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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
#116387#5094. 小 H 分糖果lgvc#0 9ms14312kbC++234.5kb2023-06-28 17:10:482023-06-28 17:10:49

Judging History

你现在查看的是测评时间为 2023-06-28 17:10:49 的历史记录

  • [2024-05-31 18:28:49]
  • 管理员手动重测本题所有提交记录
  • 测评结果:0
  • 用时:13ms
  • 内存:14404kb
  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-06-28 17:10:49]
  • 评测
  • 测评结果:0
  • 用时:9ms
  • 内存:14312kb
  • [2023-06-28 17:10:48]
  • 提交

answer

//这回只花了114514min就打完了。
//真好。记得多手造几组。
#include <bits/stdc++.h>
#define int long long
#define pai 3.141592653589793238462643383279502884197169399375105820
#define MOD 1000000007
#define eps 0.00000001
inline int min(int a,int b) {return a<b?a:b;}
inline int max(int a,int b) {return a>b?a:b;}
#define ULL unsigned long long
#define LL long long
#define INF 0x3f3f3f3f
#define INF_LL 0x3f3f3f3f3f3f3f3f
static char buf[1000000],*paa=buf,*pd=buf;
static char buf2[1000000],*pp=buf2;
#define getchar() paa==pd&&(pd=(paa=buf)+fread(buf,1,1000000,stdin),paa==pd)?EOF:*paa++
inline void pc(char ch){
	if(pp-buf2==1000000) fwrite(buf2,1,1000000,stdout),pp=buf2;
	*pp++=ch;
}
inline void pcc(){
	fwrite(buf2,1,pp-buf2,stdout);
	pp=buf2;
}
inline int read(void) {
	int w=1;
	register int x(0);register char c(getchar());
	while(c<'0'||c>'9') {if(c=='-') w=-1;c=getchar();}
	while(c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
	return w*x;
}
void write(__int128 x) {
	if(x<0) pc('-'),x=-x;
	static int sta[40];
	int top=0;
	do {
		sta[top++]=x%10,x/=10;
	} while(x);
	while(top) pc(sta[--top]+48);
}
void we(__int128 x) {
	write(x);
	pc('\n');
}
inline bool cmp_xi(int a,int b) {return a<b;}
inline bool cmp_da(int a,int b) {return a>b;}
int N,Q,hd[100009],pa[100009],dep[100009],a[100009],nxt[200009],to[200009],k,b[100009],ls[30000009],rs[30000009],siz[100009],rt[400009],dfn[100009],
kd,kq,anc[100009][17];
__int128 su;
struct n_t{
	int s1,s2;
	__int128 s3;
} seg[30000009];
inline void l(int u,int v) {nxt[++k]=hd[u],to[k]=v,hd[u]=k;}
//step1: 建出 dfn
//step2: 线段树-个数,和,平方和
//step3:修改+查询
//step4: main 函数改掉
//step5: 卡常数。 
void dfs(int n,int f) {
	anc[n][0]=f;
	pa[n]=f;dep[n]=dep[f]+1;
	siz[n]=1;dfn[n]=++kq;
	for(int i=hd[n];i;i=nxt[i]) {
		if(to[i]==f) continue;
		dfs(to[i],n);
		siz[n]+=siz[to[i]];
	}
}
inline n_t mm(n_t a,n_t b) {
	return (n_t){a.s1+b.s1,a.s2+b.s2,a.s3+b.s3};
}
#define md ((l+r)>>1)
inline void up(int n,int p,int v,int x) {
	int l=0,r=INF;
	while(1) {
		seg[n].s1+=x;
		seg[n].s2+=v*x;
		seg[n].s3+=v*v*x;
		if(l==r) return;
		if(p<=md) {
			if(!ls[n]) ls[n]=++kd;
			n=ls[n];
			r=md;
		} else {
			if(!rs[n]) rs[n]=++kd;
			n=rs[n];
			l=md+1;
		}
	}
}
inline n_t q(int n,int x) {
	n_t ans=(n_t){0,0,0};
	int l=0,r=INF;
	while(n) {
		if(l==r) {
			ans=mm(ans,seg[n]);
			break;
		} else if(x>md) {
			n=rs[n];
			l=md+1;
		} else {
			ans=mm(ans,seg[rs[n]]);
			n=ls[n];
			r=md;			
		}
	}
	return ans;
}
#define ls (n<<1)
#define rs ((n<<1)|1)
void up2(int n,int l,int r,int L,int R,int x,int y) {
	if(r<L||R<l) return;
	if(L<=l&&r<=R) {
		if(!rt[n]) rt[n]=++kd;
		if(x!=-1) up(rt[n],x,x,-1);
		up(rt[n],y,y,1);
		return;
	}
	up2(ls,l,md,L,R,x,y);up2(rs,md+1,r,L,R,x,y);
}
n_t q2(int n,int l,int r,int p,int v) {
	n_t ans=q(rt[n],v);
	if(l==r) return ans;
	if(p<=md) ans=mm(ans,q2(ls,l,md,p,v));else ans=mm(ans,q2(rs,md+1,r,p,v));
	return ans;
}
#undef ls 
#undef rs
#undef md
inline n_t sv(int x,int y,int m,int v) {
	n_t A=q2(1,1,N,dfn[x],v),B=q2(1,1,N,dfn[y],v),C=q2(1,1,N,dfn[m],v);
	A.s1=A.s1+B.s1-2*C.s1;
	A.s2=A.s2+B.s2-2*C.s2;
	A.s3=A.s3+B.s3-2*C.s3;
	if(a[m]>=v) A.s1++,A.s2+=a[m],A.s3+=a[m]*a[m];
	return A;
}
signed main(void) {
//    freopen("m.in","r",stdin);
 //   freopen("m.out","w",stdout);
	N=read();Q=read();
	for(int i=1;i<N;i++) {
		int u=read(),v=read();
		l(u,v),l(v,u);
	} 
	dfs(1,0);
	for(int i=1;i<=N;i++) a[i]=read(),su+=a[i]*a[i];
	for(int i=1;i<=N;i++) up2(1,1,N,dfn[i],dfn[i]+siz[i]-1,-1,a[i]);
	for(int i=1;i<=16;i++) for(int j=1;j<=N;j++) anc[j][i]=anc[anc[j][i-1]][i-1];
	while(Q--) {
		int op=read();
		if(op==1) {
			int x=read(),v=read();
			up2(1,1,N,dfn[x],dfn[x]+siz[x]-1,a[x],v);
			su-=a[x]*a[x];
			su+=v*v;
			a[x]=v;
		} else {
			int m=read(),n=read(),v=read(),x=m,y=n;
			if(dep[m]<dep[n]) std::swap(m,n);
			for(int i=16;i>=0;i--) {
				if(dep[anc[m][i]]>=dep[n]) {
					m=anc[m][i];
				}
			}
			for(int i=16;i>=0;i--) {
				if(anc[m][i]!=anc[n][i]) {
					m=anc[m][i];
					n=anc[n][i];
				}
			}
			if(m!=n) m=anc[m][0];
			int l=0,r=INF,md;
			while(l<r) {
				md=(l+r)>>1;
				n_t tp=sv(x,y,m,md);
				int as=tp.s2-tp.s1*md;
				if(as<=v) r=md;else l=md+1;
			}
			n_t tp=sv(x,y,m,l);
			int as=tp.s2-tp.s1*l;
			if(l==0) {
				we(su-tp.s3);
				continue;
			}
			__int128 aq=(__int128)l*l*tp.s1-tp.s3;
			we(aq-(v-as)*(2*l-1)+su);
		}
	}
	pcc();
	printf("%lld",kd);
    return 0;
}

詳細信息

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 9ms
memory: 14312kb

input:

866 841
1 864
4 153
9 559
10 410
11 336
12 417
14 666
18 241
21 184
22 849
23 40
25 783
26 189
28 329
29 216
31 864
34 581
40 131
42 625
45 744
47 723
50 633
51 447
52 454
53 88
55 619
60 259
62 680
67 126
72 371
73 742
80 196
81 536
82 647
85 254
87 172
88 489
93 708
94 227
95 340
96 7
7 91
97 594
...

output:

285125508
285374449
285871392
285072359
284419704
284843737
284692039
284936099
285944374
285174668
285019779
284651455
287282253
287175619
284878507
285369672
284880507
285404741
284913527
286053317
288622563
286960150
287194443
288326074
286937403
287883097
288535226
288195055
288643208
288632989
...

result:

wrong output format Extra information in the output file

Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Memory Limit Exceeded

Test #6:

score: 0
Memory Limit Exceeded

input:

87080 98363
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 11
11 12
12 13
13 14
14 15
15 16
16 17
17 18
18 19
19 20
20 21
21 22
22 23
23 24
24 25
25 26
26 27
27 28
28 29
29 30
30 31
31 32
32 33
33 34
34 35
35 36
36 37
37 38
38 39
39 40
40 41
41 42
42 43
43 44
44 45
45 46
46 47
47 48
48 49
49 50
50 51
51 52...

output:


result:


Subtask #4:

score: 0
Skipped

Dependency #1:

0%