USACO月赛2017.02 铂金组T1--MINCROSS

Description

Why did the cow cross the road? We may never know the full reason, but it is certain

that Farmer John’s cows do end up crossing the road quite frequently. In fact, they

end up crossing the road so often that they often bump into each-other when their

paths cross, a situation Farmer John would like to remedy.

Farmer John raises NN breeds of cows (1≤N≤100,000), and each of his fields is

dedicated to grazing for one specific breed; for example, a field dedicated to breed

12 can only be used for cows of breed 12 and not of any other breed. A long road

runs through his farm. There is a sequence of NN fields on one side of the road (one

for each breed), and a sequence of NN fields on the other side of the road (also one

for each breed). When a cow crosses the road, she therefore crosses between the

two fields designated for her specific breed.

Had Farmer John planned more carefully, he would have ordered the fields by breed

the same way on both sides of the road, so the two fields for each breed would be

directly across the road from each-other. This would have allowed cows to cross the

road without any cows from different breeds bumping into one-another. Alas, the

orderings on both sides of the road might be different, so Farmer John observes that

there might be pairs of breeds that cross. A pair of different breeds (a,b)(a,b) is

“crossing” if any path across the road for breed aa must intersect any path across the

road for breed bb.

Farmer John would like to minimize the number of crossing pairs of breeds. For

logistical reasons, he figures he can move cows around on one side of the road so

the fields on that side undergo a “cyclic shift”. That is, for some 0≤k<N, every

cow re-locates to the field kk fields ahead of it, with the cows in the last kk fields

moving so they now populate the first kk fields. For example, if the fields on one side

of the road start out ordered by breed as 3, 7, 1, 2, 5, 4, 6 and undergo a cyclic shift

by k=2k=2, the new order will be 4, 6, 3, 7, 1, 2, 5. Please determine the minimum

possible number of crossing pairs of breeds that can exist after an appropriate cyclic

shift of the fields on one side of the road.

INPUT FORMAT (file mincross.in):

The first line of input contains NN. The next NN lines describe the order, by breed ID,

of fields on one side of the road; each breed ID is an integer in the range 1…N1…N.

The last NN lines describe the order, by breed ID, of the fields on the other side of

the road.

OUTPUT FORMAT (file mincross.out):

Please output the minimum number of crossing pairs of breeds after a cyclic shift of

the fields on one side of the road (either side can be shifted).

题解

额，太水了。先树状数组求出不移动的答案，然后一个一个枚举推一下就好。（看到有人写的主席树？沃不会啊）

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100006
#define lowbit(x) (x&-x)
#define LL long long
using namespace std;
inline char nc(){
static char buf[100000],*i=buf,*j=buf;
return i==j&&(j=(i=buf)+fread(buf,1,100000,stdin),i==j)?EOF:*i++;
}
inline int _read(){
char ch=nc();int sum=0;
while(!(ch>='0')&&ch<='9')ch=nc();
while(ch>='0'&&ch<='9')sum=sum*10+ch-48,ch=nc();
return sum;
}
int n,a[maxn],b[maxn],p1[maxn],p2[maxn];
LL ans,ans0,ans1,f[maxn];
void put(int x,int y){for(;x<=n;x+=lowbit(x))f[x]+=y;}
int get(int x){
int sum=0;
for(;x;x-=lowbit(x))sum+=f[x];
return sum;
}
int main(){
freopen("mincross.in","r",stdin);
freopen("mincross.out","w",stdout);
n=_read();
for(int i=1;i<=n;i++)a[i]=_read(),p1[a[i]]=i;
for(int i=1;i<=n;i++)b[i]=_read(),p2[b[i]]=i;
for(int i=n;i>=1;i--)ans0+=get(p1[b[i]]),put(p1[b[i]],1);
ans=ans1=ans0;
for(int i=1;i<=n;i++){
ans0=ans0-p1[b[i]]+1+n-p1[b[i]];
ans=min(ans,ans0);
}
for(int i=1;i<=n;i++){
ans1=ans1-p2[a[i]]+1+n-p2[a[i]];
ans=min(ans,ans1);
}
printf("%lld",ans);
return 0;
}