POJ 2387 Til the Cows Come Home(Dijkstra简单题)

POJ 2387 Til the Cows Come Home(Dijkstra简单题)

http://poj.org/problem?id=2387

题意:

        有N个顶点和T条边的无向图,现在要问你从1号顶点到N号顶点的最短距离是多少?

分析:  

        直接通刘汝佳的Dijkstra模板,不过注意题中可能有重边.(不过不影响)

AC代码:

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>#include<queue>using namespace std;const int maxn=1000+10;const int maxm=2000+10;int n,m;struct Edge{    int from,to,dist;    Edge(int f,int t,int d):from(f),to(t),dist(d){}};struct HeapNode{    int d,u;    HeapNode(int d,int u):d(d),u(u){}    bool operator < (const HeapNode &rhs) const    {        return d > rhs.d;    }};struct Dijkstra{    int n,m;    vector<Edge> edges;    vector<int> G[maxn];    bool done[maxn];    int d[maxn];    void init(int n)    {        this->n = n;        for(int i=0;i<n;i++) G[i].clear();        edges.clear();    }    void AddEdge(int from,int to,int dist)    {        edges.push_back(Edge(from,to,dist) );        m = edges.size();        G[from].push_back(m-1);    }    int dijkstra()    {        priority_queue<HeapNode> Q;        for(int i=0;i<n;i++) d[i]=1e8;        d[0]=0;        Q.push(HeapNode(0,0));        memset(done,0,sizeof(done));        while(!Q.empty())        {            HeapNode x= Q.top(); Q.pop();            int u =x.u;            if(done[u]) continue;            done[u]=true;            for(int i=0;i<G[u].size();i++)            {                Edge &e=edges[G[u][i]];                if(d[e.to] > d[u]+e.dist)                {                    d[e.to] = d[u]+e.dist;                    Q.push(HeapNode(d[e.to],e.to));                }            }        }        return d[n-1];    }}DJ;int main(){    while(scanf("%d%d",&m,&n)==2)    {        DJ.init(n);        for(int i=0;i<m;i++)        {            int u,v,d;            scanf("%d%d%d",&u,&v,&d);            u--,v--;            DJ.AddEdge(u,v,d);            DJ.AddEdge(v,u,d);        }        printf("%d\n",DJ.dijkstra());    }    return 0;}