[POJ1804]Brainman

来源：互联网发布：java 余弦相似度算法编辑：程序博客网时间：2024/05/17 03:37

假如我们要尽量少的次数交换元素那么我们每次交换逆序对数一定是最优的只有这样才能尽可能的去掉逆序对数的数量

因此就是求逆序对数

原谅我不会动态内存……只拿了G++里的第二不过好歹混到第一页了

#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<cstring>#include<set>#include<cstring>#include<iomanip>#define mid ((l+r)>>1)#define debug(x) cerr<<#x<<"="<<x<<endlusing namespace std;typedef pair<int,int> pii;const int INF = 0x7f7f7f7f;const double pi=acos(-1.0);int a[1001],ans=0;int temp[1001];inline int init(){int now=0,ju=1;char c;bool flag=false;while(1){c=getchar();if(c=='-')ju=-1;else if(c>='0'&&c<='9'){now=now*10+c-'0';flag=true;}else if(flag)return now*ju;}}void Mergesort(int l,int r){int pointer=l;int i=l,j=mid+1;while(i<=mid&&j<=r){if(a[i]>a[j]){temp[pointer++]=a[j++];ans+=(mid-i+1);}else{temp[pointer++]=a[i++];}}while(i<=mid){temp[pointer++]=a[i++];}while(j<=r){temp[pointer++]=a[j++];}for(int k=l;k<=r;k++){a[k]=temp[k];}return;}void Merge(int l,int r){if(l<r){Merge(l,mid);Merge(mid+1,r);Mergesort(l,r);}}int main(){int cas;short int n;cas=init();for(int i=1;i<=cas;i++){ans=0;n=init();memset(temp,0,sizeof(temp));for(int j=1;j<=n;j++){a[j]=init();}Merge(1,n);printf("Scenario #%d:\n%d\n\n",i,ans);}return 0;}

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