POJ - 3253 - Fence Repair（哈夫曼树 + 优先队列）

题意：把一块木板分成n块（1 <= n <= 20000），锯完后的这n块已经给你，你每次可以把一块木板锯成两块，开销为两块之和，求最小的开销。

哈夫曼树直接每次选最小的两块木板即可。

代码如下：

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;

#define NDEBUG
#include<cassert>
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;

int main(){
    int n;
    while(scanf("%d", &n) == 1){
        priority_queue<int, vector<int>, greater<int> > q;
        int tmp;
        for(int i = 0; i < n; ++i){
            scanf("%d", &tmp);
            q.push(tmp);
        }
        ll ans = 0;
        while(--n){
            int u = q.top();  q.pop();
            int v = q.top();  q.pop();
            int sum = u + v;
            ans += sum;
            q.push(sum);
        }
        printf("%lld\n", ans);
    }
    return 0;
}