POJ 2387Til the Cows Come Home(最短单源路径)(dijkstra)

Til the Cows Come Home

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 59684 Accepted: 20312

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 51 2 202 3 303 4 204 5 201 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.

分析：最近在看图论，刚开始就是最短路，第一个是单源最短路，就看了四个算法中的dijkstra算法，也是最简单的一个。

这个题目是dijkstra算法的模板题目，在题目中注解我对于这个算法理解。

这个图片将dijkstra算法很形象的表达出来了。

#include<stdio.h>#include<iostream> #include <algorithm>#include<string.h>#include<vector>#include<math.h>#include<queue>#include<set>#define LL long long#define INF 0x3f3f3f3fusing namespace std;int KGCD(int a,int b){if(a==0)return b;if(b==0)return a;if(~a&1){ if(b&1) return KGCD(a>>1,b);else return KGCD(a>>1,b>>1) <<1; } if(~b & 1)  return KGCD(a, b>>1);  if(a > b) return KGCD((a-b)>>1, b);return KGCD((b-a)>>1, a);}  int LCM(int a,int b){ return a/KGCD(a,b)*b; } int dir[5][2]={0,1,0,-1,1,0,-1,0};using namespace std;int n,m;int map[1005][1005];//地图 int map1[1005];//单元最短路 int vis[1005];//记录访问 void dijkstra(){int k;memset(vis,0,sizeof(vis));vis[1]=1;//起点标记下for(int i=1;i<m;i++)//搜索点的个数次{int min=INF;k=1;for(int j=1;j<=m;j++)//每次搜索离出发点距离最短的点{if(!vis[j] && min > map1[j]){min=map1[j];k=j;}}vis[k]=1;//找到这个点之后标记一下 这个点已经用过了for(int j=1;j<=m;j++){//借助当前最短的那个点  开始判断借助这个点可以到达那些直接从源点到不了的点的距离if(!vis[j] && map1[j] > map1[k] + map[k][j])//判断所有的点借助这个点是否有更快的方法到达map1[j]=map1[k]+map[k][j];}} printf("%d\n",map1[m]);//一直存的单源最短  所有最后一个也就是源头到m点的距离}int main(){int x,y,z;while(scanf("%d%d",&n,&m)!=EOF){for(int i=1;i<=m;i++)//先整理出来地图 {map[i][i]=0;for(int j=1;j<i;j++)map[i][j]=map[j][i]=INF;}for(int i=0;i<n;i++){scanf("%d%d%d",&x,&y,&z);if(z<map[x][y])//避免重边  只要最小的 map[x][y]=map[y][x]=z;}for(int i=1;i<=m;i++){map1[i]=map[1][i];//从1到任何点的最短距离 }dijkstra();}return 0;}