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QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#603155 | #8718. 保区间最小值一次回归问题 | UESTC_DECAYALI# | WA | 474ms | 37504kb | C++20 | 3.0kb | 2024-10-01 14:59:46 | 2024-10-01 14:59:47 |
Judging History
answer
#include<cstdio>
#include<iostream>
#include<vector>
#include<algorithm>
#define int long long
#define RI register int
#define CI const int&
using namespace std;
const int N=500005,INF=1e18;
struct ifo
{
int l,r,v;
friend inline bool operator < (const ifo& A,const ifo& B)
{
return A.v>B.v;
}
}O[N]; int t,n,m,tl,tr,a[N],fa[N],mxp[N];
inline int getfa(CI x)
{
return fa[x]!=x?fa[x]=getfa(fa[x]):x;
}
class Segment_Tree
{
private:
int mn[N<<2];
public:
#define TN CI now=1,CI l=tl,CI r=tr
#define LS now<<1,l,mid
#define RS now<<1|1,mid+1,r
inline void build(TN)
{
if (l==r) return (void)(mn[now]=INF);
int mid=(l+r)/2; build(LS); build(RS);
}
inline int query(CI beg,CI end,TN)
{
if (beg>end) return INF;
if (beg<=l&&r<=end) return mn[now]; int mid=(l+r)/2,ret=INF;
if (beg<=mid) ret=min(ret,query(beg,end,LS));
if (end>mid) ret=min(ret,query(beg,end,RS));
return ret;
}
inline void update(CI pos,CI mv,TN)
{
if (l==r) return (void)(mn[now]=mv); int mid=(l+r)/2;
if (pos<=mid) update(pos,mv,LS); else update(pos,mv,RS);
mn[now]=min(mn[now<<1],mn[now<<1|1]);
}
#undef TN
#undef LS
#undef RS
}SEG;
inline int solve(CI l,CI r,CI v)
{
vector <int> vec;
for (RI i=l;i<=r;++i)
{
int x=getfa(O[i].l);
while (x<=O[i].r)
{
vec.push_back(x);
fa[x]=x+1; x=getfa(x);
}
}
sort(vec.begin(),vec.end());
for (RI i=0;i<vec.size();++i) mxp[i]=0;
for (RI i=l;i<=r;++i)
{
int L=lower_bound(vec.begin(),vec.end(),O[i].l)-vec.begin();
int R=lower_bound(vec.begin(),vec.end(),O[i].r+1)-vec.begin()-1;
if (L>R) return -1; else mxp[R]=max(mxp[R],L);
}
tl=0; tr=vec.size(); SEG.build(); SEG.update(0,0);
int lst=-1;
for (RI i=0;i<vec.size();++i)
{
if (i-1>=0) lst=max(lst,mxp[i-1]);
int tmp=SEG.query(lst+1,i)+abs(a[vec[i]]-v);
SEG.update(i+1,tmp);
if (a[vec[i]]<=v) lst=i;
}
lst=max(lst,mxp[vec.size()-1]);
return SEG.query(lst+1,vec.size());
}
signed main()
{
//freopen("F.in","r",stdin);
for (scanf("%lld",&t);t;--t)
{
scanf("%lld%lld",&n,&m);
for (RI i=1;i<=n;++i) scanf("%lld",&a[i]);
for (RI i=1;i<=n+1;++i) fa[i]=i;
for (RI i=1;i<=m;++i) scanf("%lld%lld%lld",&O[i].l,&O[i].r,&O[i].v);
sort(O+1,O+m+1); bool flag=1; int ans=0;
for (RI i=1;i<=m;)
{
int j=i;
while (j+1<=m&&O[j+1].v==O[i].v) ++j;
int tmp=solve(i,j,O[i].v);
if (tmp==-1) { flag=0; break; }
ans+=tmp; i=j+1;
}
if (!flag) puts("-1"); else printf("%lld\n",ans);
}
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 2ms
memory: 12264kb
input:
1 3 2 2023 40 41 1 1 2022 2 3 39
output:
2
result:
ok 1 number(s): "2"
Test #2:
score: -100
Wrong Answer
time: 474ms
memory: 37504kb
input:
1000 100 100 1 35141686 84105222 84105220 7273527 178494861 178494861 112519027 77833654 77833656 261586535 278472336 278472336 261586536 416361017 416361017 426649080 323519513 278472337 420127821 420127823 420127823 482516531 434108818 420127821 631535744 615930922 546346921 546346920 546346920 70...
output:
53 48 44 44 50 49 51 49 48 52 49 60 957 54 50 57 50 66 57 48 50 54 43 58 52 59 53 44 44 46 42 56 57 48 61 49 48 44 39 51 45 52 47 52 909 957 48 50 51 53 48 44 56 53 43 48 58 57 54 51 41 50 44 43 55 42 52 47 38 53 54 50 47 48 46 49 43 48 42 47 48 38 46 49 54 49 48 48 42 46 49 41 51 57 48 44 49 50 49 ...
result:
wrong answer 1st numbers differ - expected: '49', found: '53'