2602 shortest path problem Dihstra algorithm, 2602 dihstra

DescriptionDescription

There are n points (n <= 100) on the plane, and the coordinates of each point are-10000 ~ In the range of 10000. Some of these points are connected. If there is a line, it means you can reach another point from one point, that is, there is a path between two points, the distance of the path is a straight line between two points. The current task is to find the shortest path from one point to another.

Input description

Input Description
The first behavior is an integer n.

Rows 2nd to n + 1 (n rows in total), each line has two integers x and y, describing the coordinates of a point.

The n + 2 behavior is an integer m, indicating the number of connections in the graph.

Next m rows, each line describes a line consisting of two integers I and j, indicating that there is a line between the I and j.

Last line: two integers, s and t, respectively, represent the source point and the target point.

Output description

Output Description
Only one row and one real number (with two decimal places retained) indicates the shortest path length from s to t.

Sample Input

Sample Input
5

0 0

2 0

2 2

0 2

3 1

5

1 2

1 3

1 4

2 5

3 5

1 5

Sample output

Sample Output
3.41

Data range and prompt

Data Size & Hint
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 using namespace std; 6 double db_maxn=127; 7 double maxn=127; 8 struct node 9 {10 double x;11 double y;12 }a[1001];13 double dis[1001];14 int vis[1001];15 int n;16 double map[101][101];17 void Dijkstra(int u)18 {19 memset(vis,0,sizeof(vis));20 for(int i=1;i<=n;i++)21 {22 dis[i]=map[u][i];23 }24 dis[u]=0;25 vis[u]=1;26 for(int i=1;i<n;i++)27 {28 double minn=99999999;29 int k=-1;30 for(int j=1;j<=n;j++)31 {32 if((dis[j]<=minn)&&vis[j]==0)33 {34 minn=dis[j];35 k=j;36 }37 }38 vis[k]=1;39 for(int j=1;j<=n;j++)40 {41 if((dis[j]>=dis[k]+map[k][j])&&vis[j]==0)42 dis[j]=dis[k]+map[k][j];43 }44 }45 }46 int main()47 {48 memset(map,db_maxn,sizeof(map));49 memset(dis,db_maxn,sizeof(dis));50 scanf("%d",&n);51 for(int i=1;i<=n;i++)52 {53 scanf("%lf%lf",&a[i].x,&a[i].y);54 //a[i].cd=sqrt((pow(abs(x),2))+(pow(abs(y),2)));55 }56 int m;57 scanf("%d",&m);58 for(int i=1;i<=m;i++)59 {60 int p,q;61 scanf("%d%d",&p,&q);62 double y=sqrt(pow(a[p].x-a[q].x,2)+pow(a[p].y-a[q].y,2));63 map[p][q]=y; 64 map[q][p]=y;65 }66 int u,v;67 scanf("%d%d",&u,&v);68 Dijkstra(u);69 printf("%0.2lf",dis[v]);70 return 0;71 }