POJ 1804 Brainman
来源:互联网 发布:魔镜数据 编辑:程序博客网 时间:2024/06/05 00:10
Brainman
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 9094 Accepted: 4879
Description
Background
Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would love to be able to do cool things like that, too. He wants to beat his brother in a similar task.
Problem
Here's what Charlie thinks of. Imagine you get a sequence of N numbers. The goal is to move the numbers around so that at the end the sequence is ordered. The only operation allowed is to swap two adjacent numbers. Let us try an example:
Start with: 2 8 0 3
swap (2 8) 8 2 0 3
swap (2 0) 8 0 2 3
swap (2 3) 8 0 3 2
swap (8 0) 0 8 3 2
swap (8 3) 0 3 8 2
swap (8 2) 0 3 2 8
swap (3 2) 0 2 3 8
swap (3 8) 0 2 8 3
swap (8 3) 0 2 3 8
So the sequence (2 8 0 3) can be sorted with nine swaps of adjacent numbers. However, it is even possible to sort it with three such swaps:
Start with: 2 8 0 3
swap (8 0) 2 0 8 3
swap (2 0) 0 2 8 3
swap (8 3) 0 2 3 8
The question is: What is the minimum number of swaps of adjacent numbers to sort a given sequence?Since Charlie does not have Raymond's mental capabilities, he decides to cheat. Here is where you come into play. He asks you to write a computer program for him that answers the question. Rest assured he will pay a very good prize for it.
Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would love to be able to do cool things like that, too. He wants to beat his brother in a similar task.
Problem
Here's what Charlie thinks of. Imagine you get a sequence of N numbers. The goal is to move the numbers around so that at the end the sequence is ordered. The only operation allowed is to swap two adjacent numbers. Let us try an example:
swap (2 8) 8 2 0 3
swap (2 0) 8 0 2 3
swap (2 3) 8 0 3 2
swap (8 0) 0 8 3 2
swap (8 3) 0 3 8 2
swap (8 2) 0 3 2 8
swap (3 2) 0 2 3 8
swap (3 8) 0 2 8 3
swap (8 3) 0 2 3 8
So the sequence (2 8 0 3) can be sorted with nine swaps of adjacent numbers. However, it is even possible to sort it with three such swaps:
swap (8 0) 2 0 8 3
swap (2 0) 0 2 8 3
swap (8 3) 0 2 3 8
The question is: What is the minimum number of swaps of adjacent numbers to sort a given sequence?Since Charlie does not have Raymond's mental capabilities, he decides to cheat. Here is where you come into play. He asks you to write a computer program for him that answers the question. Rest assured he will pay a very good prize for it.
Input
The first line contains the number of scenarios.
For every scenario, you are given a line containing first the length N (1 <= N <= 1000) of the sequence,followed by the N elements of the sequence (each element is an integer in [-1000000, 1000000]). All numbers in this line are separated by single blanks.
For every scenario, you are given a line containing first the length N (1 <= N <= 1000) of the sequence,followed by the N elements of the sequence (each element is an integer in [-1000000, 1000000]). All numbers in this line are separated by single blanks.
Output
Start the output for every scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the minimal number of swaps of adjacent numbers that are necessary to sort the given sequence. Terminate the output for the scenario with a blank line.
Sample Input
44 2 8 0 310 0 1 2 3 4 5 6 7 8 96 -42 23 6 28 -100 655375 0 0 0 0 0
Sample Output
Scenario #1:3Scenario #2:0Scenario #3:5Scenario #4:0
Source
思路:直接用归并排序求逆序数
#include <iostream>#include <cstring>using namespace std;const int maxn=1000+5;int a[maxn],t[maxn];int ans=0;void mergesort(int l,int r){ if(l+1<r){ int m=l+(r-l)/2; int p=l,q=m,i=l; mergesort(l,m); mergesort(q,r); while(p<m || q<r){ if(q>=r || (p<m && a[p]<=a[q])) t[i++]=a[p++]; else { t[i++]=a[q++]; ans+=m-p; } } for(int i=l;i<r;i++) a[i]=t[i]; }}int main(){ int tmp; cin>>tmp; for(int j=1;j<=tmp;j++){ int n; cin>>n; ans=0; memset(a,0,sizeof(a)); memset(t,0,sizeof(t)); for(int i=0;i<n;i++) cin>>a[i]; mergesort(0,n); cout<<"Scenario #"<<j<<":"<<endl; cout<<ans<<endl<<endl; } return 0;}
0 0
- POJ 1804 Brainman
- poj 1804 Brainman
- POJ 1804 Brainman
- POJ 1804 Brainman
- poj 1804 Brainman
- poj 1804 Brainman
- poj 1804:Brainman
- POJ 1804 Brainman
- POJ 1804 Brainman
- POJ 1804 Brainman 笔记
- poj 1804 Brainman(归并)
- poj 1804Brainman(树状数组)
- [POJ 1804] Brainman · 逆序对
- POJ 1804 Brainman(归并排序)
- poj 1804 Brainman(求逆序对)
- poj 1804 Brainman(归并排序求逆序对)
- POJ 1804 Brainman (归并排序 -- 求逆序对数)
- poj 1804 (归并排序求逆序数)Brainman
- (一个刚刚结束秋招的本科生)应届生面试经验
- Electron-为21世纪打造的文本编辑器的基础
- 2015/11/21
- HDOJ-1878欧拉回路 && 九度OJ-1027欧拉回路
- UI基础-UINavigationController、界面通信
- POJ 1804 Brainman
- 浅谈Java动态代理
- Matlab绘制二维应力云图
- 唐巧总结的40个国人iOS技术博客
- 【开篇】移动互联网下的C++
- java面试宝典整理
- UIView 中常见的方法总结
- Unix环境编程学习笔记-----编程实例---- the fist exit
- ios中摄像头/相册获取图片,压缩图片,上传服务器方法总结