Replay
Dup4：
- 厕所是个换换脑子的好地方?
- 要读题啊，不要别人不做，自己就跟着不做啊
X：
- 读题很重要啊！什么时候才能读对题 不演队友啊 D题看错题， 直到最后一小时才看懂
- 很多时候要看榜单做题
A：Aqours
Solved.
考虑一个点的子树下面有多少个叶子节点汇聚，
那么这个时候就可以更新某些叶子节点的答案
并且把距离该点最近的叶子节点返回上去继续做这一步操作
再正着 考虑除子树外 的离当前点最近的叶子节点的距离，用于更新子树中叶子结点的答案
相当于树形dp起手式
但是这里$n很大，不能DFS$
但是又注意到$i <= j 有 fa[i] < fa[j]$
相当于$bfs序列是1\;2\;\cdots n$
可以直接两次遍历来做这件事情
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define N 3000010 5 #define pii pair <int, int> 6 int n, fa[N], ans[N], d[N]; 7 vector <pii> G[N]; 8 pii g[N]; 9 10 int main() 11 { 12 while (scanf("%d", &n) != EOF) 13 { 14 for (int i = 1; i <= n; ++i) G[i].clear(); 15 memset(ans, -1, sizeof ans); 16 memset(d, 0, sizeof d); 17 fa[1] = 0; 18 for (int i = 2; i <= n; ++i) 19 { 20 scanf("%d", fa + i); 21 ++d[fa[i]]; 22 } 23 for (int v = n; v >= 1; --v) 24 { 25 int u = fa[v]; 26 if (!d[v]) 27 G[u].push_back(pii(v, 1)); 28 else 29 { 30 sort(G[v].begin(), G[v].end()); 31 pii tmp = *G[v].begin(); 32 ++tmp.second; 33 G[u].push_back(tmp); 34 int Min = (*G[v].begin()).second; 35 for (int i = 1, len = G[v].size(); i < len; ++i) 36 { 37 pii it = G[v][i]; 38 ans[it.first] = Min + it.second; 39 } 40 } 41 } 42 for (int v = 1; v <= n; ++v) 43 { 44 g[v] = pii(1e9, 1e9); 45 int u = fa[v]; 46 if (u) 47 { 48 g[v] = g[u]; 49 ++g[v].second; 50 pii tmp = g[v]; 51 if (tmp.first != -1) for (int i = 0, len = G[v].size(); i < len; ++i) 52 { 53 pii it = G[v][i]; 54 if (it.first > tmp.first) 55 ans[it.first] = min(ans[it.first], it.second + tmp.second); 56 } 57 } 58 if (!G[v].empty()) 59 { 60 pii it = *G[v].begin(); 61 if (it.second < g[v].second) 62 g[v] = it; 63 } 64 } 65 for (int i = 1; i <= n; ++i) if (!d[i]) printf("%d %d\n", i, ans[i]); 66 } 67 return 0; 68 }
D：吉良吉影的奇妙计划
Solved.
n至于20
暴力打表
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 const int maxn = 110; 7 const ll MOD = 998244353; 8 9 10 int n; 11 ll ans = 0; 12 int arr[maxn]; 13 int tmp[maxn]; 14 15 bool judge() 16 { 17 for(int i = 1; i <= 2 * n; ++i) tmp[i] = arr[i]; 18 int flag = 1; 19 while(flag) 20 { 21 flag = 0; 22 for(int i = 1; i + 1 <= 2 * n; ++i) 23 { 24 if((tmp[i] == 1 && tmp[i + 1] == 0) || (tmp[i] == 0 && tmp[i + 1] == 1)) 25 { 26 tmp[i] = tmp[i + 1] = -1; 27 flag = 1; 28 break; 29 } 30 } 31 } 32 for(int i = 1; i + 3 <= 2 * n; ++i) 33 { 34 if(tmp[i] + tmp[i + 1] + tmp[i + 2] + tmp[i + 3] == -4) return true; 35 } 36 return false; 37 } 38 39 void DFS(int cnt, int cnt1, int cnt0) 40 { 41 if(cnt == 2 * n + 1) 42 { 43 ans = (ans + 1) % MOD; 44 return ; 45 } 46 if(cnt1) 47 { 48 if(cnt == 1 || (cnt > 1 && arr[cnt - 1] != 0)) 49 { 50 arr[cnt] = 1; 51 DFS(cnt + 1, cnt1 - 1, cnt0); 52 } 53 } 54 if(cnt0) 55 { 56 if(cnt == 1 || (cnt > 1 && arr[cnt - 1] != 1)) 57 { 58 arr[cnt] = 0; 59 DFS(cnt + 1, cnt1, cnt0 - 1); 60 } 61 } 62 if(cnt1 > 0 && cnt0 > 0) 63 { 64 if((cnt + 1 <= 2 * n && arr[cnt - 1] != -1)) 65 { 66 arr[cnt] = -1; 67 arr[cnt + 1] = -1; 68 DFS(cnt + 2, cnt1 - 1, cnt0 - 1); 69 arr[cnt] = -2; 70 arr[cnt + 1] = -2; 71 } 72 } 73 } 74 75 int main() 76 { 77 while(~scanf("%d", &n)) 78 { 79 ans = 0; 80 DFS(1, n, n); 81 cout << ans << endl; 82 } 83 return 0; 84 }
E：Souls-like Game
Solved.
阅读理解题
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 10010 6 const ll MOD = (ll)998244353; 7 int n, m, p[N][3][3]; 8 9 int main() 10 { 11 while (scanf("%d%d", &n, &m) != EOF) 12 { 13 for (int i = 1; i < n; ++i) 14 { 15 for (int j = 0; j < 3; ++j) 16 for (int k = 0; k < 3; ++k) 17 scanf("%d", &p[i][j][k]), p[i][j][k] %= MOD; 18 } 19 for (int i = 1, op, l, r; i <= m; ++i) 20 { 21 scanf("%d%d%d", &op, &l, &r); 22 if (op == 1) 23 { 24 int pp[3][3]; 25 for (int j = 0; j < 3; ++j) 26 for (int k = 0; k < 3; ++k) 27 scanf("%d", &pp[j][k]), pp[j][k] %= MOD; 28 for (int j = l; j <= r; ++j) 29 for (int k = 0; k < 3; ++k) 30 for (int o = 0; o < 3; ++o) 31 p[j][k][o] = pp[k][o]; 32 } 33 else 34 { 35 ll sum[2][3] = {0, 0, 0}; 36 for (int j = 0; j < 3; ++j) sum[r & 1][j] = 1; 37 for (int j = r - 1; j >= l; --j) 38 { 39 for (int k = 0; k < 3; ++k) sum[j & 1][k] = 0; 40 for (int k = 0; k < 3; ++k) 41 for (int o = 0; o < 3; ++o) 42 sum[j & 1][k] = (sum[j & 1][k] + p[j][k][o] * sum[(j & 1) ^ 1][o] % MOD) % MOD; 43 } 44 ll res = 0; 45 for (int j = 0; j < 3; ++j) 46 res = (res + sum[l & 1][j]) % MOD; 47 //for (int j = 0; j < 3; ++j) printf("%lld%c", sum[l & 1][j], " \n"[j == 2]); 48 printf("%lld\n", res); 49 } 50 } 51 } 52 return 0; 53 }
G：穗乃果的考试
Solved.
$考虑到每个(i, j)位置上的1可能被i \cdot (n - i + 1) \cdot j \cdot (m - j + 1)的矩形包含$
$即每个1都会产生i \cdot (n - i + 1) \cdot j \cdot (m - j + 1)的贡献 累加即可$
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const ll MOD = 998244353; 8 const int maxn = 2e3 + 10; 9 10 int n, m; 11 char mp[maxn][maxn]; 12 13 int main() 14 { 15 while(~scanf("%d %d", &n, &m)) 16 { 17 ll ans = 0; 18 for(int i = 1; i <= n; ++i) 19 { 20 for(int j = 1; j <= m; ++j) 21 { 22 scanf(" %c", &mp[i][j]); 23 if(mp[i][j] == '1') 24 { 25 ll tmp = i * (n - i + 1) % MOD * j % MOD * (m - j + 1) % MOD; 26 ans = (ans + tmp) % MOD; 27 } 28 } 29 } 30 printf("%lld\n", ans); 31 } 32 return 0; 33 }