POJ 3253 Fence Repair

解题思路:基于最小堆实现Haffman

#include <iostream>
using namespace std;
int n,plank[20001];
void heapAdjust(int s)
{
    int i,exp;
    for(i=s;i<=(n/2);)
    {
        exp=((i*2+1)<=n&&plank[i*2+1]<plank[i*2])?1:0;
        if(plank[i]>plank[i*2+exp])
            swap(plank[i],plank[i*2+exp]),i=i*2+exp;
        else break;
    }
}

int main()
{
    int i,t;
    __int64 ans=0;
    scanf("%d",&n);
    for(i=1;i<=n;i++)scanf("%d",&plank[i]);
    for(i=n/2;i>=1;i--)heapAdjust(i);
    while (n>1)
    {
        t=plank[1];plank[1]=plank[n];n--;
        heapAdjust(1);
        t+=plank[1];plank[1]=t;
        heapAdjust(1);
        ans+=t;
    }
    printf("%I64d\n",ans);
    return 0;
}