Codeforces Global Round 1

Posted 2021-02-08 antiquality

 说什么呢，还是自己菜啊

【打表】C. Meaningless Operations

题目大意

求$f(a) = max_{0 < b < a}{gcd(a oplus b, a > & > b)}.$

$2 le a_i le 2^{25} - 1$

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题目分析

题挺有意思的。其中对于$2^x-1$打表。

附上有理有据的tutorial.

【思维题】E. Magic Stones

题目大意

每次可选择一个三元组$(c_{i-1},c_i,c_{i+1})$，将其变为$(c_{i-1},c_{i-1}-c_i+c_{i+1},c_{i+1})$.问序列$a[]$能否变为$b[]$

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题目分析

没救了，以前做过一道$(c_{i-1}+c_i,-c_i,c_i+c_{i+1})$的题，居然考试时候没看出来是一样思路。

考虑构造差分数组：发现$a[]$能够成为$b[]$当且仅当两数组的差分数组$a‘[1]=b‘[1]$且$a‘[2...n]$和$b‘[2...n]$排序后相等。

 1 #include<bits/stdc++.h>
 2 const int maxn = 100035;
 3 
 4 int n,a[maxn],b[maxn];
 5 
 6 int read()
 7 {
 8     char ch = getchar();
 9     int num = 0, fl = 1;
10     for (; !isdigit(ch); ch = getchar())
11         if (ch==‘-‘) fl = -1;
12     for (; isdigit(ch); ch = getchar())
13         num = (num<<1)+(num<<3)+ch-48;
14     return num*fl;
15 }
16 int main()
17 {
18     n = read();
19     for (int i=1; i<=n; i++) a[i] = read();
20     for (int i=1; i<=n; i++) b[i] = read();
21     for (int i=n; i>=1; i--)
22         a[i] -= a[i-1], b[i] -= b[i-1];
23     if (a[1]!=b[1]) puts("No");
24     else{
25         std::sort(a+1, a+n+1);
26         std::sort(b+1, b+n+1);
27         for (int i=1; i<=n; i++)
28             if (a[i]!=b[i]){
29                 puts("No");
30                 return 0;
31             }
32         puts("Yes");
33     }
34     return 0;
35 }

【数据结构】F. Nearest Leaf

题目大意

有一颗编号为dfs序的树。询问q次点v到编号为l...r之间的叶子的最短距离。$3 leq n leq 500,000, 1 leq q leq 500,000$

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题目分析

感觉和[LNOI2014]LCA一定程度有些异曲同工之妙。

建立线段树表示树上每一个节点到点$x$的距离。之所以这样表示，是因为方便在树上转移这颗线段树。考虑边$(u,v,w)$，那么转移到$v$时，所有$v$一侧的节点距离都应减去$w$；同时$u$一侧的节点距离都加上$w$。（出题人非常良心地提供了编号=dfs序的数据好评）因此发现只需要一个支持区间加和区间最小值的线段树即可。

  1 #include<bits/stdc++.h>
  2 typedef long long ll;
  3 const int maxn = 500035;
  4 const int maxq = 500035;
  5 const int maxm = 1000035;
  6 const ll INF = 1e16;
  7 
  8 struct QRs
  9 {
 10     int l,r,id;
 11     QRs(int a=0, int b=0, int c=0):l(a),r(b),id(c) {}
 12 };
 13 struct node
 14 {
 15     ll mn,add;
 16 }f[maxn<<2];
 17 struct Edge
 18 {
 19     int v,val;
 20     Edge(int a=0, int b=0):v(a),val(b) {}
 21 }edges[maxm];
 22 int n,q;
 23 int size[maxn];
 24 ll ans[maxq],val[maxn];
 25 int edgeTot,head[maxn],nxt[maxm],deg[maxn];
 26 std::vector<QRs> qr[maxn];
 27 
 28 int read()
 29 {
 30     char ch = getchar();
 31     int num = 0, fl = 1;
 32     for (; !isdigit(ch); ch = getchar())
 33         if (ch==‘-‘) fl = -1;
 34     for (; isdigit(ch); ch = getchar())
 35         num = (num<<1)+(num<<3)+ch-48;
 36     return num*fl;
 37 }
 38 void write(ll x){if (x/10) write(x/10);putchar(x%10+‘0‘);}
 39 void addedge(int u)
 40 {
 41     int v = read(), c = read();
 42     ++deg[u], ++deg[v];
 43     edges[++edgeTot] = Edge(v, c), nxt[edgeTot] = head[u], head[u] = edgeTot;
 44     edges[++edgeTot] = Edge(u, c), nxt[edgeTot] = head[v], head[v] = edgeTot;
 45 }
 46 void dfs1(int x, int fa, ll d)
 47 {
 48     size[x] = 1;
 49     if (x!=1&&deg[x]==1) val[x] = d;
 50     else val[x] = INF;
 51     for (int i=head[x]; i!=-1; i=nxt[i])
 52     {
 53         int v = edges[i].v;
 54         if (v!=fa) dfs1(v, x, d+edges[i].val), size[x] += size[v];
 55     }
 56 }
 57 void pushup(int rt)
 58 {
 59     f[rt].mn = std::min(f[rt<<1].mn, f[rt<<1|1].mn);
 60 }
 61 void pushdown(int rt)
 62 {
 63     if (f[rt].add){
 64         f[rt<<1].mn += f[rt].add, f[rt<<1|1].mn += f[rt].add;
 65         f[rt<<1].add += f[rt].add, f[rt<<1|1].add += f[rt].add;
 66         f[rt].add = 0;
 67     }
 68 }
 69 void build(int rt, int l, int r)
 70 {
 71     if (l==r){
 72         f[rt].mn = val[l];
 73         return;
 74     }
 75     int mid = (l+r)>>1;
 76     build(rt<<1, l, mid);
 77     build(rt<<1|1, mid+1, r);
 78     pushup(rt);
 79 }
 80 void update(int rt, int L, int R, int l, int r, int c)
 81 {
 82     if (L > R) return;
 83     if (L <= l&&r <= R){
 84         f[rt].mn += c, f[rt].add += c;
 85         return;
 86     }
 87     int mid = (l+r)>>1;
 88     pushdown(rt);
 89     if (L <= mid) update(rt<<1, L, R, l, mid, c);
 90     if (R > mid) update(rt<<1|1, L, R, mid+1, r, c);
 91     pushup(rt);
 92 }
 93 ll query(int rt, int L, int R, int l, int r)
 94 {
 95     if (L <= l&&r <= R) return f[rt].mn;
 96     ll mid = (l+r)>>1, ret = INF;
 97     pushdown(rt);
 98     if (L <= mid) ret = query(rt<<1, L, R, l, mid);
 99     if (R > mid) ret = std::min(ret, query(rt<<1|1, L, R, mid+1, r));
100     return ret;
101 }
102 void dfs2(int x, int fa)
103 {
104     for (int i=0, mx=qr[x].size(); i<mx; i++)
105         ans[qr[x][i].id] = query(1, qr[x][i].l, qr[x][i].r, 1, n);
106     for (int i=head[x]; i!=-1; i=nxt[i])
107     {
108         int v = edges[i].v;
109         if (v==fa) continue;
110         update(1, 1, n, 1, n, edges[i].val);　　　　　　//写的时候这里v,x弄混了调了二十几分钟
111         update(1, v, v+size[v]-1, 1, n, -2*edges[i].val);
112         dfs2(v, x);
113         update(1, 1, n, 1, n, -edges[i].val);
114         update(1, v, v+size[v]-1, 1, n, 2*edges[i].val);
115     }
116 }
117 int main()
118 {
119     memset(head, -1, sizeof head);
120     n = read(), q = read();
121     for (int i=1; i<n; i++) addedge(i+1);
122     for (int i=1; i<=q; i++)
123     {
124         int x = read(), l = read(), r = read();
125         qr[x].push_back(QRs(l, r, i));
126     }
127     dfs1(1, 0, 0);
128     build(1, 1, n);
129     dfs2(1, 0);
130     for (int i=1; i<=q; i++)
131         write(ans[i]), putchar(‘
‘);
132     return 0;
133 }

 

 

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