题解 Time
首先枚举最大值,两边分别求逆序对的做法是错误的,这里是来自战神的hack数据
1 2 100 99 98 3 97 96 95 94 93 92 91
显然3应该跨过最大值到左边去,所以这个做法就没有正确性了
然后正解:
发现最小值一定会被移到最边上,而且因为它最小,就不会再有数跨过它移动
所以这东西没有后效性,且移动后会形成子问题
所以不断取最小值,贪心地取次数最小的方向移动,然后把它删了就行了
- 所以……要交换序列里的元素,使它们成某种顺序的时候,可以考虑「什么元素不可被跨越」或者「什么元素一定会被移动到什么位置」,然后设法固定这些元素,其余的转化为子问题?
Code:
#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
#define fir first
#define sec second
#define make make_pair
//#define int long long
char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
int ans=0, f=1; char c=getchar();
while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
return ans*f;
}
int n;
int x[N];
namespace task1{
int uni[N], usize, lim;
int ans=INF, a[N];
inline void upd(int i) {for (; i<=lim; i+=i&-i) ++a[i];}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=a[i]; return ans;}
void solve() {
for (int i=1; i<=n; ++i) uni[i]=x[i];
sort(uni+1, uni+n+1);
usize=unique(uni+1, uni+n+1)-uni-1; lim=usize+2;
for (int i=1; i<=n; ++i) x[i]=lower_bound(uni+1, uni+usize+1, x[i])-uni+1;
for (int i=1,cnt; i<=n; ++i) {
memset(a+1, 0, sizeof(int)*lim);
cnt=0;
for (int j=i; j; --j) {
cnt+=query(x[j]-1);
upd(x[j]);
}
memset(a+1, 0, sizeof(int)*lim);
for (int j=i+1; j<=n; ++j) {
cnt+=query(x[j]-1);
upd(x[j]);
}
ans=min(ans, cnt);
}
printf("%d\n", ans);
exit(0);
}
}
namespace task{
int a[N]; ll ans;
pair<int, int> s[N];
inline void upd(int i) {for (; i<=n; i+=i&-i) ++a[i];}
inline int query(int i) {int ans=0; for (; i; i-=i&-i) ans+=a[i]; return ans;}
inline void del(int i) {for (; i<=n; i+=i&-i) --a[i];}
void solve() {
for (int i=1; i<=n; ++i) upd(i);
for (int i=1; i<=n; ++i) s[i]=make(x[i], i);
sort(s+1, s+n+1);
for (reg pos1=1,pos2=1,top=1,t1,t2; top<=n; pos1=pos2=++top) {
while (top+1<=n && s[pos1].fir==s[top+1].fir) ++top;
pos2=top;
while (pos1<pos2) {
t1=min(query(s[pos1].sec-1), query(n)-query(s[pos1].sec));
t2=min(query(s[pos2].sec-1), query(n)-query(s[pos2].sec));
if (t1<t2) ans+=t1, del(s[pos1++].sec);
else ans+=t2, del(s[pos2--].sec);
}
ans+=min(query(s[pos1].sec-1), query(n)-query(s[pos1].sec)), del(s[pos1].sec);
}
printf("%lld\n", ans);
exit(0);
}
}
signed main()
{
n=read();
for (int i=1; i<=n; ++i) x[i]=read();
task::solve();
return 0;
}