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-
# define MOD
# define INF
# 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<<,l,mid
# define rch p<<|,mid+,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=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

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