pat 1087 （dfs）

原题：

 

代码（dfs）：

#include<iostream>
#include<vector>
#include<unordered_map>
#include<limits.h>
using namespace std;

int mindis[201];
vector <int> path;
int num_way;//number of the same ways
int destination;
string namelist[200];
int happy[200];
int dis[200][200];
vector<int> final_path;
int final_dis,final_happy,final_step;
vector<int> v[200];// its neghbor
unordered_map<string,int>city_num;

void dfs (int curcity,int curdis,int curhappy,int curstep)
{
    if( curdis> mindis[curdis]) return;//stop loss
    path.emplace_back(curcity);
    if(curcity == destination)//如果到了终点
    {
        if(curdis < mindis[destination])
        {
            mindis[destination] = curdis;
            final_path = path;
            final_dis = curdis;
            final_happy = curhappy;
            final_step = curstep;
            num_way = 1;
        }
        if(curdis == mindis[destination])   //equal case
        {
            num_way++;
            if(curhappy> final_happy || (curhappy == final_happy && curstep < final_step))
            {
                final_path = path;
                final_dis = curdis;
                final_happy = curhappy;
                final_step = curstep;
            }
        }
    }
    else // not reach the destination
    {
        if( curdis <  mindis[curdis])
               mindis[curcity] = curdis;
        for(int each:v[curcity])//find the neighbor
        {
            dfs(each,curdis+dis[curcity][each],curhappy+happy[each],curstep+1);
        }
    }
    path.pop_back();    
}


int main()
{
    int N,K; //N是城市数，K是路的数量   
    string s_city;
    cin>>N>>K>>s_city;    city_num[s_city] = 0; namelist[0] = s_city;
    for(int i =1;i<=N-1;i++) mindis[i] = INT_MAX;
    //接下来的N-1行： 城市名 幸福值
    string city;happy[0] = 0;
    for(int i = 1;i<=N-1;i++ )
    {
        cin>>city>>happy[i];
        city_num[city] = i; //将城市转化为数字序号便于dfs
        namelist[i] = city; 
    }
    
    //接下来存城市之间的距离，用城市序号表示
    string s1,s2;int d;
    int c1,c2;
    while(K--)
    {
        cin>>s1>>s2>>d;
        dis[city_num[s1]][city_num[s2]] = d;
        dis[city_num[s2]][city_num[s1]] = d;
        
        c1 = city_num[s1];c2= city_num[s2];
        v[c1].emplace_back(c2);
        v[c2].emplace_back(c1);
    }
    destination =city_num["ROM"];
    dfs(0,0,0,0);
    cout<<num_way<<" "<<final_dis<<" "<<final_happy<<" "<<final_happy/final_step<<endl;
    cout<<s_city;
    for(unsigned int i = 1;i<final_path.size();i++)
        cout<<"->"<<namelist[final_path[i]];
    return 0;
    
}

结果只是部分正确：

 如果改一下，

 if(curdis == mindis[destination])   //equal case
其实可以改成else

代码：

#include<iostream>
#include<vector>
#include<unordered_map>
#include<limits.h>
using namespace std;

int mindis[201];
vector <int> path;
int num_way;//number of the same ways
int destination;
string namelist[200];
int happy[200];
int dis[200][200];
vector<int> final_path;
int final_dis,final_happy,final_step;
vector<int> v[200];// its neghbor
unordered_map<string,int>city_num;

void dfs (int curcity,int curdis,int curhappy,int curstep)
{
    if( curdis> mindis[curdis]) return;//stop loss
    path.emplace_back(curcity);
    if(curcity == destination)//如果到了终点
    {
        if(curdis < mindis[destination])
        {
            mindis[destination] = curdis;
            final_path = path;
            final_dis = curdis;
            final_happy = curhappy;
            final_step = curstep;
            num_way = 1;
        }
        else  //equal case
        {
            num_way++;
            if(curhappy> final_happy || (curhappy == final_happy && curstep < final_step))
            {
                final_path = path;
                final_dis = curdis;
                final_happy = curhappy;
                final_step = curstep;
            }
        }
    }
    else // not reach the destination
    {
        if( curdis <  mindis[curdis])
               mindis[curcity] = curdis;
        for(int each:v[curcity])//find the neighbor
        {
            dfs(each,curdis+dis[curcity][each],curhappy+happy[each],curstep+1);
        }
    }
    path.pop_back();    
}


int main()
{
    int N,K; //N是城市数，K是路的数量   
    string s_city;
    cin>>N>>K>>s_city;    city_num[s_city] = 0; namelist[0] = s_city;
    for(int i =1;i<=N-1;i++) mindis[i] = INT_MAX;
    //接下来的N-1行： 城市名 幸福值
    string city;happy[0] = 0;
    for(int i = 1;i<=N-1;i++ )
    {
        cin>>city>>happy[i];
        city_num[city] = i; //将城市转化为数字序号便于dfs
        namelist[i] = city; 
    }
    
    //接下来存城市之间的距离，用城市序号表示
    string s1,s2;int d;
    int c1,c2;
    while(K--)
    {
        cin>>s1>>s2>>d;
        dis[city_num[s1]][city_num[s2]] = d;
        dis[city_num[s2]][city_num[s1]] = d;
        
        c1 = city_num[s1];c2= city_num[s2];
        v[c1].emplace_back(c2);
        v[c2].emplace_back(c1);
    }
    destination =city_num["ROM"];
    dfs(0,0,0,0);
    cout<<num_way<<" "<<final_dis<<" "<<final_happy<<" "<<final_happy/final_step<<endl;
    cout<<s_city;
    for(unsigned int i = 1;i<final_path.size();i++)
        cout<<"->"<<namelist[final_path[i]];
    return 0;
    
}

最终竟然全错、。。。

感觉应该是很简单的一个dfs，不知道为什么这个num_way 总是错误

找了一晚上，后来很生气地发现：

 有个地方 curdis和curcity太像了，变量名起的像，都有current 想表示现在的距离和城市

我给搞混了

 之后代码是：

#include<iostream>
#include<vector>
#include<unordered_map>
#include<limits.h>
using namespace std;

int mindis[201];
vector <int> path;
int num_way;//number of the same ways
int destination;
string namelist[200];
int happy[200];
int dis[200][200];
vector<int> final_path;
int final_dis,final_happy,final_step;
vector<int> v[200];// its neghbor
unordered_map<string,int>city_num;

void dfs (int curcity,int curdis,int curhappy,int curstep)
{
    if( curdis> mindis[curcity]) return;//stop loss
    path.emplace_back(curcity);
    if(curcity == destination)//如果到了终点
    {
        if(curdis < mindis[destination])
        {
            mindis[destination] = curdis;
            final_path = path;
            final_dis = curdis;
            final_happy = curhappy;
            final_step = curstep;
            num_way = 1;
        }
        else  //equal case
        {
            num_way++;
            if(curhappy> final_happy || (curhappy == final_happy && curstep < final_step))
            {
                final_path = path;
                final_dis = curdis;
                final_happy = curhappy;
                final_step = curstep;
            }
        }
    }
    else // not reach the destination
    {
        if( curdis <  mindis[curdis])
               mindis[curcity] = curdis;
        for(int each:v[curcity])//find the neighbor
        {
            dfs(each,curdis+dis[curcity][each],curhappy+happy[each],curstep+1);
        }
    }
    path.pop_back();    
}


int main()
{
    int N,K; //N是城市数，K是路的数量   
    string s_city;
    cin>>N>>K>>s_city;    city_num[s_city] = 0; namelist[0] = s_city;
    for(int i =1;i<=N-1;i++) mindis[i] = INT_MAX;
    //接下来的N-1行： 城市名 幸福值
    string city;happy[0] = 0;
    for(int i = 1;i<=N-1;i++ )
    {
        cin>>city>>happy[i];
        city_num[city] = i; //将城市转化为数字序号便于dfs
        namelist[i] = city; 
    }
    
    //接下来存城市之间的距离，用城市序号表示
    string s1,s2;int d;
    int c1,c2;
    while(K--)
    {
        cin>>s1>>s2>>d;
        dis[city_num[s1]][city_num[s2]] = d;
        dis[city_num[s2]][city_num[s1]] = d;
        
        c1 = city_num[s1];c2= city_num[s2];
        v[c1].emplace_back(c2);
        v[c2].emplace_back(c1);
    }
    destination =city_num["ROM"];
    dfs(0,0,0,0);
    cout<<num_way<<" "<<final_dis<<" "<<final_happy<<" "<<final_happy/final_step<<endl;
    cout<<s_city;
    for(unsigned int i = 1;i<final_path.size();i++)
        cout<<"->"<<namelist[final_path[i]];
    return 0;
    
}

 结果是有一例超时了

加个快读

#include<iostream>
#include<vector>
#include<unordered_map>
#include<limits.h>
using namespace std;

int mindis[201];
vector <int> path;
int num_way;//number of the same ways
int destination;
string namelist[200];
int happy[200];
int dis[200][200];
vector<int> final_path;
int final_dis,final_happy,final_step;
vector<int> v[200];// its neghbor
unordered_map<string,int>city_num;



void dfs (int curcity,int curdis,int curhappy,int curstep)
{
    if( curdis> mindis[curcity]) return;//stop loss
    path.emplace_back(curcity);
    if(curcity == destination)//如果到了终点
    {
        if(curdis < mindis[destination])
        {
            mindis[destination] = curdis;
            final_path = path;
            final_dis = curdis;
            final_happy = curhappy;
            final_step = curstep;
            num_way = 1;
        }
        else  //equal case
        {
            num_way++;
            if(curhappy> final_happy || (curhappy == final_happy && curstep < final_step))
            {
                final_path = path;
                final_dis = curdis;
                final_happy = curhappy;
                final_step = curstep;
            }
        }
    }
    else // not reach the destination
    {
        if( curdis <  mindis[curdis])
               mindis[curcity] = curdis;
        for(int each:v[curcity])//find the neighbor
        {
            dfs(each,curdis+dis[curcity][each],curhappy+happy[each],curstep+1);
        }
    }
    path.pop_back();    
}


int main()
{
    ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int N,K; //N是城市数，K是路的数量   
    string s_city;
    cin>>N>>K>>s_city;    city_num[s_city] = 0; namelist[0] = s_city;
    for(int i =1;i<=N-1;i++) mindis[i] = INT_MAX;
    //接下来的N-1行： 城市名 幸福值
    string city;happy[0] = 0;
    for(int i = 1;i<=N-1;i++ )
    {
        cin>>city>>happy[i];
        city_num[city] = i; //将城市转化为数字序号便于dfs
        namelist[i] = city; 
    }
    
    //接下来存城市之间的距离，用城市序号表示
    string s1,s2;int d;
    int c1,c2;
    while(K--)
    {
        cin>>s1>>s2>>d;
        dis[city_num[s1]][city_num[s2]] = d;
        dis[city_num[s2]][city_num[s1]] = d;
        
        c1 = city_num[s1];c2= city_num[s2];
        v[c1].emplace_back(c2);
        v[c2].emplace_back(c1);
    }
    destination =city_num["ROM"];
    dfs(0,0,0,0);
    cout<<num_way<<" "<<final_dis<<" "<<final_happy<<" "<<final_happy/final_step<<endl;
    cout<<s_city;
    for(unsigned int i = 1;i<final_path.size();i++)
        cout<<"->"<<namelist[final_path[i]];
    return 0;
    
}

还是超时了，看来，cin是在是扶不起来啊。。。。。。

同时，这种dfs的方法本身效率太低了，递归层数太多，只能说好写，但是肯定要用其他方法改进

找最短路，完全可手撕一个迪杰斯特拉算法。

超时原因有两个：cin cout 流输入输出 和dfs的递归层数，不好处理大数据

关于最短路解决，可以看这篇博客：

https://www.cnblogs.com/ranzhong/p/14258677.html

样例全部通过 了