problem poj 1804 ati 遞歸 scenario over aps first mem

Brainman

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 12175		Accepted: 6147

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.

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.

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

4
4 2 8 0 3
10 0 1 2 3 4 5 6 7 8 9
6 -42 23 6 28 -100 65537
5 0 0 0 0 0

Sample Output

Scenario #1:
3

Scenario #2:
0

Scenario #3:
5

Scenario #4:
0
題意 兩個位置的元素可以任意交換 交換最少的數量使序列有序
解析  歸並排序求逆序對
AC代碼

#include<iostream>
#include<stdio.h>
using namespace std;
const int maxn = 1e5+50,inf=0x3f3f3f3f;
typedef long long ll;
const int mod=1e9+7;
int a[maxn],t[maxn],ans;
void merge_sort(int *a,int x,int y,int *t)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;    //劃分
        int p=x,q=m,i=x;
        merge_sort(a,x,m,t); //遞歸求解
        merge_sort(a,m,y,t); //遞歸求解
        while(p<m||q<y)    //只要有一個序列非空就要繼續合並
        {

            if(q>=y||( p<m && a[p]<=a[q]))  //當第二個序列為空直接復制 兩個都不為空進行比較
                t[i++]=a[p++];
            else
                t[i++]=a[q++],ans+=m-p;
        }
        for(int i=x;i<y;i++)    //將排好序的數組再復制回給a
            a[i]=t[i];
    }
}
int main()
{
    int n,m,kase=1;
    cin>>m;
    while(m--)
    {
        cin>>n;
        ans=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
        }
        merge_sort(a,0,n,t);//左閉右開
        printf("Scenario #%d:\n%d\n\n",kase++,ans);
    }
}

歸並排序

/*

分治三步法：

    劃分問題：把序列分成元素個數盡量相等的兩半
    遞歸求解：把兩半元素分別排序
    合並問題：把兩個有序表合並成一個
*/
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+50,inf=0x3f3f3f3f;
typedef long long ll;
const int mod=1e9+7;
int a[maxn],t[maxn];
void merge_sort(int *a,int x,int y,int *t)
{
    if(y-x>1)
    {
        int m=x+(y-x)/2;    //劃分
        int p=x,q=m,i=x;
        merge_sort(a,x,m,t); //遞歸求解
        merge_sort(a,m,y,t); //遞歸求解
        while(p<m||q<y)    //只要有一個序列非空就要繼續合並
        {

            if(q>=y||( p<m && a[p]<=a[q]))  //當第二個序列為空直接復制 兩個都不為空進行比較
                t[i++]=a[p++];
            else
                t[i++]=a[q++];
        }
        for(int i=x;i<y;i++)    //將排好序的數組再復制回給a
            a[i]=t[i];
    }
}
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
    }
    merge_sort(a,0,n,t);//左閉右開
    for(int i=0;i<n;i++)
    {
        cout<<a[i]<<" ";
    }
    cout<<endl;
}

POJ 1804 逆序對數量 / 歸並排序