题目大意不多说了
貌似每个苦逼的acmer都要做一下这个splay树的模版题目吧
还是有很多操作的，估计够以后当模版了。。。。
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <algorithm>
5 #include <cmath>
6
7 using namespace std;
8 const int N = 1000010;
9 const int INF = 0x3f3f3f3f;
10 #define ls ch[x][0]
11 #define rs ch[x][1]
12
13 struct SplayTree{
14 int ch[N][2] , pre[N];
15 int ml[N] , mr[N] , mm[N]; //区间最大
16 int val[N] , sz[N] , sum[N] , size , rt;
17 int rev[N] , to[N]; // lazy标记
18 int del[N] , top; //内部保存删除的数的位置，然后可以从其中提取位置给新进入的元素
19 int a[N];
20
21 void push_up(int x)
22 {
23 sz[x] = sz[ls]+sz[rs]+1;
24 sum[x] = sum[ls]+sum[rs]+val[x];
25 /*********************/
26 ml[x] = max(ml[ls] , sum[ls]+val[x]+max(ml[rs] , 0));
27 mr[x] = max(mr[rs] , sum[rs]+val[x]+max(mr[ls] , 0));
28 mm[x] = max(mm[ls] , max(mm[rs] , max(ml[rs],0)+max(mr[ls],0)+val[x]));
29 /*********************/
30 }
31
32 void push_down(int x)
33 {
34 if(rev[x]){
35 if(ls){
36 rev[ls]^=1 ;
37 swap(ml[ls] , mr[ls]);
38 }
39 if(rs){
40 rev[rs]^=1 ;
41 swap(ml[rs] , mr[rs]);
42 }
43 swap(ls , rs);
44 rev[x] = 0;
45 }
46 /**to[x]的初始化要注意**/
47 if(to[x]!=-INF){
48 if(ls){
49 sum[ls] = sz[ls]*to[x];
50 ml[ls] = mr[ls] = mm[ls] = max(sum[ls] , to[x]);
51 val[ls] = to[ls] = to[x];
52 }
53 if(rs){
54 sum[rs] = sz[rs]*to[x];
55 ml[rs] = mr[rs] = mm[rs] = max(sum[rs] , to[x]);
56 val[rs] = to[rs] = to[x];
57 }
58 to[x] = -INF;
59 }
60 }
61
62 void newNode(int &x , int fa , int v)
63 {
64 if(top != -1)
65 x = del[top--];
66 else x = ++size;
67 ch[x][0] = ch[x][1] = 0;
68 pre[x] = fa;
69 val[x] = v , sz[x] = 1;
70 sum[x] = ml[x] = mr[x] = mm[x] = val[x];
71 rev[x] = 0 , to[x] = -INF;
72 }
73
74 void build(int &x , int l , int r , int fa)
75 {
76 if(l>r) return;
77 int m=(l+r)>>1;
78 newNode(x , fa , a[m]);
79 build(ls , l , m-1 , x);
80 build(rs , m+1 , r , x);
81 push_up(x);
82 }
83
84 void init(int n)
85 {
86 /*****因为所有点的最终叶子节点都下标设为了0，所以0上的数据要不影响整棵树*****/
87 sz[0] = sum[0] = ch[0][0] = ch[0][1] = val[0] = pre[0] = 0;
88 to[0] = ml[0] = mr[0] = mm[0] = -INF;
89 rev[0] = 0;
90 top = -1 , rt = size = 0;
91 newNode(rt , 0 , -INF);
92 newNode(ch[rt][1] , rt , -INF);
93 build(ch[ch[rt][1]][0] , 1 , n , ch[rt][1]);
94 push_up(ch[rt][1]);
95 push_up(rt);
96 }
97
98 void Rotate(int x , int f)
99 {
100 int y=pre[x] , z=pre[y];
101 /**y要旋转到下方，下传lazy标记**/
102 push_down(y);
103 ch[y][!f] = ch[x][f] , pre[ch[x][f]] = y;
104 ch[x][f] = y , pre[y] = x;
105 ch[z][ch[z][1]==y] = x , pre[x]=z;
106 push_up(y);
107 push_up(x);
108 }
109
110 void Splay(int x , int goal)
111 {
112 while(pre[x] != goal){
113 if(pre[pre[x]] == goal) Rotate(x , ch[pre[x]][0]==x);
114 else{
115 int y = pre[x] , z = pre[y];
116 int f = ch[z][0] == y;
117 if(ch[y][f] == x) Rotate(x , !f);
118 else Rotate(y,f);
119 Rotate(x , f);
120 }
121 }
122 push_up(x);
123 if(goal == 0) rt = x;
124 }
125
126 int Select(int pos)
127 {
128 int x = rt;
129 push_down(x);
130 while(sz[ls]+1 != pos){
131 if(sz[ls]+1>pos) x=ls;
132 else{
133 pos = pos-sz[ls]-1;
134 x = rs;
135 }
136 push_down(x);
137 }
138 return x;
139 }
140 /***从pos位置之后插入cnt个值***/
141 void Insert(int pos , int cnt)
142 {
143 int x = Select(pos);
144 Splay(x , 0);
145 int y = Select(pos+1);
146 Splay(y , x);
147 build(ch[y][0] , 1 , cnt , y);
148 /***这里build将a[1~cnt]插入到splay树中，要push_up更新***/
149 push_up(y);
150 push_up(x);
151 }
152 /***回收被删除的数的位置***/
153 void Recycle(int x)
154 {
155 if(x){
156 del[++top]=x;
157 Recycle(ls);
158 Recycle(rs);
159 }
160 }
161
162 void Delete(int s , int t)
163 {
164 int x = Select(s-1) , y = Select(t+1);
165 Splay(x , 0) , Splay(y , x);
166 Recycle(ch[y][0]);
167 ch[y][0] = 0;
168 /***这里y的左子树删除，明显子树数据改变，所以要重新push_up更新数据***/
169 push_up(y);
170 push_up(x);
171 }
172 /****将s~t区间内的所有值都修改为v****/
173 void update(int s , int t , int v)
174 {
175 int x = Select(s-1) , y = Select(t+1);
176 Splay(x , 0) , Splay(y , x);
177 int ptr = ch[y][0];
178 to[ptr]=v;
179 val[ptr]=v;
180 sum[ptr]=v*sz[ptr];
181 ml[ptr] = mr[ptr] = mm[ptr] = max(sum[ptr] , v);
182 return ;
183 }
184
185 void Reverse(int s , int t)
186 {
187 int x = Select(s-1) , y = Select(t+1);
188 Splay(x , 0) , Splay(y , x);
189 int ptr = ch[y][0];
190 rev[ptr]^=1;
191 swap(ml[ptr] , mr[ptr]);
192 }
193
194 int querySum(int s , int t)
195 {
196 int x = Select(s-1) , y = Select(t+1);
197 // cout<<"x: "<<x<<" y: "<<y<<endl;
198 Splay(x , 0) , Splay(y , x);
199 return sum[ch[y][0]];
200 }
201
202 int queryMax()
203 {
204 int x = Select(1) , y = Select(sz[rt]);
205 Splay(x , 0) , Splay(y , x);
206 return mm[ch[y][0]];
207 }
208 }spt;
209 int main()
210 {
211 // freopen("a.in" , "r" , stdin);
212 int n , m;
213 char str[20] ;
214 int s , t , v , pos , k;
215 while(~scanf("%d%d" , &n , &m))
216 {
217 for(int i=1 ; i<=n ; i++) scanf("%d" , &spt.a[i]);
218 spt.init(n);
219 /*
220 for(int i=0 ; i<=spt.size ; i++){
221 cout<<"i: "<<i<<" sum: "<<spt.sum[i]<<" l: "<<spt.ch[i][0]<<" lv: "<<spt.val[spt.ch[i][0]]<<" r: "<<spt.ch[i][1]<<" rv: "<<spt.val[spt.ch[i][1]]<<endl;
222 }*/
223 for(int i=0 ; i<m ; i++){
224 scanf("%s" , str);
225 if(str[0] == ‘I‘){
226 scanf("%d%d" , &pos , &k);
227 for(int i=1 ; i<=k ; i++) scanf("%d" , &spt.a[i]);
228 spt.Insert(pos+1 , k);
229 }
230 else if(str[0]==‘D‘){
231 scanf("%d%d" , &pos , &k);
232 spt.Delete(pos+1 , pos+k);
233 }
234 else if(str[0] == ‘M‘ && str[2] == ‘K‘){
235 scanf("%d%d%d" , &pos , &k , &v);
236 spt.update(pos+1 , pos+k , v);
237 }
238 else if(str[0] == ‘R‘){
239 scanf("%d%d" , &pos , &k);
240 spt.Reverse(pos+1 , pos+k);
241 }
242 else if(str[0] == ‘G‘){
243 scanf("%d%d" , &pos , &k);
244 printf("%d\n" , spt.querySum(pos+1 , pos+k));
245 }
246 else{
247 /****这里不能直接用spt.mm[rt]作为答案，因为为了方便查询，
248 我们增加了两个不包含在本身数组的点，直接spt.mm[rt]会把这两个点也算进来***/
249 printf("%d\n" , spt.queryMax());
250 }
251 }
252 }
253 return 0;
254 }
时间： 2024-11-12 09:16:29