优先队列 POJ 3253 Fence Repair

题目传送门

题意：一块木板按照某个顺序切成a[1], a[2]...a[n]的长度，每次切都会加上该两段木板的长度，问选择什么顺序切能使得累加和最小

分析：网上说这是哈夫曼树。很容易想到先切掉最长的，反过来也就是相当于每次取最短的两块合并成一块，直到最后剩下原来的一块，优先队列实现

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/14 星期一 08:51:15
* File Name     :POJ_3253.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 2e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N];

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
        priority_queue<int, vector<int>, greater<int> > Q;
        for (int i=1; i<=n; ++i)    Q.push (a[i]);
        ll ans = 0;
        while (Q.size () > 1)   {
            int l1 = Q.top ();  Q.pop ();
            int l2 = Q.top ();  Q.pop ();
            ans += l1 + l2;
            Q.push (l1 + l2);
        }
        printf ("%I64d\n", ans);
    }

    return 0;
}