哈夫曼樹結合堆排序 POJ(3253)

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤
50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made;
you should ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will
result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer  N, the number of planks 
Lines 2.. N+1: Each line contains a single integer describing the length of a needed plank

Output

Line 1: One integer: the minimum amount of money he must spend to make  N-1 cuts

Sample Input

3858

Sample Output

34

很自然的想到的哈夫曼樹(貪心策略).

但是怎麼樣排序有效率呢？ 每次只是取出兩個最小的，然後再相加，再放進去，再排序。

經過測試，普通的排序會逾時。

下面代碼運行正確，但是逾時. 主要是sort 函數太費時了，因為每次排序只是插入一個數而已。

#include <iostream>#include <algorithm>using namespace std;typedef long long int64;int len;int64 arr[20001];int main() {while (cin >> len) {for (int i = 0; i < len; i++) {cin >> arr[i];}int tmplen = len;int sum = 0;sort(arr + (len - tmplen), arr + len);while (tmplen > 1) {sort(arr + (len - tmplen), arr + len);arr[len - tmplen + 1] = arr[len - tmplen] + arr[len - tmplen + 1];sum += arr[len - tmplen + 1];tmplen--;}cout << sum << endl;}return 0;}

由於只是插入一個數。下面自己寫插入排序：

#include <iostream>#include <algorithm>using namespace std;typedef long long int64;int len;int64 arr[20001];int main() {while (cin >> len) {for (int i = 0; i < len; i++) {cin >> arr[i];}int index = 1;int64 sum = 0,temp;sort(arr, arr + len);while (index < len-1) {temp = arr[index-1] + arr[index];int i;for(i=index + 1; i<len; i++){if(temp < arr[i]){ //插入完成退出迴圈arr[i-1] = temp;break;}elsearr[i-1] = arr[i];}if(i==len) //如果遍曆到了最後，也沒插入arr[i-1] = temp;index++;sum += temp;}sum += arr[len-1] + arr[len-2]; //把剩餘的兩個數加上cout << sum << endl;}return 0;}

已耗用時間 550ms

算是勉強通過。

下面寫更高效的堆排序，小頂推。

#include <iostream>using namespace std;typedef long long int64;int len;int64 arr[20001]; //arr[0] 存數當前數組的長度//調整堆void heap(int i) {int left = i * 2;int right = i * 2 + 1;int mins = i;if (left <= arr[0] && arr[left] < arr[mins])mins = left;if (right <= arr[0] && arr[right] < arr[mins])mins = right;if (i != mins) {swap(arr[i], arr[mins]);heap(mins);}}//想堆中插入一個數void inheap(int64 key) {arr[++arr[0]] = key;int64 i = arr[0];//i就是當前插入的最後位置. 迴圈 lgn 次即可while (i > 1 && arr[i] < arr[i / 2]) {swap(arr[i], arr[i / 2]);i = i / 2;}}//返回最小的兩個數的和int64 get() {int64 p = arr[1], q;arr[1] = arr[arr[0]]; //第一個位置 和 最後一個位置交換arr[0]--; //數組長度減1heap(1); //調整位置1q = arr[1];arr[1] = arr[arr[0]];arr[0]--;heap(1);inheap(p + q); //返回最小的兩個數的和return p + q;}int main() {while (cin >> len) {arr[0] = 0;for (int i = 1; i <= len; i++) {cin >> arr[i];inheap(arr[i]);}int64 ans = 0;for (int i = 1; i < len; i++) {ans += get();}cout << ans << endl;}return 0;}

已耗用時間70ms