废话就不多说了，开始。。。
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
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
很然自的想到的哈夫曼树(贪心略策).
但是怎么样排序有效率呢？ 每次只是出取两个最小的，然后再相加，再放进去，再排序。
经过测试，一般的排序会超时。
上面代码运行准确，但是超时. 主要是sort 数函太费时了，认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。上面自己写插入排序：
#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]){ //插入实现出退环循
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]; //把余剩的两个数加上
cout << sum << endl;
}
return 0;
}
运行时间 550ms
算是强勉通过。
上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] 存数前当数组的长度
//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就是前当插入的最后位置. 环循 lgn 次便可
while (i > 1 && arr[i] < arr[i / 2]) {
swap(arr[i], arr[i / 2]);
i = i / 2;
}
}
//回返最小的两个数的和
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); //回返最小的两个数的和
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
文章结束给大家分享下程序员的一些笑话语录：
一程序员告老还乡，想安度晚年，于是决定在书法上有所造诣。省略数字……，准备好文房4宝，挥起毛笔在白纸上郑重的写下：Hello World