atcoder abc 244
atcoder abc 244
D - swap hats
給定兩個 R,G,B
的排列
進行剛好 \(10^{18}\) 次操作,每一次選擇兩個交換
問最後能否相同
剛好 \(10^{18}\) 次
算出交換最少次數,判斷是否為偶數。
E - King Bombee
\(n\) 點 \(m\) 邊的簡單無向圖,給定 \(K,S,T\) 和 \(X\)
求滿足以下條件的路徑數 \(\;mod\;998244353\)
- 路徑 \(A\) 以長度為 \(K\) ,以 \(S\) 開使,\(T\) 結束,點 \(X\) 經過偶數次
簡單的計數,設 \(f_{i,j,0/1}\) 為第 \(i\) 步到 \(j\)
\(n\le 2000\)
#include <bits/stdc++.h> using namespace std; typedef unsigned long long uLL; typedef long double LD; typedef long long LL; typedef double db; const int N = 2005; const LL P = 998244353; int n, m, K, St, Ed, X, lst[N], Ecnt; struct Edge { int to, nxt; } e[N << 1]; LL f[N][N][2]; inline void Ae(int fr, int go) { e[++Ecnt] = (Edge){ go, lst[fr] }, lst[fr] = Ecnt; } int main() { scanf("%d%d%d%d%d%d", &n, &m, &K, &St, &Ed, &X); for (int i = 1, u, v; i <= m; i++) { scanf("%d%d", &u, &v); Ae(u, v), Ae(v, u); } f[0][St][0] = 1; for (int i = 1; i <= K; i++) for (int j = 1; j <= n; j++) { for (int k = lst[j], v, t; k; k = e[k].nxt) { v = e[k].to; t = j == X; (f[i][j][0] += f[i - 1][v][0 ^ t]) %= P; (f[i][j][1] += f[i - 1][v][1 ^ t]) %= P; } } printf("%lld", f[K][Ed][0]); }
F - Shortest Good Path
\(n\) 點 \(m\) 邊的簡單無向圖,
長度為 \(n\) 的序列 \(s_i\) 且 \(s_{i}\in\set{0,1}\) 能表示一條路徑 \(A\) 當點 \(i\) 出現次數 \(\mod 2=s_i\)
對所有 \(2^n\) 個序列,所表示的最短路徑的長度和
\(n\le 17\)
簡單的 bfs
,設 \(f_{s,i}\) 為序列 \(s\) 最後 \(i\) 的最短路
#include <bits/stdc++.h> using namespace std; typedef unsigned long long uLL; typedef long double LD; typedef long long LL; typedef double db; const int N = 20; int n, m, mx, a[N][N], l[N]; int f[1 << 18][N], res; LL ans; queue< pair<int, int> > Q; int main() { scanf("%d%d", &n, &m); for (int i = 1, u, v; i <= m; i++) { scanf("%d%d", &u, &v); u--, v--; a[u][++l[u]] = v; a[v][++l[v]] = u; } mx = (1 << n) - 1; for (int i = 0; i <= mx; i++) for (int j = 0; j < n; j++) f[i][j] = 1e9; for (int i = 0; i < n; i++) f[1 << i][i] = 1, Q.push(make_pair(i, 1 << i)); while (!Q.empty()) { int u = Q.front().first, s = Q.front().second; Q.pop(); for (int i = 1, v, ts; i <= l[u]; i++) { v = a[u][i], ts = s ^ (1 << v); if (f[s][u] + 1 >= f[ts][v]) continue; f[ts][v] = f[s][u] + 1, Q.push(make_pair(v, ts)); } } for (int i = 1; i <= mx; i++) { res = 1e9; for (int j = 0; j < n; j++) res = min(res, f[i][j]); if (res < 1e9) ans += res; } printf("%lld", ans); }
G - Construct Good Path
\(n\) 點 \(m\) 邊的簡單無向圖,和一個序列 \(s\) ,\(s_i\in\set{0,1}\)
構造一條路徑,使得點 \(i\) 的出現次數 \(\mod 2\) 為 \(s_i\)
輸出任意一個長度小於等於 \(4n\) 的路徑,可以證明存在
\(n,m\le 2\times 10^5\)
構造題
無向圖可以轉換為樹,結果不變
定義序列,\(A\) 如下:
- \(A_u=(u)+\sum_{(u,v)\in E} A_v+(u)+B_v\)
- \(+\) 表示序列的連線。
- 當 \(v\) 出現次數 \(mod\;2\ne s_i\) 時 \(B_v=(v,u)\) ,否則 \(B_v=()\)
\(A\) 滿足:
- \(A_u\) 從 \(u\) 開始,以 \(u\) 結束
- 對於 \(u\) 的所有兒子 \(v\) ,\(v\) 出現次數滿足條件
- 長度最多為 \(4n-3\) (菊花圖)
從任意 \(root\) 開始,最後判斷 \(root\) 出現次數
若不滿足條件,答案為 \(A_{root}+(v,root,v)\) ,\(v\) 為任意兒子。否則為 \(A_{root}\)
複雜度 \(O(n)\)
#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long uLL;
typedef long double LD;
typedef long long LL;
typedef double db;
const int N = 2e5 + 5;
int n, m, lst[N], Ecnt, vis[N], a[N];
int ans[N << 2], len;
char str[N];
struct Ed { int to, nxt; } e[N << 1];
inline void Ae(int fr, int go) {
e[++Ecnt] = (Ed){ go, lst[fr] }, lst[fr] = Ecnt;
}
void dfs(int u) {
vis[u] = 1, ans[++len] = u, a[u] ^= 1;
for (int i = lst[u], v; i; i = e[i].nxt)
if (!vis[v = e[i].to]) {
dfs(v), ans[++len] = u, a[u] ^= 1;
if (a[v]) {
ans[++len] = v, a[v] ^= 1;
ans[++len] = u, a[u] ^= 1;
}
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1, u, v; i <= m; i++) {
scanf("%d%d", &u, &v);
Ae(u, v), Ae(v, u);
}
scanf("%s", str + 1);
for (int i = 1; i <= n; i++) a[i] = str[i] - 48;
dfs(1);
if (a[1]) {
ans[++len] = e[lst[1]].to;
ans[++len] = a[1];
ans[++len] = e[lst[1]].to;
}
printf("%d\n", len);
for (int i = 1; i <= len; i++) printf("%d ", ans[i]);
}
Ex Linear Maximization
\(Q\) 次操作,每次給 \(x,y,a,b\) ,要求將 \((x,y)\) 插入集合 \(S\) ,再查詢 \(\max_{(x,y)\in S} ax+by\)
\(Q\le 2\times 10^5\)
[SDOI2014]向量集 弱化版。
其實是求點積最大,\(ans=ax+by\) ,變形得 \(y=-\dfrac{a}{b}x+\dfrac{ans}{b}\)
根據 \(b\) 的正負,求最大或最小截距,答案在上凸或下凸殼上
線段樹維護凸包,每次查詢對應凸包三分。
#include <bits/stdc++.h>
#define PB push_back
using namespace std;
typedef unsigned long long uLL;
typedef long double LD;
typedef long long LL;
typedef double db;
const int N = 500005;
int Ti, cnt;
LL ans;
struct vec {
LL x, y;
vec(LL _x = 0, LL _y = 0) : x(_x), y(_y) { }
bool operator < (vec a) const { return x ^ a.x ? x < a.x : y < a.y; }
vec operator - (vec a) { return vec(x - a.x, y - a.y); }
LL operator * (vec a) { return x * a.y - y * a.x; }
LL operator ^ (vec a) { return x * a.x + y * a.y; }
} a[N], b[N], c[N], s[N], now;
vector<vec> t[N << 2][2];
#define ls (rt << 1)
#define rs (rt << 1 | 1)
inline void bui(int rt) {
int n, m, top, len;
for (int o = 0; o <= 1; o++) {
n = m = len = top = 0;
for (vec x : t[ls][o]) a[++n] = x;
for (vec x : t[rs][o]) b[++m] = x;
for (int i = 1, j = 1; i <= n || j <= m; ) {
if (j > m || (i <= n && a[i] < b[j])) c[++len] = a[i++];
else c[++len] = b[j++];
}
for (int i = 1; i <= len; s[++top] = c[i++])
while (top >= 2 && ((c[i] - s[top]) * (s[top] - s[top - 1]) <= 0) ^ o) top--;
for (int i = 1; i <= top; i++) t[rt][o].PB(s[i]);
}
}
void ins(int x, int l, int r, int rt) {
if (l == r) return t[rt][0].PB(now), t[rt][1].PB(now);
register int mid = l + r >> 1;
if (x <= mid) ins(x, l, mid, ls);
else ins(x, mid + 1, r, rs);
if (x == r) bui(rt);
}
inline LL calc(int rt, vec a) {
int o = a.y <= 0;
int l = 1, r = t[rt][o].size();
LL res = -(1ll << 62), m1, m2;
while (r - l + 1 >= 4) {
m1 = (l + l + r) / 3, m2 = (l + r + r) / 3;
(t[rt][o][m1 - 1] ^ now) > (t[rt][o][m2 - 1] ^ now) ? r = m2 : l = m1;
}
for (int i = l; i <= r; i++) res = max(res, t[rt][o][i - 1] ^ a);
return res;
}
void ask(int ql, int qr, int l, int r, int rt) {
if (ql > r || l > qr) return;
if (ql <= l && r <= qr) { ans = max(ans, calc(rt, now)); return; }
register int mid = l + r >> 1;
ask(ql, qr, l, mid, ls), ask(ql, qr, mid + 1, r, rs);
}
#undef ls
#undef rs
int main() {
scanf("%d", &Ti);
int n = Ti;
for (LL i = 1, x, y, l, r; i <= Ti; i++) {
scanf("%lld%lld", &x, &y);
now.x = x, now.y = y;
ins(++cnt, 1, n, 1);
scanf("%lld%lld", &x, &y);
now.x = x, now.y = y;
ans = -(1ll << 62);
ask(1, i, 1, n, 1);
printf("%lld\n", ans);
}
}