poj 2387 Til the Cows Come Home （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

用优先队列优化的dijkstara，即使有负边也可以，相当于spfa，可以使用链式前向星来存图，时间上会优化很多。
记录模板

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <queue>#include <algorithm>#define maxn 4005#define inf 0x3f3f3f3fusing namespace std;//链式前向星部分int head[maxn];int cnt;struct Edge{    int to,next,w;}edge[maxn];void add(int u,int v,int w){    edge[cnt].to=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;}void init(){    memset(head,-1,sizeof(head));    cnt=0;}//dijkstrastruct node{    int dis,num;//顶点距s距离，定点编号    node() {};    node(int x,int d){dis=d;num=x;}    bool operator < (const node& a) const    {        if(dis==a.dis) return num<a.num;        else return dis>a.dis;  //距离小的先出    }};priority_queue<node> que;int d[maxn];void dijkstra(int s){    memset(d,0x3f,sizeof(d));    d[s]=0;    que.push(node(s,0));    while(!que.empty())    {        node now=que.top();        que.pop();        int x=now.num;        for(int i=head[x];~i;i=edge[i].next)        {            int v=edge[i].to;            if(d[v]>d[x]+edge[i].w)            {                d[v]=d[x]+edge[i].w;                que.push(node(v,d[v]));            }        }    }}int main(){    int t,n;    init();    cin>>t>>n;    for(int i=0;i<t;i++)    {        int u,v,w;        cin>>u>>v>>w;        add(u,v,w);        add(v,u,w);    }    dijkstra(1);    cout<<d[n]<<endl;    return 0;}

把优先队列里的结构体用pair代替，简化一下

int n,m;struct Edge{    int to,next,w;} edge[maxn];int head[maxn];int cnt;void add(int u,int v,int w){    edge[cnt].w=w;    edge[cnt].to=v;    edge[cnt].next=head[u];    head[u]=cnt++;}void init(){    memset(head,-1,sizeof(head));    cnt=0;}int dis[maxn];void dijkstra(int s){    memset(dis,0x3f,sizeof(dis));    dis[s]=0;    priority_queue<PII,vector<PII>,greater<PII> > que;  //first中存放距离,小的先出队    que.push(make_pair(0,s));    while(!que.empty())    {        PII now=que.top();        que.pop();        int x=now.se;        for(int i=head[x]; ~i; i=edge[i].next)        {            int v=edge[i].to;            int w=edge[i].w;            if(dis[v]>dis[x]+w)            {                dis[v]=dis[x]+w;                que.push(make_pair(dis[v],v));            }        }    }}