#include<bits/stdc++.h>
using namespace std;
map<string,int>city;
map<int,string>rcity;
map<int,vector<pair<int,int> > >edge;//对比string要比对比int慢很多,所以转换映射
int dis[207],path[207],hcount[207],happ[207],fstep[207],f[207];//源点到各点的最短距离,最短路径数量,快乐总数,快乐,经过点的个数,前一个点
int vis[207];
int main(){
    memset(dis,-1,sizeof(dis));
    memset(hcount,-1,sizeof(hcount));
    std::ios::sync_with_stdio(false);//关闭同步
    int n,k,i,d,s;
    string st,u,v;
    cin>>n>>k>>st;
    city[st]=0;//编号
    rcity[0]=st;//反编号
    happ[0]=0;
    dis[0]=0;
    hcount[0]=0;
    fstep[0]=0;
    path[0]=1;//init
    f[0]=0;
    for(i=1;i<n;++i){
        f[i]=i;
        cin>>u;
        rcity[i]=u;
        city[u]=i;
        cin>>happ[i];
    }
    for(i=0;i<k;++i){
        cin>>u>>v>>d;
        edge[city[u]].push_back(make_pair(city[v],d));//建邻接表
        edge[city[v]].push_back(make_pair(city[u],d));
    }
    s=0;
    vector<pair<int,int> >::iterator it;
    int next;
    while(s!=city["ROM"]){
        vis[s]=1;
        for(it=edge[s].begin();it!=edge[s].end();++it){
            next=it->first;
            if(dis[next]==-1||dis[next]>dis[s]+it->second){//松弛
                dis[next]=dis[s]+it->second;
                hcount[next]=hcount[s]+happ[next];
                path[next]=path[s];
                fstep[next]=fstep[s]+1;
                f[next]=s;
            }
            else{
                if(dis[next]==dis[s]+it->second){
                    path[next]+=path[s];
                    if(hcount[next]<hcount[s]+happ[next]){
                        hcount[next]=hcount[s]+happ[next];
                        fstep[next]=fstep[s]+1;
                        f[next]=s;
                    }
                    else{
                        if(hcount[next]==hcount[s]+happ[next]){
                            if(fstep[next]>fstep[s]+1){
                                fstep[next]=fstep[s]+1;
                                f[next]=s;
                            }
                        }
                    }
                }
            }
        }
        int mindis=-1,minnum;
        for(i=1;i<n;i++){
            if(dis[i]==-1)//如果当前边到不了初始点,直接pass
                continue;
            if(!vis[i]&&(mindis==-1||(dis[i]<mindis))){
                mindis=dis[i];
                minnum=i;
            }
        }
        s=minnum;//找到当前距离源点最近的一个点向下搜索
    }
    cout<<path[s]<<" "<<dis[s]<<" "<<hcount[s]<<" "<<hcount[s]/fstep[s]<<endl;
    int p=s;
    stack<int>ss;
    while(p){
        ss.push(p);
        p=f[p];
    }
    cout<<rcity[p];
    while(!ss.empty()){
        cout<<"->"<<rcity[ss.top()];
        ss.pop();
    }
    return 0;
}

/*#include<bits/stdc++.h>
using namespace std;
const int INF = 1e9 ;
typedef struct node{
    int v;//节点编号
    int cost;//边权值
};
int N;//城市数量
int K;//道路条数
string start;//起始点城市名字
int happy[207];//存放各个城市的快乐值
map<string,int>stringToInt;//城市名字->编号
map<int,string>intToString;//编号->城市名字
vector<node>graph[207];//邻接表
int d[207];//记录从起始城市到达点i的最少花费值
vector<int>pre[207];//记录前一个节点
vector<int>path,tempPath;
int optHappy=0;
double optAverageHappy=0;
int cnt=0;//记录最短路径条数
int inq[207];//记录节点是否在队列中
void spfa(int s){
    for(int i=0;i<N;++i)
        d[i]=INF;
    d[s]=0;
    queue<int>q;
    q.push(s);
    inq[s]=1;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        inq[u]=0;
        for(int j=0;j<graph[u].size();++j){
            int v=graph[u][j].v;
            int cost=graph[u][j].cost;
            if(!v)
                continue;
            if(d[u]+cost<d[v]){
                d[v]=d[u]+cost;
                pre[v].clear();
                pre[v].push_back(u);
                if(!inq[v]){
                    q.push(v);
                    inq[v]=1;
                }
            }
            else if(d[u]+cost==d[v])
                pre[v].push_back(u);
        }
    }
}
void dfs(int nowVisit){
    if(!nowVisit){
        cnt++;
        tempPath.push_back(nowVisit);
        int happyValue=0;
        for(int i=tempPath.size()-2;i>=0;--i)
            happyValue+= happy[tempPath[i]];
        double averageHappyValue=1.0*happyValue/(tempPath.size()-1);
        if(happyValue>optHappy){
            optHappy=happyValue;
            optAverageHappy=averageHappyValue;
            path=tempPath;
        }
        else if(happyValue==optHappy&&averageHappyValue>optAverageHappy){
            optAverageHappy=averageHappyValue;
            path=tempPath;
        }
        tempPath.pop_back();
        return;
    }
    tempPath.push_back(nowVisit);
    for(auto&it:pre[nowVisit])
  dfs(it);
    tempPath.pop_back();
}
int main(){
    std::ios::sync_with_stdio(false);//关闭同步
    cin>>N>>K>>start;
    stringToInt[start]=0; //起始城市编号为0
    intToString[0]=start;
    string city;
    int happyness;
    for(int i=1;i<N;++i){
        cin>>city>>happyness;
        stringToInt[city]=i;
        intToString[i]=city;
        happy[i]=happyness;
    }
    string city1,city2;
    int cost;
    for(int i=0;i<K;++i){
        cin>>city1>>city2>>cost;
        int id1=stringToInt[city1];
        int id2=stringToInt[city2];
        node x,y;
        x.v=id2;
        x.cost=cost;
        y.v=id1;
        y.cost=cost;
        graph[id1].push_back(x);
        graph[id2].push_back(y);
    }
    int destination=stringToInt["ROM"];
    spfa(0);
    dfs(destination);
    cout<<cnt<<" "<<d[destination]<<" "<<optHappy<<" "<<(int)optAverageHappy<<endl;
    for(int i=path.size()-1;i>=0;--i) {
        cout<<intToString[path[i]];
        if(i)
            cout<<"->";
    }
    return 0;
}*/

//SPFA先处理再DFS

All Roads Lead to Rome(30)(MAP【int,string】,邻接表,DFS,模拟,SPFA)(PAT甲级)的更多相关文章

  1. [图的遍历&多标准] 1087. All Roads Lead to Rome (30)

    1087. All Roads Lead to Rome (30) Indeed there are many different tourist routes from our city to Ro ...

  2. pat1087. All Roads Lead to Rome (30)

    1087. All Roads Lead to Rome (30) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...

  3. 1087. All Roads Lead to Rome (30)

    时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Indeed there are many different ...

  4. PAT甲级练习 1087 All Roads Lead to Rome (30分) 字符串hash + dijkstra

    题目分析: 这题我在写的时候在PTA提交能过但是在牛客网就WA了一个点,先写一下思路留个坑 这题的简单来说就是需要找一条最短路->最开心->点最少(平均幸福指数自然就高了),由于本题给出的 ...

  5. PAT (Advanced Level) 1087. All Roads Lead to Rome (30)

    暴力DFS. #include<cstdio> #include<cstring> #include<cmath> #include<vector> # ...

  6. 1087 All Roads Lead to Rome (30)(30 分)

    Indeed there are many different tourist routes from our city to Rome. You are supposed to find your ...

  7. 【PAT甲级】1087 All Roads Lead to Rome (30 分)(dijkstra+dfs或dijkstra+记录路径)

    题意: 输入两个正整数N和K(2<=N<=200),代表城市的数量和道路的数量.接着输入起点城市的名称(所有城市的名字均用三个大写字母表示),接着输入N-1行每行包括一个城市的名字和到达该 ...

  8. PAT 1087 All Roads Lead to Rome[图论][迪杰斯特拉+dfs]

    1087 All Roads Lead to Rome (30)(30 分) Indeed there are many different tourist routes from our city ...

  9. PAT_A1087#All Roads Lead to Rome

    Source: PAT A1087 All Roads Lead to Rome (30 分) Description: Indeed there are many different tourist ...

随机推荐

  1. Eclipse_插件_01_tomcat插件的安装

    1.Eclipse的tomcat插件下载地址: (1)https://sourceforge.net/projects/tomcatplugin/files/updatesite/plugins/ ( ...

  2. (转)python调取C/C++的dll生成方法

    本文针对Windows平台下,python调取C/C++的dll文件. 1.如果使用C语言,代码如下,文件名为test.c. __declspec(dllexport) int sum(int a,i ...

  3. Oracle忘记用户名密码

    一.oracle 11g登录服务开启 成功安装Oracle 11g后,共有7个服务,这七个服务的含义分别为:1. Oracle ORCL VSS Writer Service:Oracle卷映射拷贝写 ...

  4. QToolBox

    QToolBox类似与以前qq好友分组的那种控件.每个分组是一个Item. 一.添加分组: 其中每个分组是通过一下函数添加的: int addItem(QWidget * w, const QIcon ...

  5. WampServer 2.5设置外网访问/局域网手机访问(403 Forbidden错误解决方法)

    安装好wamp后,想用手机通过局域访问电脑上wamp下的网页,结果出现如下提示. (403 Forbidden)错误 1.打开http.conf文件 2.找到下图中红色方框中的onlineofflin ...

  6. centos下安装Mysql5.7.20

    0.环境 本文操作系统: CentOS 7.2.1511 x86_64MySQL 版本: 5.7.16 1.卸载系统自带的 mariadb-lib [root@centos-linux ~]# rpm ...

  7. NYOJ-求逆序数 ----------------待解决,WA

    描述 在一个排列中,如果一对数的前后位置与大小顺序相反,即前面的数大于后面的数,那么它们就称为一个逆序.一个排列中逆序的总数就称为这个排列的逆序数. 现在,给你一个N个元素的序列,请你判断出它的逆序数 ...

  8. BZOJ2809:[APIO2012]dispatching

    浅谈左偏树:https://www.cnblogs.com/AKMer/p/10246635.html 题目传送门:https://lydsy.com/JudgeOnline/problem.php? ...

  9. Android 开发:开源库Speex支持arm64的动态库文件

    随着处理器制造工艺的不断进步,和Android系统的不断发展,最近出了arm64-v8a的架构,由于项目中用到了speex的第三方语音编解码的动态库,其他架构的处理器暂不用说,一切正常,唯独到arm6 ...

  10. JavaScript下的进制转换

    JavaScript下的进制转换 //十进制转其他进制 var num = 99; console.log('十进制: ', num); console.log('八进制:', (num).toStr ...