poj3253 Fence Repair 贪心策略【挑战程序设计竞赛】

题目链接：http://poj.org/problem?id=3253

Fence Repair

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 72299 Accepted: 23743
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

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).
Source

USACO 2006 November Gold

思路

1.数据存储方式：使用multiset来存储所有的小木板，因为multiset是有序多重集合，会自动排序，并且插入和删除都是logn的复杂度
2.每次都选择两个最小的木板(集合的前两项)进行相加，将相加的结果累加进总开销，然后把合并的木板再插入到集合，重复同样的操作，直到集合元素个数为1停止

AC代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const ll maxN = 1e6+10; 

int N; 
multiset<int> st;//有序多重集合，删除、添加元素都是logn的时间复杂度 
int main(){
	cin>>N;
	int tmp;
	for(int i = 0;i<N;i++){
		scanf("%d",&tmp);
		st.insert(tmp);
	}
	ll total = 0; //总共开销 
	while(st.size()>1){
		ll sum = 0;
		sum += *st.begin(); //取最小的 
		st.erase(st.begin());
		sum += *st.begin();//取最小的  
		st.erase(st.begin());
		total+=sum; 
		st.insert(sum); //合并起来再加入 
	}
	cout<<total<<endl;
	return 0;
}