数N<=10^5，操作数M<=10^5，所有的颜色C为整数且在[0, 10^9]之间
题解
裸的树链剖分+线段树。
区间修改非常恶心，很多细节。
多写写应该就能好了吧。。。
#include <stdio.h>
#include <algorithm>
using namespace std;
#define lson l , mid , x << 1
#define rson mid + 1 , r , x << 1 | 1
#define N 100005
int fa[N] , deep[N] , si[N] , val[N] , bl[N] , pos[N] , tot;
int head[N] , to[N << 1] , next[N << 1] , cnt;
int sum[N << 2] , lc[N << 2] , rc[N << 2] , mark[N << 2] , n;
char str[10];
void add(int x , int y)
{
to[++cnt] = y;
next[cnt] = head[x];
head[x] = cnt;
}
void dfs1(int x)
{
int i , y;
si[x] = 1;
for(i = head[x] ; i ; i = next[i])
{
y = to[i];
if(y != fa[x])
{
fa[y] = x;
deep[y] = deep[x] + 1;
dfs1(y);
si[x] += si[y];
}
}
}
void dfs2(int x , int c)
{
int k = 0 , i , y;
bl[x] = c;
pos[x] = ++tot;
for(i = head[x] ; i ; i = next[i])
{
y = to[i];
if(fa[x] != y && si[y] > si[k])
k = y;
}
if(k != 0)
{
dfs2(k , c);
for(i = head[x] ; i ; i = next[i])
{
y = to[i];
if(fa[x] != y && y != k)
dfs2(y , y);
}
}
}
void pushup(int x)
{
lc[x] = lc[x << 1];
rc[x] = rc[x << 1 | 1];
sum[x] = sum[x << 1] + sum[x << 1 | 1];
if(rc[x << 1] == lc[x << 1 | 1])
sum[x] -- ;
}
void pushdown(int x)
{
int tmp = mark[x];
mark[x] = 0;
if(tmp)
{
sum[x << 1] = sum[x << 1 | 1] = 1;
lc[x << 1] = rc[x << 1] = lc[x << 1 | 1] = rc[x << 1 | 1] = tmp;
mark[x << 1] = mark[x << 1 | 1] = tmp;
}
}
void update(int b , int e , int v , int l , int r , int x)
{
if(b <= l && r <= e)
{
sum[x] = 1;
lc[x] = rc[x] = v;
mark[x] = v;
return;
}
pushdown(x);
int mid = (l + r) >> 1;
if(b <= mid)
update(b , e , v , lson);
if(e > mid)
update(b , e , v , rson);
pushup(x);
}
void solveupdate(int x , int y , int v)
{
while(bl[x] != bl[y])
{
if(deep[bl[x]] < deep[bl[y]])
{
swap(x , y);
}
update(pos[bl[x]] , pos[x] , v , 1 , n , 1);
x = fa[bl[x]];
}
if(deep[x] > deep[y])
swap(x , y);
update(pos[x] , pos[y] , v , 1 , n , 1);
}
int query(int b , int e , int l , int r , int x)
{
if(b <= l && r <= e)
{
return sum[x];
}
pushdown(x);
int mid = (l + r) >> 1 , ans = 0;
if(b <= mid)
ans += query(b , e , lson);
if(e > mid)
ans += query(b , e , rson);
if(b <= mid && e > mid && rc[x << 1] == lc[x << 1 | 1])
ans -- ;
return ans;
}
int getcl(int p , int l , int r , int x)
{
if(l == r)
return lc[x];
pushdown(x);
int mid = (l + r) >> 1;
if(p <= mid)
return getcl(p , lson);
else
return getcl(p , rson);
}
int solvequery(int x , int y)
{
int ans = 0;
while(bl[x] != bl[y])
{
if(deep[bl[x]] < deep[bl[y]])
swap(x , y);
ans += query(pos[bl[x]] , pos[x] , 1 , n , 1);
if(getcl(pos[bl[x]] , 1 , n , 1) == getcl(pos[fa[bl[x]]] , 1 , n , 1))
ans -- ;
x = fa[bl[x]];
}
if(deep[x] > deep[y])
swap(x , y);
ans += query(pos[x] , pos[y] , 1 , n , 1);
return ans;
}
int main()
{
int i , x , y , z , m;
scanf("%d%d" , &n , &m);
for(i = 1 ; i <= n ; i ++ )
scanf("%d" , &val[i]);
for(i = 1 ; i < n ; i ++ )
{
scanf("%d%d" , &x , &y);
add(x , y);
add(y , x);
}
dfs1(1);
dfs2(1 , 1);
for(i = 1 ; i <= n ; i ++ )
update(pos[i] , pos[i] , val[i] , 1 , n , 1);
while(m -- )
{
scanf("%s" , str);
switch(str[0])
{
case 'C': scanf("%d%d%d" , &x , &y , &z); solveupdate(x , y , z); break;
default: scanf("%d%d" , &x , &y); printf("%d\n" , solvequery(x , y));
}
}
return 0;
}
优质内容筛选与推荐>>1、二分查找（Binary Search）的递归和非递归
2、WebDev.WebServer40.EXE
3、.NET 中的十进制浮点类型(译文)
4、牛客练习赛14
5、window.open()打开页面