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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#109102#5418. Color the TreeshihoghmeanWA 30ms23908kbC++174.5kb2023-05-27 12:46:522023-05-27 12:46:54

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-05-27 12:46:54]
  • 评测
  • 测评结果:WA
  • 用时:30ms
  • 内存:23908kb
  • [2023-05-27 12:46:52]
  • 提交

answer


#include<bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fr(i,a,b) for(int i=a;i>=b;i--)
#define py puts("Yes")
#define pn puts("No")
#define pt puts("")
#define pb push_back
#define wt(x) write(x),puts("")
#define wr(x) write(x) ,putchar(' ')
#define tx printf("fds")
#define mp make_pair
#define fi first
#define se second
inline int read(){
	int x=0,k=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-') k=-1;
		ch=getchar();
	}
	while(ch>='0'&&ch<='9'){
		x=(x<<1)+(x<<3)+ch-48;
		ch=getchar();
	}
	return x*k;
}
void write(int x){
	if(x<0){
		x=-x;
		putchar('-');
	}
	if(x>9) write(x/10);
	putchar(x%10+'0');
}
int power(int x,int y,int mod){
	int num=1;
	while(y){
		if(y&1) num=(num*x)%mod;
		x=x*x%mod;
		y>>=1;
	}
	return num;
}
int mul(int x,int y,int mod){
	int num=0;
	while(y){
		if(y&1) num=(num+x)%mod;
		x=(x+x)%mod;
		y>>=1;
	}
	return num;
}
const int N=1e6+7,mod=998244353;
int n,tot,cnt;
int a[N];
int head[N];
struct edge{
    int to,next;
}e[N];
void add(int u,int v){
    e[++tot]={v,head[u]};
    head[u]=tot;
}
struct tree{
    int ls,rs,lazy;
    int num;
}t[N];
int dep[N],son[N],mdep[N],top[N];
int rt[N];
int rmq[N][20],rmq1[N][20];
int newnode(int x){
    t[x]={0,0,0,0};
    return x;
}
void pu(int id){
    int x=0,y=0;
    if(t[id].ls) x=t[t[id].ls].num;
    if(t[id].rs) y=t[t[id].rs].num;
    t[id].num=x+y;
}
void update(int id,int l,int r,int x,int k,int h){
    if(l==r){
        t[id].num=k;
        t[id].lazy=h;
        return ;
    }
    int mid=(l+r)>>1;
    if(x<=mid){
        if(!t[id].ls) t[id].ls=newnode(++cnt);
        update(t[id].ls,l,mid,x,k,h);
    }
    else{
        if(!t[id].rs) t[id].rs=newnode(++cnt);
        update(t[id].rs,mid+1,r,x,k,h);
    }
    pu(id);
}
int RMQ(int l,int r){
    int x=log2(r-l+1);
    return min(rmq[l][x],rmq[r-(1<<x)+1][x]);
}
int RMQ1(int l,int r){
    int x=log2(r-l+1);
    if(rmq[l][x]<rmq[r-(1<<x)+1][x]){
        return rmq1[l][x];
    }
    else{
        return rmq1[r-(1<<x)+1][x];
    }
}
int Merge(int ddp,int dp,int x,int y,int l,int r){
    if(!y||!x) return x+y;
    if(l==r){
        if(t[x].lazy>dp){
            if(RMQ(l-dp,l-ddp)<=t[x].num+t[y].num){
                t[x].num=RMQ(l-dp,l-ddp);
                t[x].lazy=l-RMQ1(l-dp,l-ddp);
            }
            else{
                t[x].num=t[x].num+t[y].num;
            }
        }
        return x;
    }
    int mid=(l+r)>>1;
    t[x].ls=Merge(ddp,dp,t[x].ls,t[y].ls,l,mid);
    t[x].rs=Merge(ddp,dp,t[x].rs,t[y].rs,mid+1,r);
    pu(x);
    return x;
}
void dfs(int u,int fa){
    dep[u]=mdep[u]=dep[fa]+1;
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa) continue;
        dfs(v,u);
        mdep[u]=max(mdep[u],mdep[v]);
        if(mdep[v]>mdep[son[u]]){
            son[u]=v;
        }
    }
}
void dfs1(int u,int fa,int tp){
    top[u]=tp;
    if(tp==u) rt[u]=newnode(++cnt);
    update(rt[top[u]],1,n,dep[u],RMQ(0,dep[u]-dep[top[u]]),dep[u]-RMQ1(0,dep[u]-dep[top[u]]));
    if(son[u]) dfs1(son[u],u,tp);
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa||v==son[u]) continue;
        dfs1(v,u,v);
    }
}
void dfs2(int u,int fa){
    for(int i=head[u];i;i=e[i].next){
        int v=e[i].to;
        if(v==fa) continue;
        dfs2(v,u);
        if(top[u]==top[v]) continue;
        rt[top[u]]=Merge(dep[top[u]],dep[u],rt[top[u]],rt[top[v]],1,n);
    }
}
signed main(){
	int tt=read();
	while(tt--){
		tot=0;
        n=read();
        fo(i,0,n) head[i]=0,rt[i]=0,son[i]=0,dep[i]=0,mdep[i]=0,top[i]=0;
        fo(i,1,cnt) t[i]={0,0,0,0};
        fo(i,0,n-1) a[i]=read();
        fo(i,0,n-1)
            fo(j,0,19) rmq[i][j]=1e9,rmq1[i][j]=0;
        fo(i,0,n-1){
            rmq[i][0]=a[i];
            rmq1[i][0]=i;
        }
        fo(i,1,19){
            fo(j,0,n-(1<<i)){
                if(rmq[j][i-1]<rmq[j+(1<<(i-1))][i-1]){
                    rmq[j][i]=rmq[j][i-1];
                    rmq1[j][i]=rmq1[j][i-1];
                }
                else{
                    rmq[j][i]=rmq[j+(1<<(i-1))][i-1];
                    rmq1[j][i]=rmq1[j+(1<<(i-1))][i-1];
                }
            }
        }
        fo(i,2,n){
            int x=read(),y=read();
            add(x,y);
            add(y,x);
        }
        cnt=0;
        dfs(1,0);
        dfs1(1,0,1);
        dfs2(1,0);
        wt(t[rt[1]].num);
	}
	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 23908kb

input:

3
4
10 15 40 1
1 2
2 3
2 4
5
10 5 1 100 1000
1 2
2 3
2 4
4 5
4
1000 200 10 8
1 2
2 3
3 4

output:

35
17
1218

result:

ok 3 number(s): "35 17 1218"

Test #2:

score: -100
Wrong Answer
time: 30ms
memory: 17900kb

input:

3000
54
43 44 11 49 17 14 7 30 35 12 34 14 15 8 29 47 30 31 39 17 26 23 26 45 34 50 49 13 35 18 29 15 13 44 47 5 22 20 19 46 30 22 26 13 47 46 43 27 48 48 13 14 30 44
1 2
1 3
2 4
3 5
4 6
2 7
4 8
1 9
3 10
1 11
8 12
2 13
9 14
1 15
1 16
15 17
15 18
7 19
1 20
19 21
13 22
10 23
17 24
9 25
9 26
24 27
8 28...

output:

180
142
222
230
156
240
225
126
100
81
155
73
154
127
149
122
204
230
132
187
153
170
78
282
195
286
191
211
119
197
211
233
88
252
239
205
173
180
195
121
109
114
180
170
226
210
182
97
199
59
56
31
115
204
203
118
137
208
53
133
189
170
173
137
233
90
163
273
80
350
156
133
146
159
240
269
137
217...

result:

wrong answer 2nd numbers differ - expected: '168', found: '142'