分类目录归档：树

CSP 202012-3 带配额的文件系统

import copy

file_template = {
"isDirectory": False,
"fileSize": 0,
"childSize": 0,
"childQuota": 0,
"descendantQuota": 0,
"descendants": {}
}
root_directory = {
"isDirectory": True,
"fileSize": 0,
"childSize": 0,
"childQuota": 0,
"descendantQuota": 0,
"descendants": {}
}

def filepath_resolver(filepath: str):
fpArr = filepath.split('/')
fpArr.pop(0)
return fpArr

def visit(filepath):
if filepath == '/':
return root_directory
fpArr = filepath_resolver(filepath)
fileIter = root_directory
try:
for file in fpArr:
fileIter = fileIter["descendants"][file]
except KeyError:
return None
return fileIter

def checkQuotaLimit(fileIter, filesize, descendants = True, child = True):
if descendants:
if fileIter["descendantQuota"] != 0: # 检查后代文件配额
if fileIter["fileSize"] + filesize > fileIter["descendantQuota"]:
return False
if child:
if fileIter["childQuota"] != 0: # 检查孩子文件配额
if fileIter["childSize"] + filesize > fileIter["childQuota"]:
return False
return True

def create(filepath, filesize, isDirectory = False):
dstIter = visit(filepath)
update_filesize = filesize
if dstIter is not None:
update_filesize -= dstIter["fileSize"]

fpArr = filepath_resolver(filepath)
fileIter = root_directory
for i in range(len(fpArr)):
file = fpArr[i]
if file not in fileIter["descendants"].keys(): # 无该目录/文件
curFile = copy.deepcopy(file_template)
if i != len(fpArr) - 1: # 非末端文件
if not checkQuotaLimit(fileIter, update_filesize, True, False): return False
curFile["isDirectory"] = True
else: # 末端文件
if not checkQuotaLimit(fileIter, update_filesize): return False
curFile["isDirectory"] = isDirectory
curFile["fileSize"] = filesize
fileIter["descendants"][file] = curFile
else: # 存在该目录/文件
curFile = fileIter["descendants"][file]
if i != len(fpArr) - 1: # 非末端文件
if not curFile["isDirectory"]: # 欲进入的新目录文件与普通文件重名
return False
else:
if not checkQuotaLimit(fileIter, update_filesize, True, False): return False
else: # 末端文件
if curFile["isDirectory"] != isDirectory: # 欲创建的同名新文件与旧文件类型冲突
return False
else: # 文件已存在，替换文件大小
if not checkQuotaLimit(fileIter, update_filesize): return False
curFile["fileSize"] = filesize
fileIter = fileIter["descendants"][file]

fileIter = root_directory
for i in range(len(fpArr)):
file = fpArr[i]
fileIter["fileSize"] += update_filesize
if i == len(fpArr) - 1:
fileIter["childSize"] += update_filesize
fileIter = fileIter["descendants"][file]
return True

def remove(filepath):
fpArr = filepath_resolver(filepath)
dstIter = visit(filepath)
if dstIter is None:
return True

fileIter = root_directory
for file in fpArr:
fileIter["fileSize"] -= dstIter["fileSize"]
if fileIter["descendants"][file] == dstIter:
if not dstIter["isDirectory"]:
fileIter["childSize"] -= dstIter["fileSize"]
break
fileIter = fileIter["descendants"][file]
fileIter["descendants"].pop(fpArr[-1])
return True

def quota(filepath, child_quota, descendant_quota):
dstIter = visit(filepath)
if dstIter is None:
return False
if not dstIter["isDirectory"]:
return False
child_flag = child_quota == 0 or dstIter["childSize"] <= child_quota
descendant_flag = descendant_quota == 0 or dstIter["fileSize"] <= descendant_quota
if child_flag and descendant_flag:
dstIter["childQuota"] = child_quota
dstIter["descendantQuota"] = descendant_quota
return True
else:
return False

def main():
n = int(input())
for loop in range(n):
cmd = input().split()
result = False
if cmd[0] == "C":
result = create(cmd[1], int(cmd[2]))
elif cmd[0] == "R":
result = remove(cmd[1])
else:
result = quota(cmd[1], int(cmd[2]), int(cmd[3]))
if result:
print("Y")
else:
print("N")
# print(root_directory)

if __name__ == '__main__':
main()

洛谷 P3369 【模板】普通平衡树

发表评论

手写Splay成功。庆祝能写出来的第一个BBST！
代码如下：

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#include<iostream>
#include<cassert>
using namespace std;
class Splay {
private:
struct Node {
int v; //值
Node* f, * l, * r; //父，左孩子，右孩子
int cnts; //本子树下包含自身的所有**元素**数目
int rep; //该值重复了多少次
Node(int v, Node* f) :v(v), f(f), l(NULL), r(NULL), cnts(1), rep(1) {}
};
Node* _root;
#define root _root->r
void update(Node* p) { //更新cnts值
p->cnts = p->rep;
if (p->l)p->cnts += p->l->cnts;
if (p->r)p->cnts += p->r->cnts;
}
bool identify(Node* s, Node* f) { //确定本节点是左孩子还是右孩子，左孩子false，右孩子true
return !f || s == f->r;
}
void connect(Node* s, Node* f, bool r) { //儿子地址,父亲地址,连接到左儿子还是右儿子
if (f)(r ? f->r : f->l) = s; //父亲更新新儿子
if (s)s->f = f; //儿子更新新父亲
}
void rotate(Node* s) { //儿子节点的旋转
Node* f = s->f, * gf = s->f->f;
if (!identify(s, f)) { //左孩子
connect(s->r, f, false);
connect(f, s, true);
connect(s, gf, identify(f, gf));
}
else { //右孩子
connect(s->l, f, true);
connect(f, s, false);
connect(s, gf, identify(f, gf));
}
update(f);
update(s);
}
void splay(Node* s, Node* e) { //伸展操作，将节点s旋转到节点e所在的位置
e = e->f;
while (s->f != e) {
Node* f = s->f;
if (f->f == e)rotate(s); //s是e的直系儿子，只需做单旋
else if (identify(f, f->f) == identify(s, f)) { //Zig-Zig或Zag-Zag，需要先旋父亲节点，再旋儿子节点
rotate(f);
rotate(s);
}
else { //Zig-Zag或Zag-Zig
rotate(s);
rotate(s);
}
}
}

public:
Splay() {
_root = new Node(0, NULL);
}
~Splay() {
delete _root;
}
Node* find(int v) {
Node* cur = root;
if (!cur)return NULL;
while (1) {
if (cur->v == v) break;
Node* next = v < cur->v ? cur->l : cur->r;
if (!next)break;
cur = next;
}
splay(cur, root);
root = cur;
if (cur->v == v)return cur;
else return NULL;
}
void del(int v) {
Node* cur = find(v);
if (!cur)return;
if (cur->rep > 1) { //节点个数出现多于1次
cur->rep--;
cur->cnts--;
return;
}
if (!cur->l && !cur->r) { //删除最后一个仅剩的节点
root = NULL;
}
else if (!cur->l) { //无左子树时直接把右子树拼到根
root = cur->r;
root->f = _root;
}
else { //有左子树时，把左子树的最大值旋到根的左子，将根的右子放到根的左子的右子，删根后补左子树
Node* l = cur->l;
while (l->r)l = l->r;
splay(l, cur->l);
Node* r = cur->r;
connect(r, l, true);
root = l;
root->f = _root;
update(l);
}
delete cur;
}
void insert(int v) {
Node* cur = find(v);
if (!root) { //特判空树
root = new Node(v, _root);
return;
}
if (cur && cur->v == v) { //元素存在，直接把次数+1
cur->rep++;
cur->cnts++;
return;
}
Node* newNode = new Node(v, _root);
if (root->v < v) { //将v接入右侧
connect(root, newNode, false);
connect(root->r, newNode, true);
root->r = NULL;
}
else { //将v接入左侧
connect(root, newNode, true);
connect(root->l, newNode, false);
root->l = NULL;
}
update(root);
update(newNode);
root = newNode;
newNode->f = _root;
}
int rank(int v) {
Node* cur = find(v);
if (!cur)return -1;
int lCnts = cur->l ? cur->l->cnts : 0;
return lCnts + 1;
}
int atrank(int rank) {
Node* cur = root;
while (cur) {
int lCnts = cur->l ? cur->l->cnts : 0;
if (lCnts < rank && rank <= lCnts + cur->rep) {
splay(cur, root);
return cur->v;
}
if (rank <= lCnts)cur = cur->l;
else {
rank -= lCnts + cur->rep;
cur = cur->r;
}
}
return -1;
}
int upper(int v) {
Node* cur = find(v);
int lCnts = root->l ? root->l->cnts : 0;
if (root->v <= v)return atrank(lCnts + root->rep + 1);
return root->v;
}
int lower(int v) {
Node* cur = find(v);
int lCnts = root->l ? root->l->cnts : 0;
if (root->v >= v)return atrank(lCnts);
return root->v;
}
#undef root
};
int main() {
Splay splay;
int n;
cin >> n;
while (n--) {
int op1, op2;
cin >> op1 >> op2;
if (op1 == 1) {
splay.insert(op2);
}
else if (op1 == 2) {
splay.del(op2);
}
else if (op1 == 3) {
cout << splay.rank(op2) << endl;
}
else if (op1 == 4) {
cout << splay.atrank(op2) << endl;
}
else if (op1 == 5) {
cout << splay.lower(op2) << endl;
}
else if (op1 == 6) {
cout << splay.upper(op2) << endl;
}
}
}

参考资料：
https://www.luogu.com.cn/blog/user19027/solution-p3369
非常感谢rentenglong同学

洛谷 P2580 于是他错误的点名开始了

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Trie树练手题，然而空间不够用最后一个点MLE，90

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#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int MaxN = 5000000;
int trie[MaxN][26];
int cnts[MaxN];
int tot;
int trieInsert(string s) {
int node = 0;
for (int i = 0; i < s.size(); i++) {
if (!trie[node][s[i] - 'a'])node = trie[node][s[i] - 'a'] = ++tot;
else node = trie[node][s[i] - 'a'];
}
return node;
}
int main() {
int n,m;
string s;
cin >> n;
while (n--) {
cin >> s;
int node=trieInsert(s);
cnts[node]=1;
}
cin >> m;
while (m--) {
cin >> s;
int node = trieInsert(s);
if (cnts[node] == 1){
cout << "OK" << endl;
cnts[node]=2;
}
else if (cnts[node] == 0)cout << "WRONG" << endl;
else if (cnts[node] == 2)cout << "REPEAT" << endl;
}
}

洛谷 P1481 魔族密码

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今学习Trie树：https://blog.csdn.net/weixin_43792276/article/details/98977397

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#include<iostream>
#include<algorithm>
#include<string>
using namespace std;
const int MaxN = 5000;
int trie[MaxN][26];
int cnts[MaxN];
int tot;
void trieInsert(string s) {
int node = 0;
for (int i = 0; i < s.size(); i++) {
if (!trie[node][s[i] - 'a'])node = trie[node][s[i] - 'a'] = ++tot;
else node = trie[node][s[i] - 'a'];
}
cnts[node]++;
}
void trieAccumulate(int node) {
for (int i = 0; i < 26; i++) {
if (trie[node][i]) {
cnts[trie[node][i]] += cnts[node];
trieAccumulate(trie[node][i]);
}
}
}
int main() {
int n;
string s;
cin >> n;
while (n--) {
cin >> s;
trieInsert(s);
}
trieAccumulate(0);
int result = 0;
for (int i = 0; i <= tot; i++)result = max(result, cnts[i]);
cout << result;
}

Openjudge BJUTACM 19g系列代码

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题目链接：
http://bjutacm.openjudge.cn/lianxi/?page=2

继续阅读 →

洛谷 P1972 [SDOI2009]HH的项链

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参考题解：https://www.luogu.org/blog/skylee/solution-p1972

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#include<iostream>
#include<algorithm>
using namespace std;
int n,m;
int arr[500005];
int c[500000];
int last[1000001];
int flag=0;
struct question{
int id,l,r,ans;
bool operator < (const question& q2) const{
if(flag==0)return r<q2.r;
else return id<q2.id;
}
}qs[200005];
inline int lowbit(int i){
return i&(-i);
}
int query(int x){
int sum=0;
while(x>0){
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
void update(int i,int v){
while(i<=n){
c[i]+=v;
i+=lowbit(i);
}
}
int main(){
ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++){
cin>>arr[i];
}
cin>>m;
for(int i=1;i<=m;i++){
cin>>qs[i].l>>qs[i].r;
qs[i].id=i;
}
sort(qs+1,qs+m+1);
int ptr=1;
for(int i=1;i<=m;i++){
for(;ptr<=qs[i].r;ptr++){
if(last[arr[ptr]]!=0)update(last[arr[ptr]],-1);
update(last[arr[ptr]]=ptr,1);
}
qs[i].ans=query(qs[i].r)-query(qs[i].l-1);
}
flag=1;
sort(qs+1,qs+m+1);
for(int i=1;i<=m;i++){
cout<<qs[i].ans<<endl;
}
}

洛谷 P2023 [AHOI2009]维护序列

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没啥说的，和上一道题一模一样。。只是把输入改掉就可以了。。

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#include<iostream>
using namespace std;
struct segmentTree{
typedef long long ll;
const static int MAX=100000;
ll MOD;
ll v[4*MAX],aTg[4*MAX],mTg[4*MAX];
inline static ll lson(ll r){return r*2;}
inline static ll rson(ll r){return r*2+1;}
void build(ll r,ll* a,ll s,ll e){
aTg[r]=0;mTg[r]=1;
if(s==e){v[r]=a[s]%MOD;return;}
ll m=(s+e)>>1;
build(lson(r),a,s,m);
build(rson(r),a,m+1,e);
v[r]=(v[lson(r)]+v[rson(r)])%MOD;
}
void pushDown(ll r,ll s,ll e){
if(aTg[r]==0&&mTg[r]==1)return;
ll m=(s+e)>>1;
v[lson(r)]=(v[lson(r)]*mTg[r]%MOD+aTg[r]*(m-s+1)%MOD)%MOD;
v[rson(r)]=(v[rson(r)]*mTg[r]%MOD+aTg[r]*(e-m)%MOD)%MOD;
aTg[lson(r)]=(aTg[lson(r)]*mTg[r]%MOD+aTg[r]%MOD)%MOD;
aTg[rson(r)]=(aTg[rson(r)]*mTg[r]%MOD+aTg[r]%MOD)%MOD;
mTg[lson(r)]=(mTg[lson(r)]*mTg[r])%MOD;
mTg[rson(r)]=(mTg[rson(r)]*mTg[r])%MOD;
aTg[r]=0;mTg[r]=1;
}
ll query(ll r,ll cs,ll ce,ll qs,ll qe){
if(qe<cs||ce<qs)return 0;
if(qs<=cs&&ce<=qe)return v[r]%MOD;
pushDown(r,cs,ce);
ll cmd=(cs+ce)>>1;
return (query(lson(r),cs,cmd,qs,qe)+query(rson(r),cmd+1,ce,qs,qe))%MOD;
}
void update(ll r,ll cs,ll ce,ll us,ll ue,ll addv=0,ll mulv=1){
if(ue<cs||ce<us)return;
if(us<=cs&&ce<=ue){
v[r]=(v[r]*mulv%MOD+addv*(ce-cs+1))%MOD;
aTg[r]=(aTg[r]*mulv%MOD+addv)%MOD;
mTg[r]=(mTg[r]*mulv)%MOD;
return;
}
pushDown(r,cs,ce);
ll cmd=(cs+ce)>>1;
update(lson(r),cs,cmd,us,ue,addv,mulv);
update(rson(r),cmd+1,ce,us,ue,addv,mulv);
v[r]=(v[lson(r)]+v[rson(r)])%MOD;
}
}ST;
typedef long long ll;
ll n,m,p;
ll a[100005];
int main(){
ios::sync_with_stdio(false);
cin>>n>>p;
for(int i=1;i<=n;i++)cin>>a[i];
cin>>m;
ST.MOD=p;
ST.build(1,a,1,n);
for(int i=1;i<=m;i++){
ll op,x,y,k;
cin>>op;
if(op==1||op==2){
cin>>x>>y>>k;
if(op==1)ST.update(1,1,n,x,y,0,k);
else ST.update(1,1,n,x,y,k);
}else{
cin>>x>>y;
cout<<ST.query(1,1,n,x,y)%p<<endl;
}
}
}

洛谷 P3373 【模板】线段树 2

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总算写出来了。。。线段树同时维护乘法和加法，欢迎体验！

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#include<iostream>
using namespace std;
struct segmentTree{
typedef long long ll;
const static int MAX=100000;
ll MOD;
ll v[4*MAX],aTg[4*MAX],mTg[4*MAX];
inline static ll lson(ll r){return r*2;}
inline static ll rson(ll r){return r*2+1;}
void build(ll r,ll* a,ll s,ll e){
aTg[r]=0;mTg[r]=1;
if(s==e){v[r]=a[s]%MOD;return;}
ll m=(s+e)>>1;
build(lson(r),a,s,m);
build(rson(r),a,m+1,e);
v[r]=(v[lson(r)]+v[rson(r)])%MOD;
}
void pushDown(ll r,ll s,ll e){
if(aTg[r]==0&&mTg[r]==1)return;
ll m=(s+e)>>1;
v[lson(r)]=(v[lson(r)]*mTg[r]%MOD+aTg[r]*(m-s+1)%MOD)%MOD;
v[rson(r)]=(v[rson(r)]*mTg[r]%MOD+aTg[r]*(e-m)%MOD)%MOD;
aTg[lson(r)]=(aTg[lson(r)]*mTg[r]%MOD+aTg[r]%MOD)%MOD;
aTg[rson(r)]=(aTg[rson(r)]*mTg[r]%MOD+aTg[r]%MOD)%MOD;
mTg[lson(r)]=(mTg[lson(r)]*mTg[r])%MOD;
mTg[rson(r)]=(mTg[rson(r)]*mTg[r])%MOD;
aTg[r]=0;mTg[r]=1;
}
ll query(ll r,ll cs,ll ce,ll qs,ll qe){
if(qe<cs||ce<qs)return 0;
if(qs<=cs&&ce<=qe)return v[r]%MOD;
pushDown(r,cs,ce);
ll cmd=(cs+ce)>>1;
return (query(lson(r),cs,cmd,qs,qe)+query(rson(r),cmd+1,ce,qs,qe))%MOD;
}
void update(ll r,ll cs,ll ce,ll us,ll ue,ll addv=0,ll mulv=1){
if(ue<cs||ce<us)return;
if(us<=cs&&ce<=ue){
v[r]=(v[r]*mulv%MOD+addv*(ce-cs+1))%MOD;
aTg[r]=(aTg[r]*mulv%MOD+addv)%MOD;
mTg[r]=(mTg[r]*mulv)%MOD;
return;
}
pushDown(r,cs,ce);
ll cmd=(cs+ce)>>1;
update(lson(r),cs,cmd,us,ue,addv,mulv);
update(rson(r),cmd+1,ce,us,ue,addv,mulv);
v[r]=(v[lson(r)]+v[rson(r)])%MOD;
}
}ST;
typedef long long ll;
ll n,m,p;
ll a[100005];
int main(){
ios::sync_with_stdio(false);
cin>>n>>m>>p;
for(int i=1;i<=n;i++)cin>>a[i];
ST.MOD=p;
ST.build(1,a,1,n);
for(int i=1;i<=m;i++){
ll op,x,y,k;
cin>>op;
if(op==1||op==2){
cin>>x>>y>>k;
if(op==1)ST.update(1,1,n,x,y,0,k);
else ST.update(1,1,n,x,y,k);
}else{
cin>>x>>y;
cout<<ST.query(1,1,n,x,y)%p<<endl;
}
}
}

洛谷 P3368 【模板】树状数组 2

发表评论

看起来实际上也挺好做的，就是只要把树状数组维护的对象改成原序列的差分数组即可。这样每次树状数组所谓的query求1到n的前缀和也就变成了就单点n的值。

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#include<iostream>
#include<cstring>
using namespace std;
struct BTA{//Update Section,Query Single Point
const static int MAX=500000;
int C[MAX+5];
BTA(){
memset(C,0,sizeof(C));
}
int lowbit(int i){
return i&(-i);
}
int query(int x){
int sum=0;
while(x>0){
sum+=C[x];
x-=lowbit(x);
}
return sum;
}
void update(int i,int v){
while(i<=MAX){
C[i]+=v;
i+=lowbit(i);
}
}
void updateSection(int l,int r,int v){
update(l,v);
update(r+1,-v);
}
}TA;
int n,m;
int main(){
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1;i<=n;i++){
int v;
cin>>v;
TA.updateSection(i,i,v);
}
for(int i=1;i<=m;i++){
int op,v1,v2,v3;
cin>>op;
if(op==1){
cin>>v1>>v2>>v3;
TA.updateSection(v1,v2,v3);
}
else{
cin>>v1;
cout<<TA.query(v1)<<endl;
}
}
}

洛谷 P1197 [JSOI2008]星球大战

发表评论

总算是没看题解自己A了一道题，呜呜呜。

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#include<iostream>
#include<vector>
#include<cstring>
#include<stack>
#include<queue>
using namespace std;
int father[400005];
int _find(int node){
if(father[node]==-1)return node;
return father[node]=_find(father[node]);
}
bool _union(int x,int y){
if(_find(x)==_find(y))return false;
father[_find(x)]=_find(y);
return true;
}
int n,m,k;
bool alive[400005];
vector<int> edges[400005];
stack<int> ans;
stack<int> q;
int cnt,cntn;
int main(){
ios::sync_with_stdio(false);
memset(father,-1,sizeof(father));
memset(alive,true,sizeof(alive));
cin>>n>>m;
cnt=0;cntn=n;
for(int i=1;i<=m;i++){
int x,y;
cin>>x>>y;
edges[x].push_back(y);
edges[y].push_back(x);
}
cin>>k;
for(int i=1;i<=k;i++){
int v;
cin>>v;
alive[v]=false;
q.push(v);
cntn--;
}
for(int i=0;i<n;i++){
if(!alive[i])continue;
for(int j=0;j<edges[i].size();j++){
if(!alive[edges[i][j]])continue;
if(_union(i,edges[i][j]))cnt++;
}
}
ans.push(cntn-cnt);
while(!q.empty()){
int cur=q.top();q.pop();
cntn++;
alive[cur]=true;
for(int i=0;i<edges[cur].size();i++){
if(!alive[edges[cur][i]])continue;
if(_union(cur,edges[cur][i]))cnt++;
}
ans.push(cntn-cnt);
}
while(!ans.empty()){
cout<<ans.top()<<endl;
ans.pop();
}
}

杰克部落

梦想本不会发光，发光的是追梦的你

分类目录归档：树

CSP 202012-3 带配额的文件系统

洛谷 P3369 【模板】普通平衡树

洛谷 P2580 于是他错误的点名开始了

洛谷 P1481 魔族密码

Openjudge BJUTACM 19g系列代码

洛谷 P1972 [SDOI2009]HH的项链

洛谷 P2023 [AHOI2009]维护序列

洛谷 P3373 【模板】线段树 2

洛谷 P3368 【模板】树状数组 2

洛谷 P1197 [JSOI2008]星球大战

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