Til the Cows Come Home POJ - 2387 （最短路）

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1到点n的最短路

代码一：79ms

#include
#include
#include
#include
#define Inf 0x3f3f3f3f
using namespace std;
const int N = 1005;
struct node{
	int to;
	int val;
};
vectorG[N];
int vis[N],dis[N];
int m,n;
void Dijkstra(int s){
	memset(vis,0,sizeof(vis));
	memset(dis,Inf,sizeof(dis));
	dis[s]=0;
	for(int k=0;kdis[i]){
				mint=dis[i];
				u=i;
			}
		}
		vis[u]=1;
		for(int i=0;idis[u]+G[u][i].val){
				dis[v]=dis[u]+G[u][i].val;
			}
		}
	}
	printf("%d\n",dis[n]);
}
int main(){
	scanf("%d%d",&m,&n);
	for(int i=0;i

代码二：47ms 

优先队列优化

#include
#include
#include
#include
#define Inf 0x3f3f3f3f
using namespace std;
const int N = 1005;
struct node{
	int to;
	int val;
};
struct Node{
	int u;
	int val;
	bool operator < (const Node other) const{
		return val>other.val;
	}
};
vectorG[N];
int vis[N],dis[N];
int m,n;
void Dijkstra(int s){
	memset(vis,0,sizeof(vis));
	memset(dis,Inf,sizeof(dis));
	priority_queuep;
	p.push((Node){s,0});
	dis[s]=0;
    while(!p.empty()){
    	Node temp=p.top();
    	p.pop();
    	int u=temp.u;
    	if(vis[u]) continue;
    	vis[u]=1;
    	for(int i=0;itemp.val+val){
    			dis[v]=temp.val+val;
    			p.push((Node){v,dis[v]});
			}
		}
 	}
	printf("%d\n",dis[n]);
}
int main(){
	scanf("%d%d",&m,&n);
	for(int i=0;i

代码三：79ms 

#include
#include
#include
#define Inf 0x3f3f3f3f
using namespace std;
const int N = 1005;
struct node{
	int u;
	int v;
	int val;
	int next;
}edge[N*4];
int cnt,head[N];
int dis[N];
int n,m;
void add(int a,int b,int c){
	edge[cnt].u=a;  //起点 
	edge[cnt].v=b;  //终点 
	edge[cnt].val=c;//权值 
	edge[cnt].next=head[a];
	head[a]=cnt++;
}
void init(){
	memset(head,-1,sizeof(head));
}
void BellmanFord(int s){    //遍历每条边 
	memset(dis,Inf,sizeof(dis));
	dis[s]=0;
	for(int i=1;i<=n;i++){
		for(int k=0;kdis[u]+val){
				dis[v]=dis[u]+val;
			}
		}
	}
	printf("%d\n",dis[n]);
}
int main(){
	scanf("%d%d",&m,&n);
	init();
	for(int i=0;i