A题，水题。
B题，一开始看题目觉得蛮复杂的，其实很简单。存好图后找可以储存的点，再遍历这些点附近可以开店的点，维护一下答案的最小值即可。
C题，推不出公式= =。见代码好了：
1 #include <stdio.h> 2 #include <algorithm> 3 #include <string.h> 4 #include <iostream> 5 #include <vector> 6 using namespace std; 7 const int N = 100000 + 5; 8 typedef long long ll; 9 10 ll n; 11 12 int main() 13 { 14 cin >> n; 15 if(n <= 2) cout << "-1" << endl; 16 else 17 { 18 if(n % 2 == 0) printf("%I64d %I64d ",n/2*n/2-1,n/2*n/2+1); 19 else printf("%I64d %I64d ",(n*n)/2,(n*n)/2+1); 20 } 21 return 0; 22 }
D题，以前用dfs+离线写过。看别人写的可持久化。内存消耗蛮大的= =。我不是很清楚节点的内存到底要开多大。。下面是铭神的代码：
1 #include <bits/stdc++.h> 2 3 struct Node 4 { 5 int lc, rc; 6 bool flag; 7 int val; 8 }; 9 10 Node node[100000 * 10 * 5]; 11 12 const int kQ = 100000 + 1; 13 const int kN = 1000 + 1; 14 15 int shelves[kQ][kN]; 16 17 int n, m, nq; 18 19 int tot; 20 21 int query(int o, int p, int l, int r) 22 { 23 if (o == 0) 24 return 0; 25 if (l == r) 26 return node[o].val; 27 int mid = l + r >> 1; 28 if (p <= mid) 29 return query(node[o].lc, p, l, mid); 30 else 31 return query(node[o].rc, p, mid + 1, r); 32 } 33 34 int modify(int o, int p, int l, int r) 35 { 36 int w = tot ++; 37 node[w] = node[o]; 38 39 if (l == r) { 40 node[w].val ^= 1; 41 return w; 42 } 43 int mid = l + r >> 1; 44 if (p <= mid) 45 node[w].lc = modify(node[o].lc, p, l, mid); 46 else 47 node[w].rc = modify(node[o].rc, p, mid + 1, r); 48 node[w].val = node[node[w].lc].val + node[node[w].rc].val; 49 return w; 50 } 51 52 int main() 53 { 54 tot = 1; 55 std::ios::sync_with_stdio(false); 56 std::cin >> n >> m >> nq; 57 for (int i = 1; i <= nq; ++ i) { 58 int op; 59 std::cin >> op; 60 if (op == 4) { 61 int k; 62 std::cin >> k; 63 for (int j = 1; j <= n; ++ j) { 64 shelves[i][j] = shelves[k][j]; 65 } 66 } else { 67 for (int j = 1; j <= n; ++ j) { 68 shelves[i][j] = shelves[i - 1][j]; 69 } 70 if (op == 3) { 71 int x; 72 std::cin >> x; 73 int w = tot ++; 74 node[w] = node[shelves[i][x]]; 75 node[w].flag ^= 1; 76 shelves[i][x] = w; 77 } else { 78 int x, y; 79 std::cin >> x >> y; 80 bool aim = (op == 1) ^ node[shelves[i][x]].flag; 81 if (query(shelves[i][x], y, 1, m) != aim) { 82 shelves[i][x] = modify(shelves[i][x], y, 1, m); 83 } 84 } 85 } 86 int answer = 0; 87 for (int x = 1; x <= n; ++ x) { 88 int tmp = node[shelves[i][x]].val; 89 if (node[shelves[i][x]].flag == false) 90 answer += tmp; 91 else 92 answer += m - tmp; 93 } 94 printf("%d ", answer); 95 } 96 }