题意

切割木板, 比如一根长21的木板要切割成5, 8, 8的三块, 每次切割花费的金额为两断的长度. 比如先把21切成16和5, 花费21元, 再把16切成8和8, 花费16元, 总计消费37元. 若先把21切成13和8, 花费21元, 再把13切成8和5, 花费13元, 总计消费34元, 花销更少
现在就要求出最小开销

思路

这道题可以用贪心, 这里主要记录一下白书上提供的优先队列 这个非常好的思路
把切割逆向看作拼接过程, 定义一个从小到大排列的优先队列

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

每次取出最短的两根木板拼接, 再将这个拼接后的长木板推入优先队列
直到fen.size() == 1, 这时木板被拼接完毕

记得看数据范围需要用long long !!

AC代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>using namespace std;
const int maxn = 20000 + 100;
priority_queue<int, vector<int>, greater<int> > fen;
typedef long long LL;
//int s[maxn];int main()
{int n, a;while( ~scanf("%d",&n) ){LL sum = 0;while( !fen.empty() ) fen.pop();for( int i = 0; i < n; i++ ){scanf("%d",&a);fen.push(a);}while( fen.size() > 1 ){int l1 = fen.top();fen.pop();int l2 = fen.top();fen.pop();sum += l1 + l2;fen.push(l1+l2);}printf("%lld\n",sum);}return 0;
}

转载于:https://www.cnblogs.com/JinxiSui/p/9740588.html

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题意

思路

AC代码

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