用线段树处理颜色段数
记录区间内的颜色段数,区间右端点的颜色,区间右端点的颜色.
int tr[maxn<<2], lc[maxn<<2], rc[maxn<<2];懒标记,记录区间是否被覆盖
int lazy[maxn<<2];合并的方法是这样,对于某一区间
? 如果 左区间的右端点颜色 == 右区间的左端点
? 那么 这左右区间合并,左区间的最右边一段和右区间最左边一段颜色是连续的,那么区间的颜色段数为 左区间颜色段数+右区间颜色段数 - 1.
? 否则 区间的颜色段数为 左区间颜色段数+右区间颜色段数.
void pushup(int rt){
lc[rt] = lc[rt*2];
rc[rt] = rc[rt*2+1];
tr[rt] = tr[rt*2] + tr[rt*2+1];
if(rc[rt*2] == lc[rt*2+1]) tr[rt]--;
}当区间的lazy不为0时, 他的左右区间都被覆盖,而且颜色段数都是1.
void pushdown(int rt){
if(lazy[rt]){
tr[rt*2] = tr[rt*2+1] = 1;
lazy[rt*2] = lazy[rt*2+1] = lazy[rt];
lc[rt*2] = lc[rt*2+1] = lc[rt];
rc[rt*2] = rc[rt*2+1] = rc[rt];
lazy[rt] = 0;
}
}建树和区间修改的操作如下
建树时,结点的初始值时字典序为l的点的颜色,rnk是字典序到点编号的映射.
void build(int rt, int l, int r){
lazy[rt] = 0;
if(l == r){
tr[rt] = 1;
lc[rt] = rc[rt] = col[rnk[l]];
return;
}
int m = (l+r)/2;
build(rt*2, l, m);
build(rt*2+1, m+1, r);
pushup(rt);
}void change(int rt, int l, int r, int ql, int qr, int val){
if(ql <= l && r <= qr){
lazy[rt] = 1;
tr[rt] = 1;
lc[rt] = rc[rt] = val;
return;
}
pushdown(rt);
int m = (l+r)/2;
if(ql <= m) change(rt*2, l, m, ql, qr, val);
if(m < qr) change(rt*2+1, m+1, r, ql, qr, val);
pushup(rt);
}查询的时候要注意,如果查询的区间被分成两段,那么也要判断中间是否连续:
int query(int rt, int l, int r, int ql, int qr){
if(ql <= l && r <= qr) return tr[rt];
pushdown(rt);
int m = (l+r)/2;
if(qr <= m) return query(rt*2, l, m, ql, qr);
else if(ql > m) return query(rt*2+1, m+1, r, ql, qr);
else{
int ret = query(rt*2, l, m, ql, qr) + query(rt*2+1, m+1, r, ql, qr);
if(rc[rt*2] == lc[rt*2+1]) ret--;
return ret;
}
}树链剖分
先是定义存图
vector<int> G[maxn];然后是每个点的深度,大小,重儿子,父亲.
int dep[maxn], siz[maxn], son[maxn], fa[maxn];然后是每个点所在的重链的顶点的编号,dfs顺序和点编号的映射,tot记录顺序的编号.
int top[maxn], tid[maxn], rnk[maxn], tot;第一次dfs,确定点之间的父子关系,大小,深度,重儿子:
void dfs1(int u, int f){
fa[u] = f, son[u] = 0;
dep[u] = dep[f]+1, siz[u] = 1;
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(v == fa[u]) continue;
dfs1(v, u);
siz[u] += siz[v];
if(siz[v] > siz[son[u]])
son[u] = v;
}
}第二次dfs,给每个结点安排重链编号,dfs序号:
void dfs2(int u, int tp){
tot++;
tid[u] = tot, rnk[tot] = u;
top[u] = tp;
if(son[u])
dfs2(son[u], tp);
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(v == son[u] || v == fa[u]) continue;
dfs2(v, v);
}
}修改操作
和普通树链剖分一致.
void changeTree(int u, int v, int val){
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u, v);
change(1, 1, tot, tid[top[u]], tid[u], val);
u = fa[top[u]];
}
if(dep[u] > dep[v]) swap(u, v);
change(1, 1, tot, tid[u], tid[v], val);
}查询操作
和普通的树链剖分相比,多了一步.
要判断u所在重链的顶点top[u]和顶点的父节点fa[top[u]]颜色是否相等,是的话ret要减一(代表这一条链上的颜色和上一条的连续).
int queryTree(int u, int v){
int ret = 0;
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u, v);
ret += query(1, 1, tot, tid[top[u]], tid[u]);
if(query2(1, 1, tot, tid[top[u]]) == query2(1, 1, tot, tid[fa[top[u]]])) --ret;
u = fa[top[u]];
}
if(dep[u] > dep[v]) swap(u, v);
ret += query(1, 1, tot, tid[u], tid[v]);
return ret;
}完整代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100100;
int n, m;
vector<int> G[maxn];
int col[maxn];
int dep[maxn], siz[maxn], son[maxn], fa[maxn];
int top[maxn], tid[maxn], rnk[maxn], tot;
void dfs1(int u, int f){
fa[u] = f, son[u] = 0;
dep[u] = dep[f]+1, siz[u] = 1;
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(v == fa[u]) continue;
dfs1(v, u);
siz[u] += siz[v];
if(siz[v] > siz[son[u]])
son[u] = v;
}
}
void dfs2(int u, int tp){
tot++;
tid[u] = tot, rnk[tot] = u;
top[u] = tp;
if(son[u])
dfs2(son[u], tp);
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(v == son[u] || v == fa[u]) continue;
dfs2(v, v);
}
}
int tr[maxn<<2], lc[maxn<<2], rc[maxn<<2];
int lazy[maxn<<2];
void pushup(int rt){
lc[rt] = lc[rt*2];
rc[rt] = rc[rt*2+1];
tr[rt] = tr[rt*2] + tr[rt*2+1];
if(rc[rt*2] == lc[rt*2+1]) tr[rt]--;
}
void pushdown(int rt){
if(lazy[rt]){
tr[rt*2] = tr[rt*2+1] = 1;
lazy[rt*2] = lazy[rt*2+1] = lazy[rt];
lc[rt*2] = lc[rt*2+1] = lc[rt];
rc[rt*2] = rc[rt*2+1] = rc[rt];
lazy[rt] = 0;
}
}
void build(int rt, int l, int r){
lazy[rt] = 0;
if(l == r){
tr[rt] = 1;
lc[rt] = rc[rt] = col[rnk[l]];
return;
}
int m = (l+r)/2;
build(rt*2, l, m);
build(rt*2+1, m+1, r);
pushup(rt);
}
void change(int rt, int l, int r, int ql, int qr, int val){
if(ql <= l && r <= qr){
lazy[rt] = 1;
tr[rt] = 1;
lc[rt] = rc[rt] = val;
return;
}
pushdown(rt);
int m = (l+r)/2;
if(ql <= m) change(rt*2, l, m, ql, qr, val);
if(m < qr) change(rt*2+1, m+1, r, ql, qr, val);
pushup(rt);
}
int query(int rt, int l, int r, int ql, int qr){
if(ql <= l && r <= qr) return tr[rt];
pushdown(rt);
int m = (l+r)/2;
if(qr <= m) return query(rt*2, l, m, ql, qr);
else if(ql > m) return query(rt*2+1, m+1, r, ql, qr);
else{
int ret = query(rt*2, l, m, ql, qr) + query(rt*2+1, m+1, r, ql, qr);
if(rc[rt*2] == lc[rt*2+1]) ret--;
return ret;
}
}
int query2( int rt, int l, int r, int x){
if(l == r) return lc[rt];
pushdown(rt);
int m = (l+r)/2;
if(x <= m) return query2(rt*2, l, m, x);
else return query2(rt*2+1, m+1, r, x);
}
void changeTree(int u, int v, int val){
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u, v);
change(1, 1, tot, tid[top[u]], tid[u], val);
u = fa[top[u]];
}
if(dep[u] > dep[v]) swap(u, v);
change(1, 1, tot, tid[u], tid[v], val);
}
int queryTree(int u, int v){
int ret = 0;
while(top[u] != top[v]){
if(dep[top[u]] < dep[top[v]]) swap(u, v);
ret += query(1, 1, tot, tid[top[u]], tid[u]);
if(query2(1, 1, tot, tid[top[u]]) == query2(1, 1, tot, tid[fa[top[u]]])) --ret;
u = fa[top[u]];
}
if(dep[u] > dep[v]) swap(u, v);
ret += query(1, 1, tot, tid[u], tid[v]);
return ret;
}
int main()
{
scanf("%d%d", &n,&m);
for(int i = 1; i <= n; ++i) G[i].clear();
for(int i = 1; i <= n; ++i) scanf("%d", &col[i]);
for(int i = 1; i < n; ++i){
int u, v;
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
dfs1(1, 0);
tot = 0;
dfs2(1, 1);
build(1, 1, tot);
while(m--){
char opt;
scanf(" %c", &opt);
if(opt == 'Q'){
int u, v;
scanf("%d%d", &u, &v);
printf("%d\n", queryTree(u, v));
}else if(opt == 'C'){
int u, v, val;
scanf("%d%d%d", &u, &v, &val);
changeTree(u, v, val);
}
}
}
原文地址：https://www.cnblogs.com/zzidun-pavo/p/12234978.html
时间： 2024-08-28 19:08:11