Codeforces Round #303 (Div. 2)

Codeforces Round #303 (Div. 2)

A. Simple question

/*************************************** * *********************************> File Name: a. cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: wednesday ******************************** **************************************** /# include
  
   
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                Using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
               
                 PLL; int mat [110] [110]; bool OK [110]; int main () {int n; while (cin> n) {for (int I = 1; I <= n; ++ I) {OK [I] = 1; for (int j = 1; j <= n; ++ j) {cin> mat [I] [j] ;}}for (int I = 1; I <= n; ++ I) {for (int j = 1; j <= n; ++ j) {if (mat [I] [j] =-1 | mat [I] [j] = 0) {continue ;} if (mat [I] [j] = 1) {OK [I] = 0;} else if (mat [I] [j] = 2) {OK [j] = 0;} else {OK [I] = OK [j] = 0 ;}} vector
                
                  Ans; ans. clear (); for (int I = 1; I <= n; ++ I) {if (OK [I]) {ans. push_back (I) ;}} int size = ans. size (); printf ("% d \ n", size); if (size = 0) {continue;} printf ("% d", ans [0]); for (int I = 1; I <size; ++ I) {printf ("% d", ans [I]) ;}printf ("\ n ");} return 0 ;}
                
               
              
             
            
          
         
        
       
      
     
    
   
  

B. Simple question

/*************************************** * *********************************> File Name: b. cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: on Wednesday, May 20, 2015, ******************************** **************************************** /# include
  
   
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                Using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
               
                 PLL; string s, t; int main () {while (cin> s> t) {int len = s. length (); int cnt = 0; for (int I = 0; I <len; ++ I) {if (s [I]! = T [I]) {++ cnt ;}}if (cnt & 1) {printf ("impossible \ n") ;}else {bool flag = 0; for (int I = 0; I <len; ++ I) {if (s [I] = t [I]) {printf ("% c ", s [I]);} else {if (! Flag) {printf ("% c", s [I]);} else {printf ("% c", t [I]);} flag ^ = 1 ;}} printf ("\ n") ;}} return 0 ;}
               
              
             
            
          
         
        
       
      
     
    
   
  

C. Simple dp, Dp [I] [0/1/2] Indicates not to cut down the I-th tree; cut down, down to the left; cut, down to the right

/*************************************** * *********************************> File Name: c. cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: on Wednesday 10, May 20, 2015 ******************************** **************************************** /# include
  
   
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                Using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
               
                 PLL; PLL sta [100100]; int dp [100100] [3]; int main () {int n; while (~ Scanf ("% d", & n) {memset (dp, 0, sizeof (dp); for (int I = 1; I <= n; ++ I) {scanf ("% d", & sta [I]. first, & sta [I]. second);} dp [1] [1] = 1; if (sta [1]. first + sta [1]. second <sta [2]. first) {dp [1] [2] = 1 ;}for (int I = 2; I <= n; ++ I) {dp [I] [0] = max (dp [I-1] [1], dp [I-1] [2]); dp [I] [0] = max (dp [I-1] [0], dp [I] [0]); if (sta [I]. first-sta [I]. second> sta [I-1]. first) {dp [I] [1] = max (dp [I] [1], max (dp [I-1] [1], dp [I-1] [0]) + 1);} if (sta [I]. first-sta [I]. second> sta [I-1]. first + sta [I-1]. second) {dp [I] [1] = max (dp [I] [1], dp [I-1] [2] + 1 );} if (I <n & sta [I]. first + sta [I]. second <sta [I + 1]. first | I = n) {dp [I] [2] = max (dp [I] [2], max (dp [I-1] [1], dp [I-1] [0]) + 1); if (sta [I-1]. first + sta [I-1]. second <sta [I]. first) {dp [I] [2] = max (dp [I] [2], dp [I-1] [2] + 1 );}}} printf ("% d \ n", max (dp [n] [0], max (dp [n] [1], dp [n] [2]);} return 0 ;}
               
              
             
            
          
         
        
       
      
     
    
   
  

D. Just sort it with greed.

/*************************************** * *********************************> File Name: d. cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: wednesday, 38 seconds ******************************** **************************************** /# include
  
   
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                Using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
               
                 PLL; int arr [100010]; int main () {int n; while (~ Scanf ("% d", & n) {for (int I = 1; I <= n; ++ I) {scanf ("% d ", & arr [I]);} sort (arr + 1, arr + n + 1); int sum = 0; int ans = 0; for (int I = 1; I <= n; ++ I) {if (sum <= arr [I]) {++ ans; sum + = arr [I] ;}} printf ("% d \ n", ans);} return 0 ;}
               
              
             
            
          
         
        
       
      
     
    
   
  

E. The implementation process of dijkstra algorithm is the process of finding the Minimal Path Tree.
Add a vertex to the set that finds the shortest path.
Consider finding the shortest side for each vertex to reach its edge, and removing the edge having a smaller weight when the shortest path is used, wa hasn't been changed in a place. It's also a dog.

/*************************************** * *********************************> File Name: e. cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: on Wednesday ******************************** **************************************** /# include
  
   
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                Using namespace std; const double pi = acos (-1.0); const int inf = 0x3f3f3f; const double eps = 1e-15; typedef long LL; typedef pair
               
                 PLL; static const int N = 300100; LL d [N]; bool flag [N]; bool vis [N]; int head [N]; int cost [N]; int tot; int use [N]; vector
                
                  Ans; priority_queue <PLL, vector
                 
                   , Greater
                  
                    > Qu; struct node {int w; int id; int nxt; int to;} edge [N <1]; void addedge (int u, int v, int w, int id) {edge [tot]. to = v; edge [tot]. w = w; edge [tot]. nxt = head [u]; edge [tot]. id = id; head [u] = tot ++;} void Dijkstra (int s) {d [s] = 0; cost [s] = 0; memset (flag, 0, sizeof (flag); memset (vis, 0, sizeof (vis); while (! Qu. empty () {qu. pop ();} qu. push (make_pair (d [s], s); while (! Qu. empty () {PLL tmp = qu. top (); qu. pop (); int u = tmp. second; LL dist = tmp. first; if (flag [u]) {continue;} flag [u] = 1; vis [use [u] = 1; for (int I = head [u]; ~ I; I = edge [I]. nxt) {int v = edge [I]. to; int w = edge [I]. w; int id = edge [I]. id; if (d [v]> dist + w) {use [v] = id; d [v] = dist + w; cost [v] = w; qu. push (make_pair (d [v], v);} else if (d [v] = dist + w & cost [v]> w) {cost [v] = w; use [v] = id ;}}} int main () {int n, m; scanf ("% d ", & n, & m); memset (head,-1, sizeof (head); memset (vis, 0, sizeof (vis); for (int I = 1; I <= n; ++ I) {d [I] = (1LL <60); cost [I] = d [I];} tot = 0; int u, v, w, s; for (int I = 1; I <= m; ++ I) {scanf ("% d", & u, & v, & w); addedge (u, v, w, i); addedge (v, u, w, I);} scanf ("% d", & s); Dijkstra (s); LL sum = 0; for (int I = 1; I <= n; ++ I) {for (int j = head [I]; ~ J; j = edge [j]. nxt) {if (vis [edge [j]. id]) {sum + = edge [j]. w ;}}} cout <sum/2 <endl; if (! Sum) {return 0 ;}for (int I = 1; I <= m; ++ I) {if (vis [I]) {printf ("% d ", i) ;}} printf ("\ n"); return 0 ;}