poj3253 Fence Repair
来源:互联网 发布:淘宝商家可以看到买家 编辑:程序博客网 时间:2024/06/05 15:30
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 theN 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 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 theN 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
3858
Sample Output
34
Hint
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).
#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;int main(){__int64 n,m,sum,a,b;while(~scanf("%I64d",&n)){sum=0;priority_queue<__int64,vector<__int64>,greater<__int64> >q;while(n--){scanf("%I64d",&m);q.push(m);}while(q.size()>1){a=q.top();q.pop();b=q.top();q.pop();sum=sum+a+b;q.push(a+b);}printf("%I64d\n",sum);}return 0;}
这个是数组模拟的优先队列
#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;int cmp(__int64 a,__int64 b){return a<b;}int main(){__int64 i,n,m,minc,sum;while(~scanf("%I64d",&n)){__int64 *w=(__int64 *)malloc((n+1)*sizeof(__int64));for(i=0;i<n;++i)scanf("%I64d",&w[i]);sort(w,w+n,cmp);sum=0;// 例如6 7 8 9此时i==0,j==2,8 13 9此时i==0,j从2增到3,若大于则已是最后一位跳出 minc=0;//for(int i=0;i<n-1;++i){sum=w[i]+w[i+1];//前两位相加 minc+=sum;for(int j=i+2;j<=n-1;++j){if(sum>w[j])//与下一位比较 {w[j-1]=w[j];if(j==n-1)//若此时是最大当j==n-1时 意味着它已经移到最后一位了可以跳出循环了 {w[j]=sum;break;}}else//如果不大于的话,就把它放在前面一位 {w[j-1]=sum;break;}}}printf("%I64d\n",minc);}return 0;}
- POJ3253 Fence Repair
- poj3253 Fence Repair( 贪心 )
- poj3253 Fence Repair
- poj3253 Fence Repair
- poj3253 Fence Repair---赫夫曼树
- poj3253 Fence Repair
- POJ3253 Fence Repair
- POJ3253 Fence Repair
- POJ3253--Fence Repair
- POJ3253 Fence Repair(贪心)
- POJ3253 Fence Repair
- poj3253 Fence Repair
- Poj3253—Fence Repair
- POJ3253-Fence Repair
- POJ3253-Fence Repair
- [POJ3253]-Fence Repair
- POJ3253-Fence Repair
- POJ3253 Fence Repair
- 并查集的简单例题
- 容器使用笔记(扩展篇)
- 制作svg动画
- 关于windows server 2008 连接oracle数据库响应极慢的问题
- hdoj 1272 小希的迷宫【并查集】
- poj3253 Fence Repair
- next数组介绍
- 黑马程序员———死锁
- tkinter菜单笔记=>持续更新
- 经典中的经典Unique Binary Search Trees II
- HALCON算子函数——Chapter 13 : Object
- HDU 4160 — Dolls 最小路径覆盖
- c语言中冒泡排序、插入排序、选择排序算法比较
- loadrunner解决用户名、密码不同问题