poj 3253 木板分割

题目连接

http://poj.org/problem?id=3253

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the “kerf”, the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn’t own a saw with which to cut the wood, so he mosies over to Farmer Don’s Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn’t lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.
Input

Output

Line 1: One integer N, the number of planks
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank

Sample Input

3
8
5
8

Sample Output

34

题意

用优先队列构造二叉树

代码

#include
  #include
  #include
  #include
  #include
  #include
 #include
  #include
  #include 
  #include
#define ll long long
#define maxn 20010
int num[maxn]={1};
int l[maxn];
using namespace std;
int main(int argc, char const *argv[])
{
    int n;
    scanf("%d",&n);
    priority_queue<int,vector<int>,greater<int>>que;
    for (int i = 0; i < n; ++i)
    {   int t;
        cin>>t;
        que.push(t);
    }
    ll ans=0;
    while(que.size()!=1){
        ll l1;
        ll l2;
        l1=que.top();
        que.pop();
        l2=que.top();
        que.pop();
        ans+=l1+l2;
        que.push(l1+l2);
    }
    printf("%I64d\n",ans );
    return 0;
}

Get

• 优先队列的构造
• 优先队列从小往大取数 priority_queue< int,vector< in t>,greater< int > >que