Description

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.

大致題意：有多組資料，每組資料首先輸入T和n，表示有T條邊和n個點，接下來輸入每兩個點的編號和連線他們的一條邊的長度。問從1到n的最短路程。保證1到n有一條通路。
（注意有多重邊）
思路：Dijkstra求最短路

程式碼如下

程式碼1

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>

using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=1005;
int maps[maxn][maxn];//存兩點間的距離
int dis[maxn];//表示源點到該點的最短路 
int n;//節點數 

struct node
{
    int num;
    int dis;
    friend bool operator <(node a,node b)
    {
        if(a.dis==b.dis) return a.num>b.num;
        return a.dis>b.dis;
    }
};
priority_queue<node> que;//用優先佇列優化

void init()
{
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    maps[i][j]=INF;
}
void Dijkstra(int s)
{
    int vis[maxn];//表示該點是否已經訪問過 
    for(int i=1;i<=n;i++)
    dis[i]=INF;
    memset(vis,0,sizeof(vis));

    dis[s]=0;
    node u;
    u.dis=0;
    u.num=s;
    que.push(u);
    while(!que.empty())
    {
        node u=que.top();
        que.pop();
        int x=u.num;

        if(vis[x])
            continue;
        vis[x]=1;   

        for(int i=1;i<=n;i++)
        {
            if(!vis[i]&&dis[i]>dis[u.num]+maps[u.num][i])
            {
                dis[i]=dis[u.num]+maps[u.num][i];
                node v;
                v.num=i;
                v.dis=dis[i];
                que.push(v);
            }
        }
    }
    printf("%d\n",dis[n]);
 } 

 int main()
 {

    int T,x,y,D;
    while(scanf("%d%d",&T,&n)!=EOF)
    {
        init();
        for(int i=1;i<=T;i++)
        {
            scanf("%d%d%d",&x,&y,&D);
            if(D<maps[x][y])//有多重邊，記錄最短的一條即可
            {
             maps[y][x]=maps[x][y]=D;
            }
         }
         Dijkstra(1);
     }
     return 0;
 }

程式碼2

#include<stdio.h>
#include<string.h>
#include<stdio.h>
#include<cstdio>
#include<iostream>
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=1005;
int maps[maxn][maxn];
int vis[maxn];//表示該點是否已經遍歷過 
int dis[maxn];//表示源點到該點的最短路 
int n;//節點數 
void Dijkstra(int s,int e)
{

    memset(dis,INF,sizeof(dis));
    dis[s]=0;
    for(int i=1;i<=n;i++)
    {
        int mini=INF;
        int k=-1;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]<mini)
            {
                k=j;
                mini=dis[j];
            }
        }
        vis[k]=1;
        if(k==e)
        {
            printf("%d\n",dis[k]);
            return;
        }
        for(int j=1;j<=n;j++)
        {
            if(!vis[j])
            {
                dis[j]=min(dis[k]+maps[k][j],dis[j]);
            }
        }
    }
 } 

 int main()
 {

    int T,x,y,D;

    while(scanf("%d%d",&T,&n)!=EOF)
    {
        memset(vis,0,sizeof(vis));
        memset(maps,INF,sizeof(maps));
        for(int i=1;i<=T;i++)
        {

            scanf("%d%d%d",&x,&y,&D);
            if(D<maps[x][y])
            {
             maps[x][y]=D;
             maps[y][x]=D;
            }
         }
         Dijkstra(1,n);
     }
     return 0;
 }