• pat倒数3天

    1148
    解题三步走：
    希望今天能写 5题 在看两题

    Werewolf（狼人杀） is a game in which the players are partitioned into two parties: the werewolves and the human beings. Suppose that in a game,

    player #1 said: “Player #2 is a werewolf.”;
    player #2 said: “Player #3 is a human.”;
    player #3 said: “Player #4 is a werewolf.”;
    player #4 said: “Player #5 is a human.”; and
    player #5 said: “Player #4 is a human.”.
    Given that there were 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. Can you point out the werewolves?

    Now you are asked to solve a harder version of this problem: given that there were N players, with 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. You are supposed to point out the werewolves.

    1.找到说谎者 真话假话题目
    2. 第一个想法是用扩展的并查集去解决这个问题
    3. 第二个想法是什么？ 用朴素的逻辑推理看看能不能解决 毕竟只有20分
    4. 时间给的很宽
    5. 首先我们需要对一堆话里面的特质关注 我们需要注意哪些特质
    6. 矛盾 枚举可以吗
    7. 枚举应该可以 两层循环 检查矛盾 矛盾发生liar++ 如果刚好两个那就是答案
    8. 还有没有个更简单的方法 O（n）级别 的
    9. 我们发现发生矛盾的 人不一定出问题，起争议时至少有一个人可以被确定不是狼
    10. 想不出来
    11. 我曹搞得我暴力也不想写了 麻了
    12. 下一题吧
    13.看一下别人的思路看看我能不能写
    13. 第一个并查集是找出来一定是狼人的人，
    14. 用并查集的是别的名字一样 的题
    15. 找不到合适的性质能两个 两个节点合并 等会在把暴力不上把 先写下一题

    1149 Dangerous Goods Packaging
    When shipping goods with containers, we have to be careful not to pack some incompatible goods into the same container, or we might get ourselves in serious trouble. For example, oxidizing agent （氧化剂） must not be packed with flammable liquid （易燃液体）, or it can cause explosion.

    1. When shipping goods with containers, we have to be carefull not to pack some incompatible
    2. goods into the same contianer
    3. we might get ourselves in serious trouble
    4. oxidizing agent must not be packed with flammable liquid
    5. it can cause explosion.

    Now you are given a long list of incompatible goods, and several lists of goods to be shipped. You are supposed to tell if all the goods in a list can be packed into the same container.
    //you are supposed to tell if all the goods in a list can be packed into the same container
    // Now you are given a long list of incompatible goods, and several lists of goods to be shipped

    看上去6 3
    20001 20002
    20003 20004
    20005 20006
    20003 20001
    20005 20004
    20004 20006
    4 00001 20004 00002 20003
    5 98823 20002 20003 20006 10010
    3 12345 67890 23333
    显示图论或者并查集 可能用不到深度和官渡，一眼看上去

    1. 读题
    2. 浪费太多时间了，不深究了 直接开始写吧 不行再搜
    3. 暴力出问题了 两个测试点拿不到分这是不能接受的
      10.所以还是要改一下算法
    4. 发现别人的代码也是O（n^2）
    5. 那就减掉一些无愿意 的循环
    6. 在这里插入代码片在这里插入代码片`
    #include
#include
#include
using namespace std;
int main(){
    int n;
    int m;
    cin>>n;
    cin>>m;
    map<string,int> map1;
    set<string> set1;
    for(int i = 0;i<n;i++){
         string u;string v;
        cin>>u;
        cin>>v;
        set1.insert(u);
        set1.insert(v);
        map1[u+v]=1;
        map1[v+u] =1;
        }
 
    for(int j = 0;j<m;j++){
        int t;
        cin>>t;
        string nums[t];
        for(int i  = 0;i<t;i++){
            cin>>nums[i];
        }
        bool h =true;
        
        for(int i = 0;i<t;i++){
              if(set1.find(nums[i]) == set1.end()){
                    continue;
                }
                
            for(int k = i+1;k<t;k++){

                if(map1[nums[i]+nums[k]]>0){
                    h = false;
                    break;
                }
            }
            if(h == false){
                break;
            }
        }
        if(h == false){
            cout<<"No"<<endl;
        }else{
            cout<<"Yes"<<endl;
        }
        }
        
    }


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    全过 并不难相反挺简单的

    第三题
    1150 Travelling Salesman Problem
    一看就是图论
    但是25分应该看题目的长度应该也算25分里面比较男的了

    The “travelling salesman problem” asks the following question: “Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?” It is an NP-hard problem in combinatorial optimization, important in operations research and theoretical computer science. (Quoted from “https://en.wikipedia.org/wiki/Travelling_salesman_problem”.)

    1. 旅行商问题 asks the following question :
    2. Given a list pf cities and the distances between each pair of cities
    3. what is the shortest possible route that visits each city
    4. and returns to the origin city
    5. it is an NP hard problem in combinatorial optimization
    6. 去每个城市然后返回的最短路径
    7. 给定的环中找到结构

    In this problem, you are supposed to find, from a given list of cycles, the one that is the closest to the solution of a travelling salesman problem.

    Input Specification:
    Each input file contains one test case. For each case, the first line contains 2 positive integers N (2

    1. Each input file contains one test case
    2. the first line contains 2 positive integers N
    3. the number of cities and M the number of edges in an undirected graph
    4. the number of edges in an undirected graph Then M lines follow each describes an edge
    5. CITY 1 CITY2 Dist
    6. cities are numbered from 1 to N
    7. 我还以为要用最短路但是好像不用， 直接建立一张图就查询边就可以找到长度了
    8. 查看一下数据范围
    9. 两百个点，可以用哪个矩阵
    10. 250ms 我绝得得用那个了，得用那个什么东西了 C++了
    11. ``
      20在这里插入代码片.

    盲猜怎么写吧

    void  check(vector<int> path,set<int>set1){
//计算路径的长度
     int psize = path.size();
     int len = 0;
     for(int i = 0;i<psize-1;i++){
         len+= graph[path.get(i)][path.get(i+1)];
     }
     cout<<len<<" ";
   if(set1.end() != n){
      cout<<"(Not a TS cycle)"<<endl;
      return;
   }
   if(path.size() > n+1){
      cout<<"(TS cycle)"<<endl;
   }
   if(path.size() == n+1){
     cout<<"(TS simple cycle)"<<endl;
   } 
}
int n,int m;//全局
cin>>n;
cin>>m;
int graph[300][300];//全局
for(int i  =0;i<m;i++){
   int u; int v;cin>>u;cin>>v;int w;cin>>w;
   graph[u][v] =w;
   graph[v][u] = w; 
}
int q = 0;
cin>>q;
for(int i = 0;i<q;i++){
  int k = 0;
  cin>>k;
  vector<int>path;
  set<int> set1;
  for(int  j = 0;j<k;j++){
  cout<<"path "<<i+1<<": ";
     check(path,set1);
  }
}

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    有时候盲写还是很爽的
    修改后通过代码是这样的

    #include
#include
#include
#include
using namespace std;
int n;int m;//全局
int graph[300][300];//全局
  int len = 0;
bool  check1(vector<int> path,set<int>set1){
//计算路径的长度
     int psize = path.size();
   len = 0;
     for(int i = 0;i<psize-1;i++){
         if(graph[path[i]][path[i+1]] == 0){
             len = -1;
             break;
         }
         len+= graph[path[i]][path[i+1]];
     }
    if(len == -1){
      cout<<"NA "<<"(Not a TS cycle)"<<endl;
        return 0;
    }else{
     cout<<len<<" ";
    }
    if(path[0] != path[path.size()-1]){
        cout<<"(Not a TS cycle)"<<endl;
    }else{
   if(set1.size() != n){
      cout<<"(Not a TS cycle)"<<endl;
      return 0;
   }else if(path.size() > n+1 && set1.size() == n){
      cout<<"(TS cycle)"<<endl;
       return 1;
   }else if(path.size() == n+1 && set1.size() == n){
     cout<<"(TS simple cycle)"<<endl;
       return 1;
   }else{
       cout<<"(Not a TS cycle)"<<endl;
   }

    }
  return 0;
}


int main(){
cin>>n;
cin>>m;
for(int i  =0;i<m;i++){
   int u; int v;cin>>u;cin>>v;int w;cin>>w;
   graph[u][v] =w;
   graph[v][u] = w; 
}
int q = 0; cin>>q;
  int min1 = 99999999;
    int index = 0;
for(int i = 0;i<q;i++){
  int k = 0;
  cin>>k;
  vector<int>path;
  set<int> set1;
  for(int j = 0;j<k;j++){
      int v; cin>>v; path.push_back(v); set1.insert(v);
  }
     cout<<"Path "<<i+1<<": ";
    if(check1(path,set1) == 1){
      if(len < min1){
          min1  = len;
          index = i+1;
      }
    }
}
    cout<<"Shortest Dist("<<index<<")"<<" = "<<min1;
}



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    第四题

    1151 LCA in a Binary Tree

    The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

    1. The lowest common ancestor(LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants;
      Given any two nodes in a binary tree, you are supposed to find their LCA.
      Given any tow nodes in a binary tree, you are supposed to find their LCA;

    Input Specification:
    Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the binary tree, respectively. In each of the following two lines, N distinct integers are given as the inorder and preorder traversal sequences of the binary tree, respectively.
    It is guaranteed that the binary tree can be uniquely determined by the input sequences. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

    Output Specification:
    For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y. where X is A and Y is the other node. If U or V is not found in the binary tree, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found…

    LCA是经典的问题

    1. 首先我们可以通过前序和中序建立一棵树 然后通过LCA 的方法来好到它的根节点
    2. LCA的方法第一是递归来找
    3. 第二个方法是倍增法 但是我感觉我都不是特别喜欢
    4. 主要是写的少，而且首先他需要先建立一棵树，那个区间问题，就有一点恶心，
    5. 然后还要再去找根，说实话对我来说有一点累了，
    6. 我先或许可以盲写但是然后一点点调试 现在状态差花个一个半小时45分钟看状态应该能写
    7. 我先把思路写出来把
    8. 首先是不建树，建树很麻烦因为
    9. 先序遍历时什么 是根左右
    10. 中序遍历时什么 是左根右
    11. 首先我认为建树是没必要的
    12. 你可以计算出所有节点的 深度，递归虽然其实也相当建树了
    13. 那么你找到中序遍历两个节点之间包括自己深度最小的节点就是lca
    14. 但是我们也可以想一下不用递归计算深度
    15. 一般上我们认为先序比遍历再前面的深度一定是、
    16. 在这里插入代码片
    #include 
#include
using namespace std;
map<int,int> map1;
map<int,int> map2;
int findlca(int u,int v){
    int s = map1[u];
    int e = map1[v];
    int min1 = 99999;
    for(int i = s;i<e;i++){
       min1 = min(map2[inorder[i]],min1);
    }
    return preorder[min1];
}
int main(){
    int n;
    int m;
    cin>>n;
    cin>>m;
    int preorder[m];
    int inorder[m];
    for(int i = 0;i<m;i++){
        cin>>inorder[i];
        map1[include[i]] = i;
    }
    for(int j = 0;j<m;j++){
        cin>>preorder[i]
        map2[preorder[i]] = i;
    }
    for(int i = 0;i<n;i++){
        int u;int v;
        cin>>u;cin>>v;
        if(u>m || u<1 && v>=1 && v<=m){
           cout<<"ERROR: "<<u<<" is not found.";
        }else if(v>m||v<1 && u>=1 && u<=m){
            cout<<"ERROR: "<<v<<" is not found.";
        }else if(u>=1 && u<= m && v>=1 && v<=m){
           int root =  findlca(u,v);
            cout<<"LCA of "<<u<<" and "<<v<<" is "<<root<<".";
        }else{
            cout<<"ERROR: "<<u<<" and "<<v<<" are not found.";
        }
        
    }
}






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    修修补补后正确的代码

    #include 
#include
using namespace std;
map<int,int> map1;
map<int,int> map2;
int preorder[10100];
    int inorder[10100];
    int n;
    int m;
int findlca(int u,int v){
    int s = min(map1[u],map1[v]);
    int e = max(map1[v],map1[u]);
    int min1 = 1000000;
    for(int i = s-1;i<=e-1;i++){
       min1 = min(map2[inorder[i]],min1);
    }
    return preorder[min1-1];
}
int main(){

    cin>>n;
    cin>>m;
    
    for(int i = 0;i<m;i++){
        cin>>inorder[i];
        map1[inorder[i]] = i+1;
    }
    for(int j = 0;j<m;j++){
        cin>>preorder[j];
        map2[preorder[j]] = j+1;
    }
    for(int i = 0;i<n;i++){
        int u;int v;
        cin>>u;cin>>v;
        
        if( map1[v] ==0 && map1[u] ==0){
            cout<<"ERROR: "<<u<<" and "<<v<<" are not found.";
            
        }else{
        if(map1[u] == 0&&map1[v] > 0){
           cout<<"ERROR: "<<u<<" is not found.";
        }else `在这里插入代码片`if(map1[v] == 0 && map1[u] > 0){
            cout<<"ERROR: "<<v<<" is not found.";
        }else if(map1[u] > 0 && map1[v] > 0){
           int root =  findlca(u,v);
            if(root == u){
             cout<<u<<" is an ancestor of "<<v<<".";   
            }else if(root==v){
             cout<<v<<" is an ancestor of "<<u<<".";   
            }else{
            cout<<"LCA of "<<u<<" and "<<v<<" is "<<root<<".";
            }
        }
        }
        cout<<endl;
    }
}



又休息了好久


1152 



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    1152 Google Recruitment
    分
    下一题
    In July 2004, Google posted on a giant billboard along Highway 101 in Silicon Valley (shown in the picture below) for recruitment. The content is super-simple, a URL consisting of the first 10-digit prime found in consecutive digits of the natural constant e. The person who could find this prime number could go to the next step in Google’s hiring process by visiting this website.

    1. In July 2004 google posted on a giant billboard along Highway 101 in Silicon Valley
    2. for recruitment The content is super-simple a URLconsisting of the first 10_ digit prime found in consecutive digits of th enatural constatne e the person hwo find this prime number
    3. could go to the next step in Googles’ s hiring pricess by visiting this website

    直接 写吧

    int string2int(string s){
    int x = 0;
    int len = s.length();
    for(int i = 0;i<len;i++){
      x = (s[i] - '0') +x*10;
}
return x;
}
bool is_Prime(int x){
    int  up = sqrt(x);
    if(x<=1){
    return false;}
    if(x <= 2){
    return true;
    }
    for(int i = 2;i<=up;i++){
      if(x%i== 0){
       return false;
}
}
return true;
}
//需要检查不确定是不是这样大概是这样
int main(){
string s; 
cin>>s;
int k = 0;
cin>>k;
for(int i = 0;i+k<s.length();i++){
    //
    string temp = substr(i,i+k);
    if(is_Prime(string2int(temp))){
       cout<<string_temp;
       return;
    }
}
cout<<404;
}

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    去编辑器里面修改一下
    调试一侠

    #include
using namespace std;
int string2int(string s){
    int x = 0;
    int len = s.length();
    for(int i = 0;i<len;i++){
      x = (s[i] - '0') +x*10;
}
return x;
}
bool is_Prime(int x){
    int  up = sqrt(x);
    if(x<=1){
    return false;}
    if(x <= 2){
    return true;
    }
    for(int i = 2;i<=up;i++){
      if(x%i== 0){
       return false;
}
}
return true;
}
//需要检查不确定是不是这样大概是这样
int main(){
int n;cin>>n;  int k = 0;cin>>k;
string s; 
cin>>s;
for(int i = 0;i+k<s.length();i++){
    //
    string temp = s.substr(i,k);

    if(is_Prime(string2int(temp))){
       cout<<temp;
       return 0;
    }
}
cout<<404;
}


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    #include
    using namespace std;

    #include
    #include
    #include
    #include
    #include
    #include
    #include
    #include
    #include
    #include
    #include
    常用头文件别忘记啦
    芝士

    休息一会儿
    虽然这题不值得休息就是了

    休息回来吧狼人杀傻吊
    还是先盲打

    5
    -2
    +3
    -4
    +5
    +4

    int n; cin>>n; 
int speak[n];
bool wfs[n];
for(int i = 0;i<n;i++){
   int a;cin>>a;
   speak[i] =a;
   if(a<0){
     wfs[abs(a)] = 1;
}else{
    wfs[abs(a)] = 0;
}
}

for(int i = 0 ;i<n;i++){
bool wfs[n];
   for(int j = 0;j<n;j++){
       //先来个双重循环
       if(i == j) continue;
       //然后需要干什么呢？ 
       //假设i和j是狼人
       
     if(check(wfs,speak)){
         return 0;
     } 
  
}
}


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    哈市稍微有点细节的啊

    
```cpp
#include
using namespace std;
int lair;
 int n;
 int temp [1010];
bool check1(int i,int a,int b){
   if(temp[i]<0){
             if(abs(temp[i]) == a || abs(temp[i]) == b){
                 return true;//说的是真话
             }else{
                 return false;
             }
          }else{
             if(abs(temp[i]) == a ||abs(temp[i]) == b){
                 return false;
                 //说的是谎话
             }else{
                 //说的是真话
                 return true;
             } 
}
}
bool check(int  temp[],int i,int j){
    if(check1(i,i,j) == true && check1(j,i,j) == true || check1(i,i,j) == false && check1(j,i,j) == false){
        return false;
    }
    lair = 0;
      for(int index = 1;index<n+1;index++){
           if(check1(index,i,j) == false){
               lair++;
           }      
          }
    if(lair == 2){
        return true;
    }
    return false;
      } 

int main(){
   cin>>n;

    for(int i = 1;i<n+1;i++){
        cin>>temp[i];
    }
   for(int i = 1;i<n+1;i++){
       for(int j = i+1;j<n+1;j++){
     // cout<
         if(check(temp,i,j)){
             cout<<i<<" "<<j;
             return 0;
         }  
       }
   } 
    cout<<"No Solution";
    
    
}


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    这样就全对了
休息下一下  等会研究哈希表的查找长度
我认为还算比较麻烦的题目

算了算了明天写吧
没有发挥到极致 
没办法我认为我的潜力没有完全发挥
但是  怎么说呢？ 
有些人就是能更加抵抗劳累把
明天 把第二页的题目清一清把
后天清一下第一页的   看那一下规则
大后天去考尊卑环境，记得发

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  • 原文地址：https://blog.csdn.net/weixin_43401039/article/details/126637813