Codeforces Round #169 (Div. 2)補題記錄
阿新 • • 發佈：2021-01-12
前三題沒什麼可說的
CF - 276C.Little Girl and Maximum Sum
記錄每個索引詢問的次數，將陣列內的數從大到小排序，詢問次數從大到小排序，相乘求和即可。
CF - 276D.Little Girl and Maximum XOR
求\([l,r]\)區間內兩個數異或的最大值。
我們考慮\(l\)和\(r\)異或所得的數\(x\)，我們的答案就是從這個數\(x\)的最高位\(1\)開始，後面所有位全部置為\(1\)。
注意\(long~long\)
#if __cplusplus >= 201103L #pragma comment(linker, "/STACK:102400000,102400000") #pragma GCC optimize(3, "Ofast", "inline") #include <bits/stdc++.h> #else #include <algorithm> #include <bitset> #include <cmath> #include <iostream> #include <map> #include <string.h> #include <vector> #endif using namespace std; #define inf __INT_MAX__ #define enf INT_MIN #define INF LLONG_MAX #define ENF LLONG_MIN const int MAXN = 2e5 + 10; const double pi = acos(-1.0); const double eps = 1e-6; typedef long long ll; typedef unsigned long long ull; #define zhengfu(x) ((x > eps) - (x < -eps)) int main() { ll l, r; cin >> l >> r; if (l == r) cout << 0 << endl; else { ll tmp = l ^ r; for (ll i = 63; i >= 0; i--) if (tmp & (1LL << i)) { cout << (1LL << (i + 1)) - 1LL << endl; break; } } }
CF - 276E.Little Girl and Problem on Trees
給你一棵樹，其中，除了根節點\(1\)以外，其他的節點度數不大於2。
也就是說，如果把\(1\)從中刪除，那麼就會變成一條條鏈。
兩個操作，一個操作是對於節點\(v\)，令與其距離不超過\(d\)的節點加\(x\)，包括自己。
另一個是詢問節點\(v\)上的加和是多少。
考慮題面：
區間更新，單點查詢\(\longrightarrow\)顯然線段樹\(or\)樹狀陣列
考慮區間更新：
-
當距離\(d\)只侷限於節點\(v\)所在的鏈上，即\(depth[v]-d\)不包括根節點\(1\)時
只需要對於該鏈的節點進行簡單的區間更新即可。
-
當距離\(d\)不侷限於\(v\)所在鏈時，顯然，更新包括了根節點\(1\)
-
那麼， 我們可以根據樹的深度建一個樹狀陣列，我們只需要更新\([1,d-depth[v]]\)之間的節點即可
-
所以，我們的更新過程就成了這樣：
- 首先在鏈上區域性更新新\(x\)，範圍是\([depth[v]-d,depth[v]+d]\)
- 然後在深度樹狀陣列更新\(x\)，範圍是\([1,d-depth[v]]\)
- 然後在鏈上區域性更新\(-x\)，範圍是\([1,d-depth[v]]\)
-
再考慮單點查詢，答案顯然就變成了，深度樹狀陣列\(+\)鏈上樹狀陣列的和。
#if __cplusplus >= 201103L #pragma comment(linker, "/STACK:102400000,102400000") #pragma GCC optimize(3, "Ofast", "inline") #include <bits/stdc++.h> #else #include <algorithm> #include <bitset> #include <cmath> #include <iostream> #include <map> #include <string.h> #include <vector> #endif using namespace std; #define inf __INT_MAX__ #define enf INT_MIN #define INF LLONG_MAX #define ENF LLONG_MIN const int MAXN = 2e5 + 10; const double pi = acos(-1.0); const double eps = 1e-6; typedef long long ll; typedef unsigned long long ull; #define zhengfu(x) ((x > eps) - (x < -eps)) /* 每條鏈一個樹狀陣列,記錄鏈上不過根的更新 一個總樹狀陣列記錄過根的深度的更新 */ vector<int> e[MAXN]; struct node { vector<int> nd; int n; void init() { n = nd.size() - 1; } inline int lowbit(int x) //核心操作,取x這個數的二進位制最後一位1 { return -x & x; } inline void update(int x, int k) //單點修改 { for (; x <= n; x += lowbit(x)) nd[x] += k; } void update(int l, int r, int k) { //區間更新 update(l, k); if (r < n) update(r + 1, -k); } inline int query(int x) //求x的字首和 { int sum(0); for (; x > 0; x -= lowbit(x)) sum += nd[x]; return sum; } } tree[MAXN]; int depmax = enf; int dp[MAXN]; int nm[MAXN]; void dfs(int anc, int x, int dep, int num) { tree[num].nd.push_back(0); dp[x] = dep, nm[x] = num; if (e[x].size() == 1) { depmax = max(dep, depmax); tree[num].init(); return; } for (auto it : e[x]) if (it != anc) dfs(x, it, dep + 1, num); } int main() { int n, q; cin >> n >> q; for (int i = 1, x, y; i < n; i++) { cin >> x >> y; e[x].push_back(y); e[y].push_back(x); } for (int i = 1; i <= e[1].size(); i++) { tree[i + 1].nd.push_back(0); tree[i + 1].nd.push_back(0); dfs(1, e[1][i - 1], 2, i + 1); } dp[1] = nm[1] = 1; for (int i = 0; i <= depmax; i++) tree[1].nd.push_back(0); tree[1].init(); for (int i = 1, j, v, x, d; i <= q; i++) { cin >> j >> v; int pos = dp[v], tmp = nm[v]; if (!j) { scanf("%d%d", &x, &d); tree[tmp].update(max(2, pos - d), pos + d, x); if (pos - d < 2) { int len = d - pos + 2; tree[tmp].update(2, len, -x); tree[1].update(1, len, x); } } else { if (v == 1) cout << tree[1].query(pos) << endl; else cout << tree[tmp].query(pos) + tree[1].query(pos) << endl; } } }
六花保佑程式碼不\(WA\)
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