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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#599708 | #7108. Couleur | xuzhihaodedie | AC ✓ | 2902ms | 76832kb | C++20 | 3.3kb | 2024-09-29 09:30:27 | 2024-09-29 09:30:27 |
Judging History
answer
#include<bits/stdc++.h>
using namespace std;
#define int long long
//#define x first
//#define y second
#define PII pair<int,int>
const int N=1e5+10;
struct node {
int l,r,sum;
}t[N*25];
int root[N],idx,a[N];
int z[N];
void build(int &x,int l,int r) {
x=++idx;
if(l==r) return ;
int mid=l+r>>1;
build(t[x].l,l,mid);
build(t[x].r,mid+1,r);
}
void insert(int x,int &y,int l,int r,int k) {
y=++idx;
t[y]=t[x];
t[y].sum++;
if(l==r) return ;
int mid=l+r>>1;
if(k<=mid) insert(t[x].l,t[y].l,l,mid,k);
else insert(t[x].r,t[y].r,mid+1,r,k);
}
int query(int x,int y,int l,int r,int val) {//找比val大的数有几个
if(l==r) {
if(l>=val) return t[y].sum-t[x].sum;
return 0;
}
int mid=l+r>>1;
if(val<=l) return t[y].sum-t[x].sum;
if(val>r) return 0;
if(val<=mid) return query(t[x].l,t[y].l,l,mid,val)+query(t[x].r,t[y].r,mid+1,r,val);
else return query(t[x].r,t[y].r,mid+1,r,val);
}
int query1(int x,int y,int l,int r,int val) {//找比val小的数有几个
if(l==r) {
if(l<=val) return t[y].sum-t[x].sum;
return 0;
}
int mid=l+r>>1;
if(val>=r) return t[y].sum-t[x].sum;
if(val<l) return 0;
if(val<=mid) return query1(t[x].l,t[y].l,l,mid,val);
else return query1(t[x].r,t[y].r,mid+1,r,val)+query1(t[x].l,t[y].l,l,mid,val);
}
void solve() {
int n;
cin>>n;
int maxn=0;
idx=0;
for(int i=1;i<=n;i++) cin>>a[i],maxn=max(maxn,a[i]);
int ans=0;
build(root[0],1,maxn);
for(int i=1;i<=n;i++) {
ans+=query(root[0],root[i-1],1,maxn,a[i]+1);
insert(root[i-1],root[i],1,maxn,a[i]);
}
multiset<int> s;
set<int> st;
st.insert(0);
st.insert(n+1);
s.insert(ans);
vector<int> w;
for(int i=1;i<=n;i++) cin>>z[i];
map<PII,int> mp;
mp[{0,n+1}]=ans;
for(int i=1;i<=n;i++) {
auto p=s.end();
p--;
ans=*p;
w.push_back(ans);
int x=z[i];
x^=ans;
auto q=st.upper_bound(x);
int r=*q;
q--;
int l=*q;
int res1=0,res2=0,res3=0;
if(x-l<=r-x) {
for(int j=l+1;j<x;j++) {
res1+=query(root[l],root[j-1],1,maxn,a[j]+1);//左区间逆序对数
}
for(int j=l+1;j<x;j++) {
res2+=query1(root[x],root[r-1],1,maxn,a[j]-1);//横跨两区间的逆序对数
}
int ret1=0,ret2=0;
ret1+=query(root[l],root[x-1],1,maxn,a[x]+1);
ret2+=query1(root[x],root[r-1],1,maxn,a[x]-1);
res3=ret1+ret2;
int res=mp[{l,r}];
int ret=res-res1-res2-res3;//右区间逆序对个数
s.erase(s.lower_bound(res));
s.insert(ret);
s.insert(res1);
mp[{l,x}]=res1;
mp[{x,r}]=ret;
} else {
for(int j=x+1;j<r;j++) {
res1+=query(root[x],root[j-1],1,maxn,a[j]+1);//右区间逆序对数
}
for(int j=x+1;j<r;j++) {
res2+=query(root[l],root[x-1],1,maxn,a[j]+1);//横跨两区间的逆序对数
}
int ret1=0,ret2=0;
ret1+=query(root[l],root[x-1],1,maxn,a[x]+1);
ret2+=query1(root[x],root[r-1],1,maxn,a[x]-1);
res3=ret1+ret2;
int res=mp[{l,r}];
int ret=res-res1-res2-res3;//左区间逆序对个数
s.erase(s.lower_bound(res));
s.insert(ret);
s.insert(res1);
mp[{l,x}]=ret;
mp[{x,r}]=res1;
}
st.insert(x);
}
for(int i=0;i<w.size();i++) {
if(i<w.size()-1) cout<<w[i]<<" ";
else cout<<w[i]<<endl;
}
}
signed main() {
int T=1;
cin>>T;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
while(T--) {
solve();
}
}
这程序好像有点Bug,我给组数据试试?
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 1ms
memory: 5640kb
input:
3 5 4 3 1 1 1 5 4 5 3 1 10 9 7 1 4 7 8 5 7 4 8 21 8 15 5 9 2 4 5 10 6 15 4 8 8 1 12 1 10 14 7 14 2 9 13 10 3 37 19 23 15 7 2 10 15 2 13 4 5 8 7 10
output:
7 0 0 0 0 20 11 7 2 0 0 0 0 0 0 42 31 21 14 14 4 1 1 1 0 0 0 0 0 0
result:
ok 3 lines
Test #2:
score: 0
Accepted
time: 2902ms
memory: 76832kb
input:
11116 10 10 5 10 3 6 4 8 5 9 8 31 27 24 11 12 3 0 2 3 1 10 8 2 7 2 8 10 1 10 9 10 6 5 2 13 2 1 0 1 3 1 10 7 10 7 6 1 3 10 6 7 9 21 18 10 1 6 5 4 8 9 10 10 2 10 4 8 8 5 7 2 6 7 20 10 9 1 15 0 4 2 9 7 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 10 1 2 3 4 5 6 7 8 9 10 6 3 5 2 7 10 9 1 4 8 10 1 10 1 3...
output:
21 18 16 12 10 6 4 1 1 0 12 12 10 10 4 4 4 2 1 0 20 16 9 5 3 3 3 0 0 0 22 14 8 8 5 5 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 19 12 7 4 4 2 2 1 0 0 20 18 8 3 1 1 0 0 0 0 45 21 21 10 3 3 3 0 0 0 17 11 8 2 1 1 1 0 0 0 13 4 1 0 0 0 0 0 0 0 29 27 22 15 9 7 4 3 1 0 26 16 9 2 1 1 1 1 1 0 0 0 0 0 0 ...
result:
ok 11116 lines
Extra Test:
score: 0
Extra Test Passed