HDU 1394 Minimum Inversion Number
2014-08-13 14:19
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题解:首先是很基础的树状数组求逆序对,然后对于每一个第一个数往后移动,对于逆序数的贡献是n-a[i]-1-a[i]。枚举然后求最小值即可。
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; int n,c[5001],x,a[5001]; int add(int x){while(x<=n)c[x]++,x+=x&-x;} int sum(int x){int s=0;while(x>0)s+=c[x],x-=x&-x;return s;} int main(){ while(~scanf("%d",&n)){ int ans=0; memset(c,0,sizeof(c)); for(int i=0;i<n;i++){ scanf("%d",&a[i]); ans+=(i-sum(a[i]+1)); add(a[i]+1); }int tmp=ans; for(int i=0;i<n-1;i++){ tmp=tmp-a[i]+n-a[i]-1; ans=min(ans,tmp); }printf("%d\n",ans); } return 0; }
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