剩下的题等礼拜天补吧。最近要赶作业什么的还要补点dp
A - Ada and List（伸展树或者vector暴力- -）
思路：
vector玄学均摊复杂度...能直接暴力A了。窝居然写了人生第一个伸展树- -。蠢到家了。
1 #include <bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5 vector<int>vec;
6 int n, m;
7 scanf("%d%d", &n, &m);
8 for(int i = 1; i <= n; i++)
9 {
10 int x;
11 scanf("%d", &x);
12 vec.push_back(x);
13 }
14
15 for(int i = 0; i < m; i++)
16 {
17 int cmd, k, x;
18 scanf("%d%d", &cmd, &k);
19 if(cmd == 1)
20 {
21 scanf("%d", &x);
22 vec.insert(vec.begin() + k - 1, x);
23 }
24 else if(cmd == 2)
25 {
26 vec.erase(vec.begin() + k - 1);
27 }
28 else if(cmd == 3)
29 {
30 printf("%d\n", vec[k - 1]);
31 }
32 }
33 return 0;
34 }
这个伸展树的板子还是挺厉害了。
1 #include<bits/stdc++.h>
2 using namespace std;
3
4 #define Key_value ch[ch[root][1]][0]
5 const int MAXN = 500010;
6 const int INF = 0x3f3f3f3f;
7 int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN];
8 int root,tot1;
9 int sum[MAXN],rev[MAXN],same[MAXN];
10 int lx[MAXN],rx[MAXN],mx[MAXN];
11 int s[MAXN],tot2;//内存池和容量
12 int a[MAXN];
13 int n,q;
14
15 void NewNode(int &r,int father,int k)
16 {
17 if(tot2) r = s[tot2--];//取的时候是tot2--,存的时候就是++tot2
18 else r = ++tot1;
19 pre[r] = father;
20 ch[r][0] = ch[r][1] = 0;
21 key[r] = k;
22 sum[r] = k;
23 rev[r] = same[r] = 0;
24 lx[r] = rx[r] = mx[r] = k;
25 size[r] = 1;
26 }
27 void Update_Rev(int r)
28 {
29 if(!r)return;
30 swap(ch[r][0],ch[r][1]);
31 swap(lx[r],rx[r]);
32 rev[r] ^= 1;
33 }
34 void Update_Same(int r,int v)
35 {
36 if(!r)return;
37 key[r] = v;
38 sum[r] = v*size[r];
39 lx[r] = rx[r] = mx[r] = max(v,v*size[r]);
40 same[r] = 1;
41 }
42 void push_up(int r)
43 {
44 int lson = ch[r][0], rson = ch[r][1];
45 size[r] = size[lson] + size[rson] + 1;
46 sum[r] = sum[lson] + sum[rson] + key[r];
47 lx[r] = max(lx[lson],sum[lson] + key[r] + max(0,lx[rson]));
48 rx[r] = max(rx[rson],sum[rson] + key[r] + max(0,rx[lson]));
49 mx[r] = max(0,rx[lson]) + key[r] + max(0,lx[rson]);
50 mx[r] = max(mx[r],max(mx[lson],mx[rson]));
51 }
52 void push_down(int r)
53 {
54 if(same[r])
55 {
56 Update_Same(ch[r][0],key[r]);
57 Update_Same(ch[r][1],key[r]);
58 same[r] = 0;
59 }
60 if(rev[r])
61 {
62 Update_Rev(ch[r][0]);
63 Update_Rev(ch[r][1]);
64 rev[r] = 0;
65 }
66 }
67 void Build(int &x,int l,int r,int father)
68 {
69 if(l > r)return;
70 int mid = (l+r)/2;
71 NewNode(x,father,a[mid]);
72 Build(ch[x][0],l,mid-1,x);
73 Build(ch[x][1],mid+1,r,x);
74 push_up(x);
75 }
76 void Init()
77 {
78 root = tot1 = tot2 = 0;
79 ch[root][0] = ch[root][1] = size[root] = pre[root] = 0;
80 same[root] = rev[root] = sum[root] = key[root] = 0;
81 lx[root] = rx[root] = mx[root] = -INF;
82 NewNode(root,0,-1);
83 NewNode(ch[root][1],root,-1);
84 for(int i = 0;i < n;i++)
85 scanf("%d",&a[i]);
86 Build(Key_value,0,n-1,ch[root][1]);
87 push_up(ch[root][1]);
88 push_up(root);
89 }
90 //旋转,0为左旋，1为右旋
91 void Rotate(int x,int kind)
92 {
93 int y = pre[x];
94 push_down(y);
95 push_down(x);
96 ch[y][!kind] = ch[x][kind];
97 pre[ch[x][kind]] = y;
98 if(pre[y])
99 ch[pre[y]][ch[pre[y]][1]==y] = x;
100 pre[x] = pre[y];
101 ch[x][kind] = y;
102 pre[y] = x;
103 push_up(y);
104 }
105 //Splay调整，将r结点调整到goal下面
106 void Splay(int r,int goal)
107 {
108 push_down(r);
109 while(pre[r] != goal)
110 {
111 if(pre[pre[r]] == goal)
112 {
113 push_down(pre[r]);
114 push_down(r);
115 Rotate(r,ch[pre[r]][0] == r);
116 }
117 else
118 {
119 push_down(pre[pre[r]]);
120 push_down(pre[r]);
121 push_down(r);
122 int y = pre[r];
123 int kind = ch[pre[y]][0]==y;
124 if(ch[y][kind] == r)
125 {
126 Rotate(r,!kind);
127 Rotate(r,kind);
128 }
129 else
130 {
131 Rotate(y,kind);
132 Rotate(r,kind);
133 }
134 }
135 }
136 push_up(r);
137 if(goal == 0) root = r;
138 }
139 int Get_kth(int r,int k)
140 {
141 push_down(r);
142 int t = size[ch[r][0]] + 1;
143 if(t == k)return r;
144 if(t > k)return Get_kth(ch[r][0],k);
145 else return Get_kth(ch[r][1],k-t);
146 }
147
148 //在第pos个数后面插入tot个数
149 void Insert(int pos,int tot)
150 {
151 for(int i = 0;i < tot;i++)scanf("%d",&a[i]);
152 Splay(Get_kth(root,pos+1),0);
153 Splay(Get_kth(root,pos+2),root);
154 Build(Key_value,0,tot-1,ch[root][1]);
155 push_up(ch[root][1]);
156 push_up(root);
157 }
158
159 //删除子树
160 void erase(int r)
161 {
162 if(!r)return;
163 s[++tot2] = r;
164 erase(ch[r][0]);
165 erase(ch[r][1]);
166 }
167 //从第pos个数开始连续删除tot个数
168 void Delete(int pos,int tot)
169 {
170 Splay(Get_kth(root,pos),0);
171 Splay(Get_kth(root,pos+tot+1),root);
172 erase(Key_value);
173 pre[Key_value] = 0;
174 Key_value = 0;
175 push_up(ch[root][1]);
176 push_up(root);
177 }
178
179
180 //得到第pos个数开始的tot个数的和
181 int Get_Sum(int pos,int tot)
182 {
183 Splay(Get_kth(root,pos),0);
184 Splay(Get_kth(root,pos+tot+1),root);
185 return sum[Key_value];
186 }
187
188
189 void InOrder(int r)
190 {
191 if(!r)return;
192 push_down(r);
193 InOrder(ch[r][0]);
194 printf("%d ",key[r]);
195 InOrder(ch[r][1]);
196 }
197
198 int main()
199 {
200 scanf("%d%d", &n, &q);
201 Init();
202
203 while(q--)
204 {
205 int cmd;
206 scanf("%d", &cmd);
207 if(cmd == 1)
208 {//1 k x means that you will add number x to position k
209 int k;
210 scanf("%d", &k);
211 Insert(k - 1, 1);
212 }
213 else if(cmd == 2)
214 {//2 k means that you will erase number from position k
215 int k;
216 scanf("%d", &k);
217 Delete(k, 1);
218 }
219 else if(cmd == 3)
220 {//3 k means that you will print number from position k
221 int k;
222 scanf("%d", &k);
223 printf("%d\n",Get_Sum(k,1));
224 }
225
226 }
227 return 0;
228 }
B - Ada and Queue（双端队列）
思路：
打个标记，记录一下现在是正着还是反着就好了。然后就是一个双端队列。
1 #include <bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5 deque<int>que;
6 deque<int>::iterator it;
7 deque<int>::reverse_iterator rit;
8
9 int n;
10 scanf("%d", &n);
11 int tig = 0;
12 while(n--)
13 {
14 char cmd[100];
15 scanf("%s", cmd);
16
17 if(cmd[0] == ‘b‘)
18 {//back - Print number from back and then erase it
19 if(que.size() == 0)
20 {
21 puts("No job for Ada?");
22 }
23 else if(tig == 0)
24 {
25 rit = que.rbegin();
26 cout << *rit << "\n";
27 que.pop_back();
28 }
29 else
30 {
31 it = que.begin();
32 cout << *it << "\n";
33 que.pop_front();
34 }
35 }
36 else if(cmd[0] == ‘f‘)
37 {//front - Print number from front and then erase it
38 if(que.size() == 0)
39 {
40 puts("No job for Ada?");
41 }
42 else if(tig == 1)
43 {
44 rit = que.rbegin();
45 cout << *rit << "\n";
46 que.pop_back();
47 }
48 else if(tig == 0)
49 {
50 it = que.begin();
51 cout << *it << "\n";
52 que.pop_front();
53 }
54 }
55 else if(cmd[0] == ‘r‘)
56 {//reverse - Reverses all elements in queue
57 tig ^= 1;
58 }
59 else if(cmd[0] == ‘p‘)
60 {//push_back N - Add element N to back
61 int x;
62 scanf("%d", &x);
63 if(tig == 0)
64 {
65 que.push_back(x);
66 }
67 else
68 {
69 que.push_front(x);
70 }
71 }
72 else if(cmd[0] == ‘t‘)
73 {//toFront N - Put element N to front
74 int x;
75 scanf("%d", &x);
76 if(tig == 1)
77 {
78 que.push_back(x);
79 }
80 else
81 {
82 que.push_front(x);
83 }
84 }
85 }
86 return 0;
87 }
C - Ada and Field
D - Ada and Cycle（bitset优化bfs）
E - Face Detection（水题暴力枚举）
思路：
暴力枚举一下2*2的小矩形的左上角就好了，wa点！！！n*m的图，两层for循环写成n*n了！！！
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 int n, m;
5 char ma[55][55];
6 #define isInside(x, y) 1 <= x && x <= n && 1 <= y && y <= m
7
8 bool judge(int x, int y)
9 {
10 if(isInside(x, y) && isInside(x + 1, y) && isInside(x, y + 1) && isInside(x + 1, y + 1))
11 {
12 map<char, int>s;
13 s[ma[x][y]] = 1;
14 s[ma[x + 1][y]] = 1;
15 s[ma[x][y + 1]] = 1;
16 s[ma[x + 1][y + 1]] = 1;
17 if(s[‘f‘] + s[‘a‘] + s[‘c‘] + s[‘e‘] == 4) return true;
18 }
19 return false;
20 }
21 int main()
22 {
23 scanf("%d%d", &n, &m);
24 for(int i = 1; i <= n; i++)
25 scanf("%s", ma[i] + 1);
26
27 int ans = 0;
28 for(int i = 1; i <= n; i++)
29 {
30 for(int j = 1; j <= m; j++)
31 {
32 if(judge(i, j)) ans++;
33 }
34 }
35 printf("%d\n", ans);
36 return 0;
37 }
F - Gears
时间： 2024-10-20 00:45:51