哈夫曼树结合堆排序 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 needsN(1 ≤N≤ 20,000) planks of wood, each having some integer lengthLi(1 ≤Li≤ 50,000) units. He then purchases a single long board just long enough to saw into theNplanks (i.e., whose length is the sum of the lengthsLi). 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 theN-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 theNplanks. 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
Lines 2.. N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3 8 5 8
Sample Output
34
很自然的想到的哈夫曼树(贪心策略).
但是怎么样排序有效率呢？ 每次只是取出两个最小的，然后再相加，再放进去，再排序。
经过测试，普通的排序会超时。
下面代码运行正确，但是超时. 主要是sort 函数太费时了，因为每次排序只是插入一个数而已。
#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;
}
由于只是插入一个数。下面自己写插入排序：
#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]){ //插入完成退出循环
arr[i-1] = temp;
break;
}else
arr[i-1] = arr[i];
}
if(i==len) //如果遍历到了最后，也没插入
arr[i-1] = temp;
index++;
sum += temp;
}
sum += arr[len-1] + arr[len-2]; //把剩余的两个数加上
cout << sum << endl;
}
return 0;
}
运行时间 550ms
算是勉强通过。
下面写更高效的堆排序，小顶推。
#include <iostream>
using namespace std;
typedef long long int64;
int len;
int64 arr[20001]; //arr[0] 存数当前数组的长度
//调整堆
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);
}
}
//想堆中插入一个数
void inheap(int64 key) {
arr[++arr[0]] = key;
int64 i = arr[0];
//i就是当前插入的最后位置. 循环 lgn 次即可
while (i > 1 && arr[i] < arr[i / 2]) {
swap(arr[i], arr[i / 2]);
i = i / 2;
}
}
//返回最小的两个数的和
int64 get() {
int64 p = arr[1], q;
arr[1] = arr[arr[0]]; //第一个位置 和 最后一个位置交换
arr[0]--; //数组长度减1
heap(1); //调整位置1
q = arr[1];
arr[1] = arr[arr[0]];
arr[0]--;
heap(1);
inheap(p + q); //返回最小的两个数的和
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;
}
运行时间70ms