TOJ 1600 Fence Obstacle Course -- 线段树,链表

转自:http://www点2cto.com/kf/201209/153025.html

用链表居然比线段树快。。。。。

#include 
   
   
    
    
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         #include 
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           #include 
          
            #include 
           
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                   #include 
                  
                    #define mp make_pair #define X first #define Y second #define MEMSET(a, b) memset(a, b, sizeof(a)) using namespace std; typedef unsigned int ui; typedef long long ll; typedef unsigned long long ull; typedef pair 
                   
                     pii; typedef vector 
                    
                      vi; typedef vi::iterator vi_it; typedef map 
                     
                       mii; typedef priority_queue 
                      
                        pqi; typedef priority_queue 
                       
                         , greater 
                        
                          > rpqi; typedef priority_queue 
                         
                           pqp; typedef priority_queue 
                          
                            , greater 
                           
                             > rpqp; const int MAX_N = 50000 + 2; int seg[MAX_N][2]; int main(int argc, char *argv[]) { // freopen("D:\\in.txt", "r", stdin); int n, s, i; list 
                            
                              coll; list 
                             
                               ::iterator it; cin >> n >> s; coll.push_back(mp(0, s)); for (i = 0; i < n; ++i) scanf("%d%d", seg[i], seg[i] + 1); for (i = n - 1; i >= 0; --i) { int lcost = INT_MAX, rcost = INT_MAX; for (it = coll.begin(); it != coll.end();) { // pii tmp = *it; if (seg[i][0] <= it->Y && seg[i][1] >= it->Y) { lcost = min(lcost, it->X + it->Y - seg[i][0]); rcost = min(rcost, it->X + seg[i][1] - it->Y); it = coll.erase(it); } else { ++it; } } if (lcost != INT_MAX) { coll.push_back(mp(lcost, seg[i][0])); coll.push_back(mp(rcost, seg[i][1])); } } int ans = INT_MAX; for (it = coll.begin(); it != coll.end(); ++it) { ans = min(ans, it->X + abs(it->Y)); } cout << ans << endl; return 0; } 
                              
                             
                            
                           
                          
                         
                        
                       
                      
                     
                    
                   
                  
                 
                
               
              
             
            
           
          
       
      
      
     
     
    
    
   
   

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