poj 2253 (dis最短路径)

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Description

Input

Output

Sample Input

2
0 0
3 4

3
17 4
19 4
18 5

0

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Source

Ulm Local 1997

#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
double G[210][210];
int n;
struct Point{
    double x,y;
}a[210];
float dist(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
void dij(){
    double dis[210];
    int vis[210];
    for(int i=1;i<=n;i++)
        vis[i]=0;
    vis[1]=1;
    dis[1]=0.0;
    for(int i=2;i<=n;i++)
        dis[i]=G[1][i];
    for(int i=1;i<n;i++){
        double min=999999.0;
        int v;
        for(int j=2;j<=n;j++)
            if(!vis[j] && dis[j]<min){
                min=dis[j];
                v=j;
            }
        vis[v]=1;
        for(int j=1;j<=n;j++){
            double tmp=(dis[v]<G[v][j] ? G[v][j] : dis[v]);     //注意这里是最短路径变形
            dis[j]= tmp<dis[j] ? tmp : dis[j];
        }
    }
    cout<<setiosflags(ios::fixed)<<setprecision(3)<<dis[2]<<endl<<endl;
}
int main(){
    int cas=1;
    while(cin>>n&&n){
        for(int i=1;i<=n;i++)
            cin>>a[i].x>>a[i].y;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                G[i][j]=dist(a[i],a[j]);
        cout<<"Scenario #"<<cas++<<endl;
        cout<<"Frog Distance = ";
        dij();
    }
    return 0;
}

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poj 2253 (dis最短路径)

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原文地址：http://blog.csdn.net/my_acm/article/details/37991205