线段树合并的一些题
线段树合并的一些题
codeforces 600E E. Lomsat gelral
传送门：https://codeforces.com/contest/600/problem/E
题意：
给你一颗n个节点的树，树上的每一个节点都有一种颜色，询问每一个节点所在的子树颜色数量最多的那些颜色的值的总和
题解：
维护子树颜色的数量和答案，线段树合并即可
代码：
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
struct node {
int l, r, sum, val;
LL ans;
} tree[maxn * 50];
int root[maxn];
int col[maxn];
int tree_cnt;
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
#define ls tree[rt].l
#define rs tree[rt].r
void push_up(int rt) {
if(tree[ls].sum == tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = tree[ls].sum;
tree[rt].ans = tree[ls].ans + tree[rs].ans;
} else if(tree[ls].sum > tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = tree[ls].val;
tree[rt].ans = tree[ls].ans;
} else {
tree[rt].sum = tree[rs].sum;
tree[rt].val = tree[rs].val;
tree[rt].ans = tree[rs].ans;
}
}
void update(int &x, int l, int r, int pos, int val) {
if(!x) x = ++tree_cnt;
if(l == r) {
tree[x].val = l;
tree[x].sum += val;
tree[x].ans = l;
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(tree[x].l, l, mid, pos, val);
else update(tree[x].r, mid + 1, r, pos, val);
push_up(x);
}
int merge(int x, int y, int l, int r) {
if(!x) return y;
if(!y) return x;
if(l == r) {
tree[x].val = l;
tree[x].sum += tree[y].sum;
tree[x].ans = l;
return x;
}
int mid = (l + r) >> 1;
tree[x].l = merge(tree[x].l, tree[y].l, l, mid);
tree[x].r = merge(tree[x].r, tree[y].r, mid + 1, r);
push_up(x);
return x;
}
int Max = 100000;
LL ans[maxn];
void dfs(int u, int fa) {
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(v == fa) continue;
dfs(v, u);
merge(root[u], root[v], 1, Max);
}
update(root[u], 1, Max, col[u], 1);
ans[u] = tree[root[u]].ans;
}
void init() {
memset(head, -1, sizeof(head));
memset(col, 0, sizeof(col));
tree_cnt = tot = 0;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n;
scanf("%d", &n);
init();
for(int i = 1; i <= n; i++) {
scanf("%d", &col[i]);
root[i] = i;
tree_cnt++;
}
for(int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs(1, 0);
for(int i = 1; i <= n; i++) {
printf("%lld%c", ans[i], i == n ? '\n' : ' ');
}
return 0;
}
P3521 [POI2011]ROT-Tree Rotations
题意：
给你一颗树，只有叶子节点有权值，你可以交换一个点的左右子树，问你最小的逆序对数
题解：
线段树维护权值个个数即可
然后左右子树合并时计算交换和不交换的贡献取一个min即可
代码：
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
struct node {
int l, r, sum;
} tree[maxn * 40];
int tree_cnt;
int root[maxn];
void update(int &x, int l, int r, int val) {
if(!x) x = ++tree_cnt;
tree[x].sum++;
if(l == r) return;
int mid = (l + r) >> 1;
if(val <= mid) update(tree[x].l, l, mid, val);
else update(tree[x].r, mid + 1, r, val);
}
int n;
LL num1, num2, ans = 0;
void merge(int &x, int y) {
if(!x || !y) {
x = x + y;
return;
}
tree[x].sum += tree[y].sum;
num1 += 1LL * tree[tree[x].l].sum * tree[tree[y].r].sum;
num2 += 1LL * tree[tree[x].r].sum * tree[tree[y].l].sum;
merge(tree[x].l, tree[y].l);
merge(tree[x].r, tree[y].r);
}
void dfs(int &x) {
int val;
scanf("%d", &val);
int ls = 0, rs = 0;
if(!val) {
dfs(ls);
dfs(rs);
num1 = num2 = 0;
x = ls;
merge(x, rs);
ans += min(num1, num2);
} else {
update(x, 1, n, val);
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
scanf("%d", &n);
int x = 0;
dfs(x);
printf("%lld\n", ans);
return 0;
}
P4556 [Vani有约会]雨天的尾巴
题意：
首先村落里的一共有n座房屋，并形成一个树状结构。然后救济粮分m次发放，每次选择两个房屋(x,y)，然后对于x到y的路径上（含x和y)每座房子里发放一袋z类型的救济粮。
然后深绘里想知道，当所有的救济粮发放完毕后，每座房子里存放的最多的是哪种救济粮。
题解：
树链剖分的写法很明显了，维护一个max即可
讲一下线段树合并的写法
区间更新用单点更新和差分来代替，求一个LCA，x->y的更新即可用在点x+1，点y+1，点lca（x,y）-1,点fa（lca(x,y)）-1 后，线段树合并来取代， 线段树维护最多的救济粮编号val，最多救济粮的数量sum，然后在合并的时候就可以统计出u节点的答案了
代码
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
struct node {
int l, r, sum, val;
} tree[maxn * 40];
int root[maxn];
int tree_cnt;
int sz[maxn], dep[maxn], fa[maxn], top[maxn], w[maxn], son[maxn], W[maxn], cnt;
void init() {
dep[1] = 1; fa[1] = 0;
memset(head, -1, sizeof(head));
tree_cnt = 0;
tot = 0;
cnt = 0;
}
#define ls tree[rt].l
#define rs tree[rt].r
void push_up(int rt) {
if(tree[ls].sum > tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = tree[ls].val;
} else if(tree[ls].sum == tree[rs].sum) {
tree[rt].sum = tree[ls].sum;
tree[rt].val = min(tree[ls].val, tree[rs].val);
} else {
tree[rt].sum = tree[rs].sum;
tree[rt].val = tree[rs].val;
}
}
void update(int &x, int l, int r, int pos, int val) {
if(!x) x = ++tree_cnt;
if(l == r) {
tree[x].sum += val;
if(tree[x].sum) {
tree[x].val = l;
} else {
tree[x].val = 0;
}
return;
}
int mid = (l + r) >> 1;
if(pos <= mid) update(tree[x].l, l, mid, pos, val);
else update(tree[x].r, mid + 1, r, pos, val);
push_up(x);
}
void merge(int &x, int y, int l, int r) {
if(!x || !y) {
x = x + y;
return;
}
if(l == r) {
tree[x].sum += tree[y].sum;
if(tree[x].sum) {
tree[x].val = l;
} else {
tree[x].val = 0;
}
return;
}
int mid = (l + r) >> 1;
merge(tree[x].l, tree[y].l, l, mid);
merge(tree[x].r, tree[y].r, mid + 1, r);
push_up(x);
}
void dfs1(int u) {
sz[u] = 1; son[u] = 0;
for (int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if (v != fa[u]) {
fa[v] = u;
dep[v] = dep[u] + 1;
dfs1(v);
sz[u] += sz[v];
if (sz[v] > sz[son[u]]) son[u] = v;
}
}
}
void dfs2(int u, int tp, int x) {
top[u] = tp; w[u] = ++cnt; W[cnt] = u;
if (son[u]) dfs2(son[u], tp, 1);
for (int i = head[u]; ~i; i = edge[i].nxt) {
int v = edge[i].v;
if (v == son[u] || v == fa[u]) continue;
dfs2(v, v, 2);
}
}
int LCA(int x, int y) {
while (top[x] != top[y]) {
if (dep[top[x]] < dep[top[y]]) std::swap(x, y);
x = fa[top[x]];
}
if (dep[x] > dep[y]) std::swap(x, y);
return x;
}
int ans[maxn];
void dfs(int u, int fa) {
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(v == fa) continue;
dfs(v, u);
merge(root[u], root[v], 1, 100000);
}
ans[u] = tree[root[u]].val;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int n, m;
init();
scanf("%d%d", &n, &m);
for(int i = 1; i < n; i++) {
int u, v;
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
dfs1(1);
dfs2(1, 1, 1);
for(int i = 1; i <= m; i++) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
int lca = LCA(x, y);
update(root[x], 1, 100000, z, 1);
// bug;
update(root[y], 1, 100000, z, 1);
update(root[lca], 1, 100000, z, -1);
if(fa[lca]) update(root[fa[lca]], 1, 100000, z, -1);
}
dfs(1, 0);
for(int i = 1; i <= n; i++) {
printf("%d\n", ans[i]);
}
return 0;
}
2016湖南省赛 I Tree Intersection（线段树合并，树链剖分）
题意：
给你一个n个结点的树，树上每个节点有自己的颜色
问你删除第i条边后形成的两颗子树有多少个相同的颜色
题解：
树链剖分写法：
对于每一种颜色来说，如果这个颜色是在单独的一颗子树中，那么就不会对其他的边产生贡献，所以我们单独对每一种颜色对边的贡献讨论，如果这个颜色只有一个，那么就不会产生贡献，否则，他就可以在两个相同颜色之间的路径上的边产生贡献，用树剖剖下来后区间加一加就可以了
线段树合并的写法
对于每一种颜色建权值线段树，线段树合并统计答案即可
代码
树链剖分写法：
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
int n;
int a[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn], w[maxn], son[maxn], W[maxn];
int sum[maxn << 2], lazy[maxn << 2], vis[maxn << 2];
struct EDGE {
int u, v, nt;
} edge[maxn << 1];
int head[maxn], summ, cnt;
int u[maxn];
int v[maxn];
void add_edge(int u, int v) {
edge[++summ].u = u;
edge[summ].v = v;
edge[summ].nt = head[u];
head[u] = summ;
}
void dfs1(int u) {
sz[u] = 1; son[u] = 0;
for (int i = head[u]; ~i; i = edge[i].nt) {
int v = edge[i].v;
if (v != fa[u]) {
fa[v] = u;
dep[v] = dep[u] + 1;
dfs1(v);
sz[u] += sz[v];
if (sz[v] > sz[son[u]]) son[u] = v;
}
}
}
void dfs2(int u, int tp, int x) {
top[u] = tp; w[u] = ++cnt; W[cnt] = u;
if (son[u]) dfs2(son[u], tp, 1);
for (int i = head[u]; ~i; i = edge[i].nt) {
int v = edge[i].v;
if (v == son[u] || v == fa[u]) continue;
dfs2(v, v, 2);
}
}
vector<int> vec[maxn];
void init() {
memset(head, -1, sizeof(head));
summ = 1; cnt = 0;
for(int i = 1; i <= n; i++) {
vec[i].clear();
}
for (int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
vec[a[i]].push_back(i);
}
for (int i = 1; i < n; i++) {
// int u, v;
scanf("%d %d", &u[i], &v[i]);
add_edge(u[i], v[i]);
add_edge(v[i], u[i]);
}
dep[1] = 1; fa[1] = 0;
dfs1(1);
dfs2(1, 1, 1);
}
void pushup(int rt) {
if(vis[ls] == vis[rs]) vis[rt] = vis[ls];
else vis[rt] = -1;
}
void pushdown(int rt, int mid) {
if (lazy[rt]) {
lazy[rt << 1] = lazy[rt];
lazy[rt << 1 | 1] = lazy[rt];
vis[rt << 1] = 0;
vis[rt << 1 | 1] = 0;
lazy[rt] = 0;
}
}
void build(int l, int r, int rt) {
lazy[rt] = 0;
sum[rt] = 0;
if (l == r) {
vis[rt] = 0;
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
}
void update(int L, int R, int val, int l, int r, int rt) {
// debug2(L, R);
if(L > R || r < L || R < l) return;
if(vis[rt] == 1) return;
if (L <= l && r <= R) {
// sum[rt] += val * (r - l + 1);
// lazy[rt] += val;
if(vis[rt] == 0) {
vis[rt] = 1;
sum[rt]++;
return;
}
}
pushdown(rt, r - l + 1);
int mid = (l + r) >> 1;
if (L <= mid) update(L, R, val, lson);
if (R > mid) update(L, R, val, rson);
pushup(rt);
}
int query(int pos, int l, int r, int rt) {
if (l == r) {
return sum[rt];
}
pushdown(rt, r - l + 1);
int mid = (l + r) >> 1;
int ans = sum[rt];
if (pos <= mid) ans += query(pos, lson);
else ans += query(pos, rson);
return ans;
}
int LCA(int x, int y) {
// debug2(x, y);
while (top[x] != top[y]) {
// bug;
if (dep[top[x]] < dep[top[y]]) swap(x, y);
update(w[top[x]], w[x], 1, 1, n, 1);
x = fa[top[x]];
}
if (dep[x] > dep[y])
std::swap(x, y);
update(w[x] + 1, w[y], 1, 1, n, 1);
return x;
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
while(scanf("%d", &n) != EOF) {
init();
build(1, n, 1);
// for(int i = 1; i <= n; i++) {
// // printf("%d\n", w[]);
// // debug1(w[i]);
// }
for(int i = 1; i <= n; i++) {
vis[1] = 0;
lazy[1] = 1;
int sz = vec[i].size();
if(sz < 2) continue;
int bb = LCA(vec[i][0], vec[i][1]);
for(int j = 2; j < sz; j++) {
// debug2(vec[i][0], vec[i][1]);
bb = LCA(bb, vec[i][j]);
}
}
for(int i = 1; i < n; i++) {
int ans;
// debug2(v[i], u[i]);
if(fa[v[i]] == u[i]) {
ans = query(w[v[i]], 1, n, 1);
} else {
ans = query(w[u[i]], 1, n, 1);
}
printf("%d\n", ans);
}
}
return 0;
}
线段树合并写法
/**
* ┏┓ ┏┓
* ┏┛┗━━━━━━━┛┗━━━┓
* ┃ ┃
* ┃ ━ ┃
* ┃ ＞ ＜ ┃
* ┃ ┃
* ┃... ⌒ ... ┃
* ┃ ┃
* ┗━┓ ┏━┛
* ┃ ┃ Code is far away from bug with the animal protecting
* ┃ ┃ 神兽保佑,代码无bug
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┃
* ┃ ┗━━━┓
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛
*/
// warm heart, wagging tail,and a smile just for you!
//
// _ooOoo_
// o8888888o
// 88" . "88
// (| -_- |)
// O\ = /O
// ____/`---'\____
// .' \| |// `.
// / \||| : |||// \
// / _||||| -:- |||||- \
// | | \ - /// | |
// | \_| ''\---/'' | |
// \ .-\__ `-` ___/-. /
// ___`. .' /--.--\ `. . __
// ."" '< `.___\_<|>_/___.' >'"".
// | | : `- \`.;`\ _ /`;.`/ - ` : | |
// \ \ `-. \_ __\ /__ _/ .-` / /
// ======`-.____`-.___\_____/___.-`____.-'======
// `=---='
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
// 佛祖保佑 永无BUG
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <bitset>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
LL gcd(LL a, LL b) {
return b ? gcd(b, a % b) : a;
}
LL lcm(LL a, LL b) {
return a / gcd(a, b) * b;
}
double dpow(double a, LL b) {
double ans = 1.0;
while(b) {
if(b % 2)ans = ans * a;
a = a * a;
b /= 2;
} return ans;
}
LL quick_pow(LL x, LL y) {
LL ans = 1;
while(y) {
if(y & 1) {
ans = ans * x % mod;
} x = x * x % mod;
y >>= 1;
} return ans;
}
int col[maxn];
int cnt[maxn];
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void init() {
memset(head, -1, sizeof(head));
tot = 0;
}
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
struct node {
int l, r, sum, val;
} tree[maxn * 20];
int root[maxn];
int rt_cnt = 0;
int n;
int ans[maxn];
void push_up(int rt) {
tree[rt].sum = tree[tree[rt].l].sum + tree[tree[rt].r].sum;
}
int build(int l, int r, int pos) {
int rt = ++rt_cnt;
tree[rt].l = 0;
tree[rt].r = 0;
tree[rt].sum = 0;
if(l == r) {
tree[rt].val = 1;
if(tree[rt].val != cnt[l]) {
tree[rt].sum = 1;
} else {
tree[rt].sum = 0;
}
return rt;
}
int mid = (l + r) >> 1;
if(pos <= mid) tree[rt].l = build(l, mid, pos);
else tree[rt].r = build(mid + 1, r, pos);
push_up(rt);
return rt;
}
void merge(int &x, int y, int l, int r) {
if(!x || !y) {
if(!x) x = y;
return;
}
if(l == r) {
tree[x].val += tree[y].val;
tree[x].sum = (tree[x].val != cnt[l]);
return;
}
int mid = (l + r) >> 1;
merge(tree[x].l, tree[y].l, l, mid);
merge(tree[x].r, tree[y].r, mid + 1, r);
push_up(x);
}
void dfs(int u, int fa, int id) {
root[u] = build(1, n, col[u]);
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(v == fa) continue;
dfs(v, u, i);
merge(root[u], root[v], 1, n);
}
if(u != 1) {
int tid = id / 2 + 1;
ans[tid] = tree[root[u]].sum;
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%d", &col[i]);
// vec[col[i]].push_back(i);
cnt[col[i]]++;
}
init();
for(int i = 1, u, v; i < n; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
// debug1(n);
dfs(1, -1, -1);
for(int i = 1; i <= n - 1; i++) {
printf("%d\n", ans[i]);
}
return 0;
}
每一个不曾刷题的日子
都是对生命的辜负
从弱小到强大，需要一段时间的沉淀，就是现在了
~buerdepepeqi
