POJ 3253 Fence Repair （堆排序）

来源：互联网发布：erdai cms 编辑：程序博客网时间：2024/05/19 00:36

题目连接：～(￣▽￣～)(～￣▽￣)～

用堆排序建立小根堆每回选取2个最小的值相加，再把得到的值放回堆中排序

code：

#include <stdio.h>int num[20009], n = 0;__int64 sum = 0, sum2 = 0;void sift(int l);void heapsort();int main(){int i = 0;while(scanf("%d",&n) != EOF){sum = sum2 = 0;for(i = 1; i<=n; i++)scanf("%d",&num[i]);heapsort();printf("%I64d\n",sum2);}return 0;}void sift(int l){int i = 0, j = 0, x = 0;x = num[l];i = l; j = 2*i;while(j<=n){if(num[j]>num[j+1] && j<n) j++;if(x>num[j]){num[i] = num[j];i = j;j = 2*j;}elsebreak;}num[i] = x;}void heapsort()//堆排序{int i = n;for(i = n/2; i>0; i--)sift(i);//从底层向上对每个节点建立小根堆while(n>=2){sum = num[1];num[1] = num[n];n--;sift(1);sum += num[1];//用sum记录两最小元素的和num[1] = sum;sift(1);//把得到的和在入堆进行排序sum2 += sum;//sum2记录当前2最小元素的和（即为要花的钱）}}