SUOI08 一收一行破 (lca+树状数组)

 1 #include 2 #define pa pair 3 #define ll long long
 4 using namespace std;
 5 const int maxn=200020;
 6 
 7 inline ll rd(){
 8     ll x=0;char c=getchar();int neg=1;
 9     while(c‘0‘||c>‘9‘){if(c==‘-‘) neg=-1;c=getchar();}
10     while(c>=‘0‘&&c‘9‘) x=x*10+c-‘0‘,c=getchar();
11     return x*neg;
12 }
13 
14 int N,M,L=1;
15 int eg[maxn*2][2],egh[maxn],ect;
16 int dep[maxn],fa[maxn],v[maxn];
17 int qu[maxn*2][2],qh[maxn],op[maxn][3];
18 ll tr[maxn];
19 bool flag[maxn];
20 
21 inline int lowbit(int x){return x&(-x);}
22 inline int getfa(int x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
23 inline void adeg(int a,int b){
24     eg[++ect][0]=b;eg[ect][1]=egh[a];egh[a]=ect;
25 }inline void adq(int a,int b,int i){
26     qu[i][0]=b;qu[i][1]=qh[a];qh[a]=i;
27 }
28 
29 inline void add(int x,int y){
30     for(;x&&xy;
31 }
32 inline ll query(int x){
33     ll re=0;for(;x;x-=lowbit(x)) re+=tr[x];return re;
34 }
35 
36 void tarjan(int x){
37     flag[x]=1;
38     for(int i=egh[x];i!=-1;i=eg[i][1]){
39         if(flag[eg[i][0]]) continue;
40         dep[eg[i][0]]=dep[x]+1;L=max(L,dep[x]+1);
41         tarjan(eg[i][0]);fa[getfa(eg[i][0])]=getfa(x);
42     }
43     for(int i=qh[x];i!=-1;i=qu[i][1]){
44         if(flag[qu[i][0]]) op[i>>1][0]=getfa(qu[i][0]);
45     }
46 }
47 
48 
49 
50 int main(){
51     int i,j,k;
52     N=rd(),M=rd();
53     memset(egh,-1,sizeof(egh));memset(qh,-1,sizeof(qh));
54     for(i=1;ird();
55     for(i=1;i){
56         int a=rd(),b=rd();
57         adeg(a,b);adeg(b,a);
58     }for(i=1;i){
59         int a=rd(),b=rd();
60         if(a==1){
61             int c=rd();adq(b,c,i1);adq(c,b,i1|1);
62             op[i][1]=b,op[i][2]=c;
63         }else op[i][0]=b;
64     }for(i=1;ii;
65     dep[1]=1;tarjan(1);//printf("ll");
66     for(i=1;i){
67         add(dep[i],v[i]);add(dep[i]+1,-v[i]);
68     }
69     for(i=1;i){
70         if(op[i][1]){
71             add(dep[op[i][0]],1);add(dep[op[i][0]]+1,-1);
72             if(op[i][0]!=op[i][1]) add(dep[op[i][0]]+1,1),add(dep[op[i][1]]+1,-1);
73             if(op[i][0]!=op[i][2]) add(dep[op[i][0]]+1,1),add(dep[op[i][2]]+1,-1);
74         }else{
75             if(op[i][0]>L) printf("0\n");
76             else printf("%lld\n",query(op[i][0]));
77         }
78     }
79     
80     return 0;
81 }