UVa-煎饼
来源:互联网 发布:java的流 编辑:程序博客网 时间:2024/04/27 13:44
Stacks of Flapjacks
Background
Stacks and Queues are often considered the bread and butter of data structures and find use in architecture, parsing, operating systems, and discrete event simulation. Stacks are also important in the theory of formal languages.
This problem involves both butter and sustenance in the form of pancakes rather than bread in addition to a finicky server who flips pancakes according to a unique, but complete set of rules.
The Problem
Given a stack of pancakes, you are to write a program that indicates how the stack can be sorted so that the largest pancake is on the bottom and the smallest pancake is on the top. The size of a pancake is given by the pancake's diameter. All pancakes in a stack have different diameters.
Sorting a stack is done by a sequence of pancake ``flips''. A flip consists of inserting a spatula between two pancakes in a stack and flipping (reversing) the pancakes on the spatula (reversing the sub-stack). A flip is specified by giving the position of the pancake on the bottom of the sub-stack to be flipped (relative to the whole stack). The pancake on the bottom of the whole stack has position 1 and the pancake on the top of a stack ofn pancakes has position n.
A stack is specified by giving the diameter of each pancake in the stack in the order in which the pancakes appear.
For example, consider the three stacks of pancakes below (in which pancake 8 is the top-most pancake of the left stack):
8 7 2 4 6 5 6 4 8 7 8 4 5 5 6 2 2 7
The stack on the left can be transformed to the stack in the middle via flip(3). The middle stack can be transformed into the right stack via the commandflip(1).
The Input
The input consists of a sequence of stacks of pancakes. Each stack will consist of between 1 and 30 pancakes and each pancake will have an integer diameter between 1 and 100. The input is terminated by end-of-file. Each stack is given as a sin
gle line of input with the top pancake on a stack appearing first on a line, the bottom pancake appearing last, and all pancakes separated by a space.
The Output
For each stack of pancakes, the output should echo the original stack on one line, followed by some sequence of flips that results in the stack of pancakes being sorted so that the largest diameter pancake is on the bottom and the smallest on top. For each stack the sequence of flips should be terminated by a 0 (indicating no more flips necessary). Once a stack is sorted, no more flips should be made.
Sample Input
1 2 3 4 55 4 3 2 15 1 2 3 4
Sample Output
1 2 3 4 505 4 3 2 11 05 1 2 3 41 2 0
// by BNU_LZM#include<cstdio>#include<cmath>#include<string>#include<sstream>#include<cstring>#include<iostream>#include<algorithm>using namespace std;const int maxn = 50;struct node{int num, id;bool operator < (const node &rhs) const{return num < rhs.num;}}a[maxn], b[maxn];int n;void init(){freopen("practicein.txt","r",stdin);//freopen("practiceout.txt","w",stdout);}void readdata(){string s;while(getline(cin, s)){stringstream ss(s);int nn; n = 0;while(ss >> nn) {a[++n].num = nn;a[n].id = n;b[n].num = nn;b[n].id = n;}sort(b+1, b+1+n);/*for(int i = 1; i <= n; i++) printf("%d ", b[i].num);printf("\n");*/for(int i = n; i > 0; i--){int q = b[i].id;for(int j = 1; j <= n; j++){if(a[j].id == q){if(j == i) break;if(j != 1) {printf("%d ", n+1-j);for(int k = j; k > j/2; k--){ if(k == j-k+1) continue;swap(a[k], a[j-k+1]);}}printf("%d ", n+1-i);for(int k = i; k > i/2; k--){ if(k == i-k+1) continue;swap(a[k], a[i-k+1]);}}}}printf("0\n");}}int main(){init();readdata();work();return 0;}
- UVa-煎饼
- UVA 120 煎饼堆
- uva 120 煎饼堆
- UVa 120 煎饼堆
- UVa 120 煎饼堆
- UVA 120 煎饼
- 例题8-1 煎饼 UVa 120
- 煎饼(Stacks of Flapjacks,UVa 120)
- UVA 120 Stacks of Flapjacks 煎饼
- uva 120 Stacks of Flapjacks 煎饼
- 水题(翻煎饼,uva 120)
- UVA 120 Stacks of Flapjacks (煎饼)
- UVa 120 Stacks of Flapjacks 摞煎饼
- 煎饼排序 (Stacks of Flapjacks, Uva 120)
- UVa Problem 120 Stacks of Flapjacks (煎饼堆)
- UVA 120 Stacks of Flapjacks(翻煎饼,啦啦啦啦~)
- HDU 1988 Flipping Burned Pancakes (同uva煎饼一样的题)
- 翻煎饼
- [LeetCode] 42. Trapping Rain Water java
- python爬虫常用第三方库
- 当编写Servlet时出现type Status report message HTTP method GET is not supported by this URL description The
- iOS开发
- MySQL推荐书籍
- UVa-煎饼
- MUI能否用于微信开发?
- 第三次c++实验 分段函数
- android EditText光标和下划线设置
- js中call和apply的区别
- 浅谈Android五大布局——LinearLayout、FrameLayout和AbsoulteLayout
- python中string的相关操作实例代码
- 蓝桥杯——算法训练 求找公倍数 (Vip试题)
- swift类和结构体(一)