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优先队列模拟一下,数据答案用long long,52行代码1A

Posted by 151210107 at 2017-07-24 10:42:49 on Problem 3253
数据结构中学过哈夫曼树,构建过程中每次都取值最小的节点和次小的节点构建左右子树,有了父母的几点就不用再考虑了,用优先队列的话,小的优先出对,如果队列中能出队两个元素,就
相加和构成新节点,如果最后优先队列只剩一个节点,则它就是树根,这个节点的值不必再累加到答案上,因为之前进队前已经累加过了。

#include <iostream>
#include <stdio.h>
#include <queue>
using namespace std;
const int maxn = 20010;
int a[maxn],n;
struct node
{
    long long weight;
    friend bool operator<(node n1,node n2)
    {
        return n1.weight>n2.weight;
    }
};
long long solve()
{
    priority_queue<node>qu;
    node cur,min1,min2;  ///min1最小值,min2次小值
    for(int i = 0; i < n; i++)
    {
        cur.weight = a[i];
        qu.push(cur);
    }
    long long ans=0;
    while(!qu.empty())
    {
        min1 = qu.top();
        qu.pop();
        if(!qu.empty())
        {
            min2 = qu.top();
            qu.pop();
            cur.weight = min1.weight + min2.weight;
            ans += cur.weight;
            qu.push(cur);
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i = 0; i < n; i++)
            scanf("%d",&a[i]);
        printf("%lld\n",solve());
    }
    return 0;
}


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