {
 63     freopen("conference.in","r",stdin);
 64     freopen("conference.out","w",stdout);
 65     cin >> n >> m;
 66     EdgeCount=0;
 67     for (int i = 1; i <= m; i++)
 68     {
 69         cin >> u >> v >> w;
 70         Insert(u, v, w);
 71         Insert(v, u, w);
 72     }
 73     memset(f, 0x3f, sizeof(f));
 74     for (int i = 0; i <= 2; i++)
 75     {
 76         cin>>a[i];
 77         Dijkstra(i,a[i]);
 78     }
 79     s=1e+8;
 80     for (int i = 1; i <= n; i++)
 81         if (s > f[0][i]+f[1][i]+f[2][i])
 82         {
 83             s = f[0][i]+f[1][i]+f[2][i];
 84             t = i;
 85         }
 86     cout << s;
 87     s = 0;
 88     for (int i = 0; i <= 2; i++)
 89     {
 90         k = t;
 91         while (pre[i][k] != 0)
 92         {
 93             b[++s] = (p){min(pre[i][k],k),max(pre[i][k],k)};
 94             k = pre[i][k];
 95         }
 96     }
 97     sort(b+1,b+(s+1));
 98     t = 1;
 99     for (int i = 2; i <= s; i++)
100         if (b[i].x != b[i-1].x || b[i].y != b[i-1].y)
101             t++;
102     cout << " " << t << endl;
103     cout << b[1].x << " " << b[1].y << endl;
104     for (int i=2;i<=s;i++)
105         if (b[i].x != b[i-1].x || b[i].y != b[i-1].y)
106             cout<<b[i].x<<" "<<b[i].y<<endl;
107     return 0;
108 }

BOI'98 DAY 2 TASK 1 CONFERENCE CALL Dijkstra/Dijkstra+priority_queue/SPFA