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sgu 203

2015-06-08 10:31 309 查看
构造哈夫曼树,开始乱搞,蛤蛤。

时间复杂度:O(n∗log2n)O(n*log_2n)

事实上正解应该是 O(n)O(n),蛤蛤。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>

const int MAXN = 500005;
typedef std::priority_queue<long long,std::vector<long long>,std::greater<long long> >  Heap_ll;

int n, p[MAXN];
Heap_ll heap;
long long ans;

int main()
{
#ifndef ONLINE_JUDGE    
    freopen("sgu203.in","r",stdin);
    freopen("sgu203.out","w",stdout);
#endif

    scanf("%d",&n);

    for(int i = 1; i <= n; i++)
        scanf("%d",&p[i]), heap.push(p[i]);

    for(int i = 1; i < n; i++)
    {
        long long a, b;
        a = heap.top(), heap.pop();
        b = heap.top(), heap.pop();
        heap.push(a + b), ans += a + b;
    }

    printf("%I64d\n",ans);  

#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif
    return 0;       
}
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