POJ

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题意:让相邻的两个数不断的交换,直到整个序列不下降,求最少交换次数.

分析:归并排序求逆序数.

参考代码:

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<iostream>using namespace std;const int maxn = 1e3+5;int n;int a[maxn];int ans;int tmp[maxn];void Merge( int l, int mid, int r){    int i = l;    int j = mid+1;    int k = l;    while( i <= mid && j <= r)    {        if( a[i] > a[j])        {            tmp[k++] = a[j++];            ans += mid-i+1;        }        else            tmp[k++] = a[i++];    }    while( i <= mid)        tmp[k++] = a[i++];    while( j <= r)        tmp[k++] = a[j++];    for( int i = l; i <= r; i++)        a[i] = tmp[i];}void MergeSort( int l, int r)//归并排序{    if( l < r)    {        int mid = (l+r)>>1;        MergeSort(l,mid);//左边排序        MergeSort(mid+1,r);//右边排序        Merge(l,mid,r);//合并    }}int main(){    int T;    scanf("%d",&T);    while( T--)    {        scanf("%d",&n);        for( int i = 0; i < n; i++)            scanf("%d",&a[i]);        ans = 0;        MergeSort(0,n-1);        static int cas = 1;        printf("Scenario #%d:\n%d\n\n",cas++,ans);    }    return 0;}