BZOJ 1103 大都市(dfs序+树状数组)

应该是一道很水的题吧。。。

显然可以用树链剖分解决这个问题，虽然不知道多一个log会不会T。但是由于问题的特殊性。

每次修改都是将边权为1的边修改为0，且询问的是点i到根节点的路径长度。

令点i到根节点的路径长度为w[i]，显然初始时w[i]=dep[i].考虑修改边为(u,v),那么令u为深度大的点。

那么u的子树的所有答案就要减1.考虑dfs序，则每次需要修改的是一段连续的区间。

树状数组维护单点查询，区间修改，美滋滋。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=250005;
//Code begin...

struct Edge{int p, next;}edge[N<<1];
int tree[N], head[N], cnt=1, DFN[N][2], dep[N], pos, n;

void add_edge(int u, int v){edge[cnt].p=v; edge[cnt].next=head[u]; head[u]=cnt++;}
void dfs(int x, int fa){
    DFN[x][0]=++pos;
    for (int i=head[x]; i; i=edge[i].next) {
        int v=edge[i].p;
        if (v==fa) continue;
        dep[v]=dep[x]+1; dfs(v,x);
    }
    DFN[x][1]=pos;
}
int query(int x){
    int res=0;
    while (x) res+=tree[x], x-=lowbit(x);
    return res;
}
void add(int x, int val){while (x<=n+1) tree[x]+=val, x+=lowbit(x);}
int main ()
{
    int m, u, v;
    char s[5];
    scanf("%d",&n);
    FO(i,1,n) scanf("%d%d",&u,&v), add_edge(u,v), add_edge(v,u);
    dfs(1,0);
    FOR(i,1,n) add(DFN[i][0],dep[i]), add(DFN[i][0]+1,-dep[i]);
    scanf("%d",&m);
    m+=n-1;
    while (m--) {
        scanf("%s%d",s,&u);
        if (s[0]==‘W‘) printf("%d\n",query(DFN[u][0]));
        else {
            scanf("%d",&v);
            if (dep[u]<dep[v]) swap(u,v);
            add(DFN[u][0],-1); add(DFN[u][1]+1,1);
        }
    }
    return 0;
}