poj3253Fence Repair 解题题解
2015-07-09 22:22
330 查看
Fence Repair
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
Sample Output
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).
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 30807 | Accepted: 9947 |
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).
#include<stdio.h> #include<stdlib.h> void Insert(int *a,int len,int Count) { int i,small,count; count = Count; a[count] = len; while(count != 0) { i = (count-1)/2; if(a[count] < a[i]) { small = a[count]; a[count] = a[i]; a[i] = small; count = i; } else break; } } int Deletemin(int *a,int Count) { int count = Count; int i = 0,bigger,j,k,t; t = a[0]; a[0] = a[count --]; while((2*i+1) <= count) { j = 2 * i + 1; if(a[j+1] < a[j]&&j < count ) j = j+1; if(a[i] > a[j]) { bigger = a[i]; a[i] = a[j]; a[j] = bigger; i = j; } else break; } return t; } int main(void) { int N,l,i = 0,count = 0; long long S = 0,sum = 0; int len[20010]; scanf("%d",&N); while(count < N) { scanf("%d",&l); Insert(len,l,count); count ++; S += l; } count --; while(count > 1) { sum = Deletemin(len,count); count --; sum += Deletemin(len,count); Insert(len,sum,count); S += sum; } printf("%lld\n",S); return 0; }
相关文章推荐
- AngularJS ui-router (嵌套路由)
- JQuery中$.ajax()方法参数详解
- 【 D3.js 入门系列 — 1 】 第一个程序 HelloWorld
- jQuery网上学习资料推荐
- 【XML】——XML与HTML异同
- extJs常用的四种Ajax异步提交
- Uva10795 - A Different Task
- JQuery——DOM操作总结
- 2015.7.9 第四课 课程重点 (css :浮动、盒子模型、绝对/相对定位)
- CSS3学习内容与心得
- css3 变形记
- #jQuery设计思想大全
- 大转盘抽奖代码 jquery
- AngularJS 简单的介绍
- Jquery easyUI的核心包导入Myeclipse报错
- OLEVARIANT的替代——FIREDAC的TFDJSONDataSets和TFDJSONDeltas
- [LeetCode][Java] Remove Nth Node From End of List
- LeetCode222 Count CompleteTree Nodes(计算完全二叉树的节点数) Java 题解
- sun.misc.unsafe类的使用
- jquery invalidation插件