A ConneR and the A.R.C. Markland-N
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define inc(i, l, r) for (int i = l; i <= r; i++) 5 6 int t, n, s, k, x; 7 8 int main() { 9 cin >> t; 10 while (t--) { 11 cin >> n >> s >> k; 12 map<int, int> m; 13 inc(i, 1, k) { 14 cin >> x; 15 m[x] = 1; 16 } 17 inc(del, 0, k) { 18 if ((s + del <= n && m[s + del] == 0) || 19 (s - del >= 1 && m[s - del] == 0)) { 20 cout << del << "\n"; 21 break; 22 } 23 } 24 } 25 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define inc(i, l, r) for (int i = l; i <= r; i++) 5 6 int n; 7 double r; 8 9 int main() { 10 cin >> n; 11 for (int i = n; i >= 1; i--) r += 1.0 / i; 12 printf("%.12f", r); 13 }
每个岩浆点的对面或斜对面如果也有岩浆点，那么这两点之间就会形成障碍，统计这种障碍的数量
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define inc(i, l, r) for (int i = l; i <= r; i++) 5 6 const int maxn = 1e6 + 5; 7 8 int a[3][maxn]; 9 int n, q, k, x, y; 10 11 int main() { 12 cin >> n >> q; 13 inc(i, 0, q - 1) { 14 cin >> x >> y; 15 x--; 16 if (a[x][y] == 0) { 17 a[x][y] = 1; 18 inc(del, -1, 1) if (a[1 - x][y + del] == 1) k++; 19 } else { 20 a[x][y] = 0; 21 inc(del, -1, 1) if (a[1 - x][y + del] == 1) k--; 22 } 23 if(k) puts("NO"); 24 else puts("YES"); 25 } 26 }
分析点集的特点，可以知道每一个数据点都在下标比它小的数据点的右上方，而且离下一个数据点的距离 大于 离第0个数据点的距离。
枚举Aroma从哪个点出发；先往第0个数据点走，然后尽可能往右上走
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define inc(i, l, r) for (int i = l; i <= r; i++) 5 6 const int maxn = 1e6 + 5; 7 8 ll a0, b0, ax, bx, ay, by; 9 ll xs, ys, t; 10 11 ll dx[maxn], dy[maxn]; 12 ll dis[maxn]; 13 14 int res; 15 16 int main() { 17 cin >> a0 >> b0 >> ax >> ay >> bx >> by; 18 cin >> xs >> ys >> t; 19 ll tx = a0, ty = b0; 20 int top; 21 for (top = 0; tx <= 2e16 && ty <= 2e16; top++) { 22 dx[top] = tx, dy[top] = ty; 23 tx = tx * ax + bx, ty = ty * ay + by; 24 } 25 inc(i, 1, top - 1) dis[i] = dx[i] - dx[0] + dy[i] - dy[0]; 26 inc(i, 0, top - 1) { 27 ll tmp = t - abs(xs - dx[i]) - abs(ys - dy[i]); 28 if (tmp < 0) continue; 29 int tot = 0; 30 if (tmp <= dis[i]) { 31 tot = i - (lower_bound(dis, dis + top, dis[i] - tmp) - dis) + 1; 32 } else { 33 tot = upper_bound(dis + i + 1, dis + top, tmp - dis[i]) - dis; 34 } 35 res = max(res, tot); 36 } 37 cout << res; 38 }
最大S的方案必然是0-L都在一条链上，这条链上的数字是“山谷”型
推出状态转移方程
dp[u][v] = sub[u][v] * sub[v][u] + max(solve(par[v][u], v), solve(par[u][v], u));
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define ll long long 4 #define inc(i, l, r) for (int i = l; i <= r; i++) 5 #define pb push_back 6 7 const int maxn = 3e3 + 5; 8 vector<int> edge[maxn]; 9 int n, u, v; 10 int sub[maxn][maxn], par[maxn][maxn]; 11 ll dp[maxn][maxn], res; 12 13 int root; 14 15 int dfs(int x, int p) { 16 sub[root][x] = 1; 17 for (int i = 0; i < edge[x].size(); i++) { 18 if (edge[x][i] != p) { 19 par[root][edge[x][i]] = x; 20 sub[root][x] += dfs(edge[x][i], x); 21 } 22 } 23 return sub[root][x]; 24 } 25 26 ll solve(int u, int v) { 27 if (u == v) return 0; 28 if (dp[u][v]) return dp[u][v]; 29 return dp[u][v] = sub[u][v] * sub[v][u] + 30 max(solve(par[v][u], v), solve(par[u][v], u)); 31 } 32 33 int main() { 34 cin >> n; 35 inc(i, 1, n - 1) { 36 cin >> u >> v; 37 edge[u].pb(v); 38 edge[v].pb(u); 39 } 40 inc(i, 1, n) { 41 root = i; 42 dfs(root, -1); 43 } 44 inc(i, 1, n) inc(j, 1, n) { 45 res = max(res, solve(i, j)); 46 } 47 printf("%lld\n", res); 48 }
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