poj 3253 Fence Repair 【哈弗曼树】+【优先队列】

Fence Repair

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 L_i (1 ≤ L_i ≤
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 L_i). 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).

第一道优先队列题。。

分析：就是找最小的权值和。。。所谓的哈弗曼树。

分析：最后一块木板肯定是每一次切一块木板都要花费这块木板的长度，我们每一次都合并两个当前最小的为一棵树，删去合并后的两个最小的，将合并后的和添加进来，继续以上步奏。。。最后就会形成一颗哈弗曼树，最后将每一棵树的父节点相加就是最小的花费。、

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int M = 20005;
#define LL __int64

LL s[M];

/*bool operator<(LL a, LL b){
    return a>b;  //优先队列优先级是较小的优先
}*/
struct cmp{
    bool operator()(LL x, LL y){
        return x>y;
    }
};

int main(){
    LL n;
    while(scanf("%I64d", &n) == 1){
        LL i = 0;
        LL temp;
        priority_queue<LL, vector<LL>, cmp>q;  //按照较小的排在前面的优先队列
        for(i = 0; i < n; i ++){
            scanf("%I64d", &temp);
            q.push(temp);
        }
        LL ans = 0;
        while(q.size() > 1){
            LL t1 = q.top(); q.pop();  //每一次出队列前两个，将和如队列
            LL t2 = q.top(); q.pop();
            ans += (t1+t2);
            q.push(t1+t2);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}