poj 3253(贪心 + 优先队列）

Fence Repair

POJ - 3253                                                    

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
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

Output
Line 1: One integer: the minimum amount of money he must spend to make
N-1 cuts
Sample Input

3
8
5
8

Sample Output

34

Hint
He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.

The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into
16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

这道题应该注意，sum应该用long long，否则会哇。

而且这道题，一开始想的太简单，每次把所要求最大的分割出来，这样的贪心策略是错误的。

这样的的分割方法，想到的树状图，用优先队列实现了树状图的构建。

该树状图满足最小的，和次小的在最深处，结果等于所有叶的值 *  深度的总和。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>

using namespace std;
struct cmp
{
bool operator()(int a, int b){
return a > b;
}
};

int main()
{
int n;
scanf("%d", &n) ;
priority_queue<int,vector<int>,cmp>q;
for(int i = 0; i < n; i ++)
{
int tem;
scanf("%d", &tem);
q.push(tem);
}
long long int sum = 0;
for(int i = 1; i < n; i ++)
{
int tem1 = q.top();
q.pop();
int tem2 = q.top();
q.pop();
int t = tem1 + tem2;
sum += t;
q.push(t);

}
printf("%lld\n", sum);
while(!q.empty())
q.pop();
return 0;
}