题目描述
一棵n个点的树，给定m条路径，q次询问包含一条路径的给定路径的个数+1
输入
The first line of input contains a single integer N（1<=N<=100000） - the number of cities in Byteland. Cities are numbered from 1 to n . Each of the next N -1 lines contains two integers Ai, Bi(1<=Ai,Bi<=N) meaning that cities Ai and Biare connected by a road.
The next line contains an integer M(1<=M<=100000) - the number of highways. Each of the next m lines contains a description of a single highway. The next line contains an integer Q (1<=Q<=500000) - the number of queries. Each of the next Q lines contains a description of a query. Both highways and queries are given in the same format as the roads.
输出
Your program should output exactly Q lines. The i-th line should contain the number of routes in the i-th query.
样例输入
9
1 2
2 3
4 2
1 5
5 6
7 5
7 8
9 7
4
2 5
3 4
6 4
8 3
4
4 9
2 5
1 6
1 7
样例输出
1
4
2
2
题解
DFS序+树状数组
咦这不是 精神污染 那道题吗？然而我那道题写得太丑了。。。
我们不妨换个思路：考虑一条路径被什么样的路径所包含。
当两个点x和y没有祖先关系时，显然包含它的路径的两个端点应该分别在x和y的子树内。
当x和y具有祖先关系时，不妨设x是y的祖先，那么设x到y路径上x的儿子为z，那么包含它的路径的两个端点应该分别在z的子树外和y的子树内。
那么就可以使用DFS序把两个端点的取值范围转化为DFS序上的一段或两段区间，其中找儿子z的过程可以使用树上倍增实现。
于是把每个路径x-y看作平面上的点(pos[x],pos[y])（pos[x]表示x在DFS序中的位置），那么包含一条给定路径的所有路径就转化为至多2个矩形。
所以每次询问要求的就是矩形内的点的数目，可以使用离线+树状数组解决。把每个矩形拆成前缀相减的4个点，把所有点按x坐标排序，然后使用树状数组维护y坐标的前缀和即可。
时间复杂度$O(nlog n)$
#include <cstdio>
#include <cctype>
#include <algorithm>
#define N 100010
using namespace std;
struct data
{
int x , y , v , id;
data() {}
data(int a , int b , int c , int d) {x = a , y = b , v = c , id = d;}
bool operator<(const data &a)const {return x < a.x;}
}a[N << 1] , q[N * 30];
int n , head[N] , to[N << 1] , next[N << 1] , cnt , fa[N][20] , deep[N] , log[N] , pos[N] , last[N] , tp , f[N] , tot , ans[N * 5];
inline char nc()
{
static char buf[100000] , *p1 , *p2;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf , 1 , 100000 , stdin) , p1 == p2) ? EOF : *p1 ++ ;
}
inline int read()
{
int ret = 0; char ch = nc();
while(!isdigit(ch)) ch = nc();
while(isdigit(ch)) ret = ((ret + (ret << 2)) << 1) + (ch ^ '0') , ch = nc();
return ret;
}
inline void add(int x , int y)
{
to[++cnt] = y , next[cnt] = head[x] , head[x] = cnt;
}
void dfs(int x)
{
int i;
pos[x] = ++tp;
for(i = 1 ; (1 << i) <= deep[x] ; i ++ ) fa[x][i] = fa[fa[x][i - 1]][i - 1];
for(i = head[x] ; i ; i = next[i])
if(to[i] != fa[x][0])
fa[to[i]][0] = x , deep[to[i]] = deep[x] + 1 , dfs(to[i]);
last[x] = tp;
}
inline int find(int x , int y)
{
int i;
for(i = log[y] ; ~i ; i -- )
if((1 << i) <= y)
x = fa[x][i] , y -= (1 << i);
return x;
}
inline void update(int x)
{
int i;
for(i = x ; i <= n ; i += i & -i) f[i] ++ ;
}
inline int query(int x)
{
int i , ans = 0;
for(i = x ; i ; i -= i & -i) ans += f[i];
return ans;
}
int main()
{
int m , k , i , x , y , z , p;
n = read();
for(i = 2 ; i <= n ; i ++ ) x = read() , y = read() , add(x , y) , add(y , x) , log[i] = log[i >> 1] + 1;
dfs(1);
m = read();
for(i = 1 ; i <= m ; i ++ ) x = read() , y = read() , a[i] = data(pos[x] , pos[y] , 0 , 0) , a[i + m] = data(pos[y] , pos[x] , 0 , 0);
k = read();
for(i = 1 ; i <= k ; i ++ )
{
x = read() , y = read();
if(deep[x] < deep[y]) swap(x , y);
if(deep[x] > deep[y] && fa[z = find(x , deep[x] - deep[y] - 1)][0] == y)
{
q[++tot] = data(pos[x] - 1 , pos[z] - 1 , -1 , i) , q[++tot] = data(pos[x] - 1 , last[z] , 1 , i) , q[++tot] = data(pos[x] - 1 , n , -1 , i);
q[++tot] = data(last[x] , pos[z] - 1 , 1 , i) , q[++tot] = data(last[x] , last[z] , -1 , i) , q[++tot] = data(last[x] , n , 1 , i);
}
else
{
q[++tot] = data(pos[x] - 1 , pos[y] - 1 , 1 , i) , q[++tot] = data(pos[x] - 1 , last[y] , -1 , i);
q[++tot] = data(last[x] , pos[y] - 1 , -1 , i) , q[++tot] = data(last[x] , last[y] , 1 , i);
}
}
sort(a + 1 , a + 2 * m + 1) , sort(q + 1 , q + tot + 1);
for(p = i = 1 ; i <= tot ; i ++ )
{
while(p <= m * 2 && a[p].x <= q[i].x) update(a[p ++ ].y);
ans[q[i].id] += q[i].v * query(q[i].y);
}
for(i = 1 ; i <= k ; i ++ ) printf("%d
" , ans[i] + 1);
return 0;
}