POJ 2387 Til the Cows Come Home BFS最短路求解

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题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=66569#problem/A

题意：给定某些点之间的距离，起终点一定，求回到目的地的最短距离。

以前用的是Djikstra写的，矩阵的最短路写多了，第一想到的就是BFS，Dijkstra的循环写法似乎有点忘记了。习惯性的还是比较喜欢用邻接矩阵来存储，练习下邻接表的写法。

邻接矩阵：

#include<iostream>#include<sstream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<set>#include<map>#include<queue>#include<vector>#include<string>#define LL long long#define INF 0x7fffffffusing namespace std;int Map[1100][1100],dis[1100],vis[1100];int T,N;void bfs(){    queue<int> Q;    memset(dis,-1,sizeof(dis));    memset(vis,0,sizeof(vis));    Q.push(N);vis[N]=1;dis[N]=0;    while(!Q.empty()){        int tp=Q.front();        Q.pop();vis[tp]=0;        for(int i=1;i<=N;i++){            if(Map[tp][i]==-1) continue;            if(dis[i]!=-1 && dis[tp]+Map[tp][i]>=dis[i]) continue;            dis[i]=dis[tp]+Map[tp][i];            if(!vis[i]) Q.push(i),vis[i]=1;        }    }    cout<<dis[1]<<endl;}int main(){    //freopen("D:\\in.txt","r",stdin);    while(cin>>T){        cin>>N;        int a,b,c;        memset(Map,-1,sizeof(Map));        for(int i=0;i<T;i++){            cin>>a>>b>>c;            if(Map[a][b]==-1) Map[a][b]=Map[b][a]=c;            else Map[a][b]=Map[b][a]=min(Map[a][b],c);        }        bfs();    }    return 0;}

邻接表：

#include<iostream>#include<sstream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<set>#include<map>#include<queue>#include<vector>#include<string>#define LL long long#define INF 0x7fffffffusing namespace std;struct edge{    int u,v,w;};int N,T,dis[1100],vis[1100];vector<int> G[1100];vector<edge> E;void bfs(){    memset(dis,-1,sizeof(dis));memset(vis,0,sizeof(vis));    queue<int> Q;    Q.push(N);vis[N]=1;dis[N]=0;    while(!Q.empty()){        int tp=Q.front();Q.pop();vis[tp]=0;        for(int p=0;p<(int)G[tp].size();p++){            edge &e=E[G[tp][p]];            int u=e.u,v=e.v,w=e.w;            if(dis[v]!=-1 && dis[u]+w>=dis[v]) continue;            dis[v]=dis[u]+w;            if(!vis[v]) Q.push(v),vis[v]=1;        }    }    cout<<dis[1]<<endl;}int main(){    //freopen("D:\\in.txt","r",stdin);    while(cin>>T){        cin>>N;        for(int i=0;i<=N;i++) G[i].clear();E.clear();        int a,b,c;        for(int i=0;i<T;i++){            cin>>a>>b>>c;            E.push_back((edge){a,b,c});            G[a].push_back((int)E.size()-1);            E.push_back((edge){b,a,c});            G[b].push_back((int)E.size()-1);        }        bfs();    }    return 0;}

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