poj 2253 Frogger [dijkstra]

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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

思路：最短路or最小生成树，松弛过程中，取点时收入集合边的最大值，即Answer；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

typedef long long LL;
const int N=1002;

int x[N],y[N];
double gra[N][N];
double dist[N];
int vis[N];
double ans;

void dijkstra(int n)
{
    for(int i=1;i<=n;i++)
    {
        dist[i]=INF;
        vis[i]=0;
    }
    dist[1]=0;
    for(int i=1;i<=n;i++)
    {
        int pos=-1;
        double minn=INF;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dist[j]<minn)
            {
                minn=dist[j];
                pos=j;
            }
        }
        if(ans<minn)
            ans=minn;
        if(pos==2) break;
        vis[pos]=1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dist[j]>gra[pos][j])
            {
                dist[j]=gra[pos][j];
            }
        }
    }

}

int main()
{
    int n;
    int cas=0;
    while(scanf("%d",&n)==1)
    {
        if(n==0) break;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",x+i,y+i);
        }
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                gra[i][j]=gra[j][i]=(double)sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
            }
        }
        ans=0;
        dijkstra(n);
        printf("Scenario #%d\nFrog Distance = %.3f\n\n",++cas,ans);
    }
    return 0;
}

poj 2253 Frogger [dijkstra]

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原文地址：http://blog.csdn.net/code_or_code/article/details/43793353