LeetCode 179 Zhou race problem solution

5352. Each generates a string of characters is an odd number

 1 class Solution {
 2 public:
 3     string generateTheString(int n) {
 4         string ans="";
 5         if(n%2==1){
 6             for(int i=0;i<n;i++){
 7                 ans+='a';
 8             }
 9         }else{
10             for(int i=0;i<n-1 ; i ++ ) {
 11                  years + = ' a ' ;
12              }
 13              years + = ' b ' ;
14          }
 15          return year;
16      }
 17 };

Lamp switch 5353. III

Solution one: to find the law: When the position of the largest lighting with the same amount of time lights, then light blue:

. 1  class Solution {
 2  public :
 . 3      /// when an equal number of positions and the maximum turn-on time of lighting, then it turned blue light 
. 4      int numTimesAllBlue (Vector < int > & Light) {
 . 5          int n-= Light .size ();
 . 6          int MAX_NUM = 0 ; // record the maximum position of the bright light 
. 7          int CNT = 0 ; // number of recording lights 
. 8          int ANS = 0 ;
 . 9          for ( int I = 0 ; I <n- ; I ++ ) {
 10              MAX_NUM =max(max_num,light[i]);
11             cnt++;
12             if(cnt==max_num){
13                 ans++;
14             }
15         }
16         return ans;
17     }
18 };

Solution two: disjoint-set:

. 1  const  int MAXN = . 5 * 1E4 + 100 ;
 2  class Solution {
 . 3  public :
 . 4      int Father [MAXN];
 . 5      int Find ( int X) {
 . 6          // return Father [X] = (Father [X] == X? X: Find (Father [X])); 
. 7          the while ! (Father [X] = X) {
 . 8              X = Father [X];
 . 9          }
 10          return X;
 . 11      }
 12 is      // not need to write this function is combined, because there will be following the merger, and the merger is small to large, that is, when a large father. . 
13      int numTimesAllBlue(vector<int>& light) {
14         int n=light.size();
15         int ans=0;
16         int op[n+1];
17         for(int i=1;i<=n;i++){
18             father[i]=i;
19             op[i]=0;
20         }
21         int cnt=0;
22         for(int i=0;i<n;i++){
23             int NUM = Light [I];
 24              OP [NUM] = . 1 ;
 25              
26 is              IF (NUM> . 1 && OP [num- . 1 ]) [num-Father . 1 ] = NUM; // represented by one lamp before the lamp current so is the amount, which is incorporated into the current lamp ranks. 
27              IF (NUM + . 1 <= n-&& OP [NUM + . 1 ]) Father [NUM] = NUM + . 1 ; // large copy to his father 
28              
29              CNT ++ ;
 30              IF (OP [ . 1 ] && Find ( . 1 ) == CNT) ANS ++ ;
 31 is          }
 32          returnyears;
33      }
 34 };

5354. notification time required for all employees

 1 const int maxn=1e5+50;
 2 //vector<int>dist(maxn);
 3 
 4 class Solution {
 5 public:
 6     
 7     int numOfMinutes(int n, int headID, vector<int>& manager, vector<int>& informTime) {
 8         vector<int>dist(maxn,0);
 9         vector<int>edge[maxn];
10         int ans=0;
11         IF (n-<= . 1 ) return  0 ;
 12 is          // traverse the level bit, and bob:
 13 is          
14          // int = n-manager.size (); 
15          for ( int I = 0 ; I <n-; I ++ ) {
 16              dist [ I] = 0 ;
 . 17              IF (Manager [I] = -! . 1 ) Edge [Manager [I]] push_back (I);.
 18 is          }
 . 19          Queue < int > que;
 20 is          que.push (headID);
 21 is          the while ( ! que.empty ()) { // BFS template 
22             int x=que.front();
23             que.pop();
24             ans=max(ans,dist[x]);
25             for(int i=0;i<edge[x].size();i++){
26                 int v=edge[x][i];
27                 dist[v]=dist[x]+informTime[x];
28                 que.push(v);
29             }
30         }
31         return ans;
32     }
33 };

5355. T-second position after the frog

 1 const int maxn=1e4;
 2 class Solution {
 3 public:
 4     double frogPosition(int n, vector<vector<int>>& edges, int t, int target) {
 5         if(edges.size()==0)return 1.0;
 6         vector<int>edge[maxn];
 7         double dist[maxn];
 8         int cnt[n+1][2];
 9         for(int i=1;i<=n;i++){
10             dist[i]=0.0;
11             //cnt[i].clear();
12             cnt[i][0]=0;
13             cnt[i][1]=0;
14         }
15         for(int i=0;i<edges.size();i++){
16             if(edges[i][0]>edges[i][1])swap(edges[i][0],edges[i][1]);
17             edge[edges[i][0]].push_back(edges[i][1]);
18             //edge[edges[i][1]].push_back(edges[i][0]);
19         }
20         
21         queue<int>que;
22         que.push(1);
23         dist[1]=1.0;
24         int start=0;int end=1;
25         int level=1;
26         while(!que.empty()){
27             start++;
28             int min_num=level;
29             
30             int x=que.front();
31             que.pop();
32             int num=edge[x].size();
33             double percent=1.0/(double)num;
34             for(int i=0;i<num;i++){
35                 int v=edge[x][i];
36                 if(edge[v].size()==0)cnt[v][1]=1;
37                 cnt[v][0]=min_num;
38                 dist[v]=(double)percent*dist[x];
39                 que.push(v);
40             }
41             //cout<<level<<endl;
42             if(start==end){
43                 level++;
44                 start=0;
45                 end=que.size();
46             }
47         }
48         //cout<<cnt[target][0]<<" "<<level<<" "<<cnt[target][1]<<endl;
49         if((cnt[target][1]==0&&t==cnt[target][0])||(cnt[target][1]&&t>=cnt[target][0])){
50             // cout<<"ok"<<endl;
51             // cout<<dist[target]<<endl;
52             return dist[target];
53         }else{
54             return 0.0;
55         }
56         
57     }
58 };