poj 1862 Stripies/优先队列

原题链接：http://poj.org/problem?id=1862

简单题，贪心+优先队列主要练习一下stl大根堆

写了几种实现方式写成类的形式还是要慢一些。。。

手打的heap：

1：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 class Solution{
 public:
     static const int Max_N = ;
     int sz;
     double heap[Max_N];
     inline void init(){
         sz = ;
     }
     inline void push(double x){
         int i = sz++;
         while (i > ){
             int p = (i - ) >> ;
             if (heap[p] >= x) break;
             heap[i] = heap[p];
             i = p;
         }
         heap[i] = x;
     }
     inline double pop(){
         double ret = heap[], x = heap[--sz];
         int i = ;
         while ((i << ) +  < sz){
             int a = (i << ) + , b = (i << ) + ;
             if (b < sz && heap[a] <= heap[b]) a = b;
             if (heap[a] <= x) break;
             heap[i] = heap[a];
             i = a;
         }
         heap[i] = x;
         return ret;
     }
 };
 int main(){
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int n;
     double a, b;
     Solution ans;
     ans.init();
     scanf("%d", &n);
     while (n--){
         scanf("%lf", &a);
         ans.push(a);
     }
     while (ans.sz > ){
         a = ans.pop(), b = ans.pop();
         ans.push( * sqrt(a * b));
     }
     printf("%.3lf", ans.pop());
     return ;
 }

2：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 struct Node{
     static const int Max_N = ;
     int sz;
     double heap[Max_N];
     Node() :sz(){}
     inline void push(double x){
         int i = sz++;
         while (i > ){
             int p = (i - ) >> ;
             if (heap[p] >= x) break;
             heap[i] = heap[p];
             i = p;
         }
         heap[i] = x;
     }
     inline double pop(){
         double ret = heap[], x = heap[--sz];
         int i = ;
         while ((i << ) +  < sz){
             int a = (i << ) + , b = (i << ) + ;
             if (b < sz && heap[a] <= heap[b]) a = b;
             if (heap[a] <= x) break;
             heap[i] = heap[a];
             i = a;
         }
         heap[i] = x;
         return ret;
     }
 }ans;
 int main(){
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int n;
     double a, b;
     scanf("%d", &n);
     while (n--){
         scanf("%lf", &a);
         ans.push(a);
     }
     while (ans.sz > ){
         a = ans.pop(), b = ans.pop();
         ans.push( * sqrt(a * b));
     }
     printf("%.3lf", ans.pop());
     return ;
 }

3：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<iostream>
 const int Max_N = ;
 double heap[Max_N];
 int sz = ;
 void push(double x){
     int i = sz++;
     while (i > ){
         int p = (i - ) >> ;
         if (heap[p] >= x) break;
         heap[i] = heap[p];
         i = p;
     }
     heap[i] = x;
 }
 double pop(){
     double ret = heap[], x = heap[--sz];
     int i = ;
     while ((i << ) +  < sz){
         int a = (i << ) + , b = (i << ) + ;
         if (b < sz && heap[a] <= heap[b]) a = b;
         if (heap[a] <= x) break;
         heap[i] = heap[a];
         i = a;
     }
     heap[i] = x;
     return ret;
 }
 int main(){
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int n;
     double a, b;

     scanf("%d", &n);
     while (n--){
         scanf("%lf", &a);
         push(a);
     }
     while (sz > ){
         a = pop(), b = pop();
         push( * sqrt(a * b));
     }
     printf("%.3lf", pop());
     return ;
 }

4：

std::priority_queue<double> ans

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<queue>
 #include<iostream>
 int main(){
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     int n;
     double a, b;
     scanf("%d", &n);
     std::priority_queue<double> ans;
     while (n--){
         scanf("%lf", &a);
         ans.push(a);
     }
     while (ans.size() > ){
         a = ans.top(), ans.pop();
         b = ans.top(), ans.pop();
         ans.push( * sqrt(a * b));
     }
     printf("%.3lf", ans.top());
     return ;
 }