Til the Cows Come Home POJ - 2387 （迪杰斯特拉！模板！）

 

Til the Cows Come Home

 POJ - 2387 

 

 

 

 

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. 

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it. 

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N 

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90

Hint

INPUT DETAILS: 

There are five landmarks. 

OUTPUT DETAILS: 

Bessie can get home by following trails 4, 3, 2, and 1.

 

 

题意：求从某点（本题为点1）到点n的最短路径

思路：迪杰斯特拉算法。

注意：本题需要考虑重边。

AC代码（已优化模板）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cmath>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn = 10010, INF = 0xfffffff;
int t,n;
int vis[maxn];
int d[maxn];

struct edge{  //定义边：起点u、终点v、长度d
    int u,v;
    int d;
    edge(int u,int v,int d)
    {
        this->u = u;
        this->v = v;
        this->d = d;

    }
};

vector<edge> Edge;
vector<int> G[maxn];

struct node{  //定义以u为起点距离为d的集合
    int u;
    int d;
    node(int d,int u)
    {
        this->d = d;
        this->u = u;
    }
    bool operator < (const node& a) const {
        return d > a.d;
    }
};

void add(int u,int v,int d)
{
    Edge.push_back(edge(u,v,d));    //存放所有边
    G[u].push_back(Edge.size()-1);  //以u为起点的边在Edge中的序号
}
void dijkstra(int s)
{
    priority_queue<node> Q;
    for(int i=0;i<=n;i++) d[i] = INF;
    d[s] = 0;
    mem(vis,0);
    Q.push(node(0,s)); //长度为0，点s为起点
    while(!Q.empty())
    {
        node x = Q.top();Q.pop();
        int u = x.u;
        if(vis[u]) continue;
        vis[u] = 1;
        for(int i=0;i<G[u].size();i++)
        {
            edge e = Edge[G[u][i]];
            if(d[e.v] > d[u] + e.d)
            {
                d[e.v] = d[u] + e.d;
                Q.push(node(d[e.v],e.v));
            }
        }
    }
}
int main()
{
    int x,y,len;
    int cnt = 0;
    while(cin>>t>>n)
    {
        for(int i=0;i<=n;i++) G[i].clear();
        Edge.clear();
        for(int i=1;i<=t;i++)
        {
            cin>>x>>y>>len;
            add(x,y,len);
            add(y,x,len);
        }
        dijkstra(1);
        printf("%d\n",d[n]);
    }
    return 0;
}

 

AC代码（c++）：

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x7fffffff
#define maxn 2010
using namespace std;

int Map[maxn][maxn];
int vis[maxn];
int dis[maxn];
int n,m;
void Dijs()
{
	int i,j;
	memset(dis,INF,sizeof(dis));
	dis[1]=0;
	for(i=2;i<=n;++i){
		dis[i]=Map[1][i];
	}
	memset(vis,0,sizeof(vis));
	vis[1]=1;
	int v,Min;
	for(i=1;i<=n;++i){                 //找出从前一点出发走最短路径能到达的点  
		Min=INF;
		for(j=1;j<=n;++j){
			if(!vis[j]&&dis[j]<Min){
				v=j;
				Min=dis[j];
			}
		}
		vis[v]=1;
		for(j=1;j<=n;++j){             //比较出从起点到达点j的最短路径
			if(!vis[j]&&Map[v][j]<INF){
				dis[j]=min(dis[j],dis[v]+Map[v][j]);
			}
		}
	}
	printf("%d\n",dis[n]);
}
int main()
{
	int a,b,c;
	int i,j;
	while(scanf("%d%d",&m,&n)!=EOF){
		for(i=0;i<=n;++i){
			for(j=0;j<=n;++j){
				if(i==j)Map[i][j]=0;
				Map[i][j]=INF;
			}
		}
		for(i=1;i<=m;++i){
			scanf("%d%d%d",&a,&b,&c);
			if(Map[a][b]>c){     //当出现重边时，留取最短的一条 
				Map[a][b]=c;
				Map[b][a]=c;
			}
		}
		Dijs();
	}
	return 0;
}

AC代码（Java）：

import java.math.*;
import java.util.*;
public class Main {
	final static int INF=0x7fffffff;
	static int [][]Map=new int[2010][2010];
	static int []vis=new int[2010];
	static int []dis=new int[2010];
	static int n,m,a,b,c;
	public static void main(String[] args) {
		Scanner cin=new Scanner(System.in);
		while(cin.hasNext()){
			m=cin.nextInt();
			n=cin.nextInt();
			for(int i=0;i<=n;++i){
				for(int j=0;j<=n;++j){
					if(i==j)Map[i][j]=0;
					Map[i][j]=INF;
				}
			}
			for(int i=1;i<=m;++i){
				a=cin.nextInt();
				b=cin.nextInt();
				c=cin.nextInt();
				if(Map[a][b]>c){
					Map[a][b]=c;
					Map[b][a]=c;
				}
			}
			Dijs();
		}
	}
	private static void Dijs(){
		Arrays.fill(dis, 0);
		dis[1]=0;
		for(int i=2;i<=n;++i){
			dis[i]=Map[1][i];	//记录从点1到各个点的距离
		}
		Arrays.fill(vis, 0);
		vis[1]=1;	//第一个点已经访问过了，从第二步开始将不再访问第一个点
		int v=0,Min;	//java需要先赋值才能用
		for(int i=1;i<=n;++i){
			Min=INF;
			for(int j=1;j<=n;++j){
				if(vis[j]!=1&&dis[j]<Min){
					v=j;	//v点代表从前一点出发能到达的一点,循环结束后找到最近的一点
					Min=dis[j];
				}
			}
			vis[v]=1;
			for(int j=1;j<=n;++j){
				if(vis[j]!=1&&Map[v][j]<INF){
					dis[j]=Math.min(dis[j],dis[v]+Map[v][j]);//java中的min函数
				}
			}
		}
		System.out.println(dis[n]);
	}
}