POJ--3253 Fence Repair(贪心/优先队列)

记录
23:57 2023-1-25

http://poj.org/problem?id=3253

reference:《挑战程序设计竞赛(第2版)》2.2.4 p47

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

3
8
5
8

Sample Output

34

Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

贪心算法，不断的寻找当前最短的和次短的木板进行合成，将合成的木板加入继续合成直到得到一块木板。找最短和次短的木板可以用最小(大)堆(c++中的优先队列)。
找资料说贪心+优先队列 = 哈夫曼 。。。哈哈哈，数据结构毕设后哈夫曼全忘完了。。。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
#define MAX_N 100000
typedef long long ll;

int N, L[MAX_N];

void solve() {
    ll ans = 0;

    while (N > 1) {
        // get the shortest and the second shortest
        int mii1 = 0, mii2 = 1;
        if(L[mii1] > L[mii2]) swap(mii1, mii2);

        for(int i = 2; i < N; i++) {
            if(L[i] < L[mii1]) {
                mii2 = mii1;
                mii1 = i;
            } else if (L[i] < L[mii2]) {
                mii2 = i;
            }
        }

        // merge the two
        int t = L[mii1] + L[mii2];
        ans += t;

        if(mii1 == N - 1) swap(mii1, mii2);
        L[mii1] = t;
        L[mii2] = L[N - 1];
        N--;
    }
    printf("%d\n", ans);
}

//使用优先队列
void solve1() {
    ll ans = 0;

    priority_queue<int, vector<int>, greater<int> > que;

    for(int i = 0; i < N; i++) {
        que.push(L[i]);
    }

    while (que.size() > 1) {
        int l1, l2;
        l1 = que.top();
        que.pop();
        l2 = que.top();
        que.pop();

        ans += l1 + l2;
        que.push(l1 + l2);
    }
    std::cout << ans << std::endl;
}

int main() {
    scanf("%d", &N);
    for (int i = 0; i < N; i++) {
        scanf("%d", &L[i]);
    }
    solve1();
}