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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
#221870#7608. Cliquesucup-team1525#TL 0ms24436kbC++172.7kb2023-10-21 14:56:212023-10-21 14:56:22

Judging History

你现在查看的是最新测评结果

  • [2023-10-21 14:56:22]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:24436kb
  • [2023-10-21 14:56:21]
  • 提交

answer

 #include<bits/stdc++.h>
using namespace std;
const int N=4e5;
const int mod=1e9+7,iv=(mod+1)/2;
int inc(int x,int y){ return (x+=y)>=mod?x-mod:x; }
int sub(int x,int y){ return (x-=y)<0?x+mod:x; }
int n;
vector<int> e[N+5];
int fa[N+5];
int sz[N+5],son[N+5],top[N+5];
int dfn[N+5],rdfn[N+5],dfns;
struct Segtr{
    int tr[N*4+5],tag[N*4+5];
    #define lc id<<1
    #define rc id<<1|1
    void build(int l,int r,int id){
        tag[id]=1;
        if(l==r){
            tr[id]=rdfn[l]<=n?1:mod-1;
            return;
        }
        int mid=l+r>>1;
        build(l,mid,lc);
        build(mid+1,r,rc);
        tr[id]=inc(tr[lc],tr[rc]);
    }
    void push_down(int id){
        if(tag[id]==1) return;
        int v=tag[id]; tag[id]=1;
        tr[lc]=1ll*tr[lc]*v%mod; tag[lc]=1ll*tag[lc]*v%mod;
        tr[rc]=1ll*tr[rc]*v%mod; tag[rc]=1ll*tag[rc]*v%mod;
    }
    void add(int l,int r,int st,int en,int x,int id){
        if(st<=l&&en>=r){
            tr[id]=1ll*tr[id]*x%mod;
            tag[id]=1ll*tag[id]*x%mod;
            return;
        }
        int mid=l+r>>1;
        push_down(id);
        if(st<=mid) add(l,mid,st,en,x,lc);
        if(en>mid) add(mid+1,r,st,en,x,rc);
        tr[id]=inc(tr[lc],tr[rc]);
    }
}T;
void dfs1(int u,int tf){
    fa[u]=tf; sz[u]=1;
    for(auto v:e[u]){
        if(v!=tf){
            dfs1(v,u);
            sz[u]+=sz[v];
            if(sz[v]>sz[son[u]]) son[u]=v;
        }
    }
}
void dfs2(int u,int tf){
    top[u]=tf;
    dfn[u]=++dfns; rdfn[dfns]=u;
    if(!son[u]) return;
    dfs2(son[u],son[u]);
    for(auto v:e[u]){
        if(v!=fa[u]&&v!=son[u])
            dfs2(v,v);
    }
}
int LCA(int u,int v){
    while(top[u]!=top[v]){
        if(dfn[top[u]]<dfn[top[v]]) swap(u,v);
        u=fa[top[u]];
    }
    return dfn[u]<dfn[v]?u:v;
}
void work(int u,int v,int x){
    int lca=LCA(u,v);
    while(top[u]!=top[lca]){
        T.add(1,dfns,dfn[top[u]],dfn[u],x,1);
        u=fa[top[u]];
    }
    if(u!=lca) T.add(1,dfns,dfn[lca]+1,dfn[u],x,1);
    while(top[v]!=top[lca]){
        T.add(1,dfns,dfn[top[v]],dfn[v],x,1);
        v=fa[top[v]];
    }
    T.add(1,dfns,dfn[lca],dfn[v],x,1);
}
int main(){
    scanf("%d",&n);
    for(int i=1,u,v;i<n;i++){
        scanf("%d %d",&u,&v);
        e[u].push_back(i+n);
        e[v].push_back(i+n);
        e[i+n].push_back(u);
        e[i+n].push_back(v);
    }
    dfs1(1,0);
    dfs2(1,1);
    T.build(1,dfns,1);
    int q;
    scanf("%d",&q);
    while(q--){
        char s[5];
        int u,v;
        scanf("%s %d %d",s,&u,&v);
        if(s[0]=='+') work(u,v,2);
        else work(u,v,iv);
        printf("%d\n",sub(T.tr[1],1));
    }
    return 0;
}


詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 24436kb

input:

5
1 2
5 1
2 3
4 2
6
+ 4 5
+ 2 2
+ 1 3
- 2 2
+ 2 3
+ 4 4

output:

1
3
7
3
7
9

result:

ok 6 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 21948kb

input:

20
8 7
19 10
12 14
3 16
17 13
7 10
5 6
1 9
15 12
5 2
16 13
3 11
20 14
18 6
1 14
16 20
11 10
3 4
20 6
30
+ 10 20
+ 14 20
+ 12 17
- 14 20
- 12 17
+ 4 6
+ 8 20
+ 3 6
- 10 20
+ 2 17
+ 1 16
+ 6 10
+ 9 10
+ 5 15
+ 7 8
- 7 8
+ 2 5
+ 3 18
+ 1 20
+ 8 16
- 3 18
- 5 15
+ 4 20
+ 14 16
- 4 6
+ 8 19
+ 4 7
- 1 16
...

output:

1
3
7
3
1
3
7
15
7
15
31
63
127
255
257
255
259
515
1027
1283
643
385
769
1537
769
785
865
481
737
369

result:

ok 30 lines

Test #3:

score: -100
Time Limit Exceeded

input:

131072
3641 72511
10338 18718
8949 15478
79108 62887
42154 28540
65359 102645
93744 48493
3277 103211
43542 61315
93315 118634
24021 57937
31034 436
30411 6208
1388 25159
93424 128520
119820 87281
5860 97559
59648 8225
57244 58766
119685 13716
130165 60958
79806 116338
97486 80167
101963 95499
51263...

output:


result: