储能表
将n, m分解为二进制，考虑一个log(n)层的trie树，n会在这颗trie树上走出了一个路径，因为 行数 $ \le n$，所以在n的二进制路径上，每次往1走的时候，与m计算贡献，m同样处理，$O(Tlog(n)log(m))$
当然可以数位dp, $f_{i, n, m, k}$分别代表考虑到第i位，与n, m, k的大小关系，不是很好写，就写了上面的方法。
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (int i = a; i <= b; i++)
3 #define drep(i, a, b) for (int i = a; i >= b; i--)
4 #define REP(i, a, b) for (int i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define xx first
8 #define yy second
9 using namespace std;
10 typedef long long ll;
11 typedef pair<int, int> pii;
12 typedef unsigned long long ull;
13 const int inf = 0x3f3f3f3f;
14 const ll INF = 0x3f3f3f3f3f3f3f3fll;
15 template <typename T> void Max(T& a, T b) { if (b > a) a = b; }
16 //*********************************
17
18 ll n, m, k, mod;
19
20 ll POW(ll base, ll num) {
21 ll ret = 1;
22 while (num) {
23 if (num & 1) ret = ret * base % mod;
24 base = base * base % mod;
25 num >>= 1;
26 }
27 return ret;
28 }
29
30 ll calc(ll st, ll num) {
31 if (st < 0) { num += st; st = 0; }
32 if (num <= 0) return 0;
33 ll t1 = 2 * st + num - 1;
34 if (t1 & 1) num >>= 1;
35 else t1 >>= 1;
36 return t1 % mod * (num % mod) % mod;
37 }
38 ll solve(ll x, ll y, ll i, ll j) {
39 ll ret(0);
40 if (j > i) swap(i, j), swap(x, y);
41 ll z = x ^ (y ^ (y & ((1ll << i) - 1)));
42 ret = calc(z - k, 1ll << i);
43 return (ret * ((1ll << j) % mod)) % mod;
44 }
45 int main() {
46 /*
47 freopen("table.in", "r", stdin);
48 freopen("table.out", "w", stdout);
49 */
50 int T; scanf("%d", &T);
51 while (T--) {
52 scanf("%lld%lld%lld%lld", &n, &m, &k, &mod);
53
54 ll x(0);
55 ll ans(0);
56 drep(i, 62, 0) if (n >> i & 1) {
57 ll y(0);
58 drep(j, 62, 0) if (m >> j & 1) {
59 (ans += solve(x, y, i, j)) %= mod;
60 y |= 1ll << j;
61 }
62 x |= 1ll << i;
63 }
64
65 printf("%lld\n", ans);
66 }
67 }
table
数字配对
网络流，首先这个图不是奇环，这个图如果有环，那么结构应该是对称的？【我也不是非常清楚，求教。。。】
把费用和容量连成边，进行二分图染色，一直增广到费用加上当前费用小于0为止，别忘了考虑最后一次的流量。
#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define pb push_back
#define mp make_pair
#define xx first
#define yy second
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
template <typename T> void Max(T& a, T b) { if (b > a) a = b; }
//*********************************
const int maxn = 205;
struct Ed {
int u, v, c; ll w; int nx; Ed() {}
Ed(int _u, int _v, int _c, ll _w, int _nx) :
u(_u), v(_v), c(_c), w(_w), nx(_nx) {}
} E[maxn * maxn << 1];
int G[maxn], edtot = 1;
void addedge(int u, int v, int c, ll w) {
E[++edtot] = Ed(u, v, c, w, G[u]);
G[u] = edtot;
E[++edtot] = Ed(v, u, 0, -w, G[v]);
G[v] = edtot;
}
bool judge(int num) {
int sqr = sqrt(num);
if (num == 1) return 0;
rep(i, 2, sqr) if (num % i == 0) return 0;
return 1;
}
int s, t;
int knd[maxn];
int n;
int a[maxn], b[maxn], c[maxn];
void dfs(int x, int col) {
knd[x] = col;
for (int i = 1; i <= n; i++) if (knd[i] == -1 && (a[i] % a[x] == 0 && judge(a[i] / a[x]) || a[x] % a[i] == 0 && judge(a[x] / a[i]))) dfs(i, col ^ 1);
}
ll dis[maxn]; int pre[maxn];
bool spfa() {
rep(i, 1, n + 2) dis[i] = -INF;
static int que[maxn]; int qh(0), qt(0);
static int vis[maxn]; memset(vis, 0, sizeof(vis));
vis[que[++qt] = s] = 1; dis[s] = 0;
while (qh != qt) {
int x = que[++qh]; if (qh == maxn - 1) qh = 0;
for (int i = G[x]; i; i = E[i].nx) if (E[i].c && dis[E[i].v] < dis[x] + E[i].w) {
dis[E[i].v] = dis[x] + E[i].w; pre[E[i].v] = i;
if (!vis[E[i].v]) {
vis[que[++qt] = E[i].v] = 1;
if (qt == maxn - 1) qt = 0;
}
}
vis[x] = 0;
}
return dis[t] != -INF;
}
int ans; ll cost;
bool mcf() {
int flow = inf; ll nm(0);
for (int i = pre[t]; i; i = pre[E[i].u]) { flow = min(flow, E[i].c); nm += E[i].w; }
if (nm * flow + cost < 0) {
ans += cost / -nm;
return 0;
}
ans += flow, cost += nm * flow;
for (int i = pre[t]; i; i = pre[E[i].u]) E[i].c -= flow, E[i ^ 1].c += flow;
return 1;
}
int main() {
/*
freopen("pair.in", "r", stdin);
freopen("pair.out", "w", stdout);
*/
scanf("%d", &n);
rep(i, 1, n) scanf("%d", a + i);
rep(i, 1, n) scanf("%d", b + i);
rep(i, 1, n) scanf("%d", c + i);
memset(knd, -1, sizeof(knd));
rep(i, 1, n) if (knd[i] == -1) dfs(i, 0);
rep(i, 1, n) rep(j, 1, n) if (knd[i] == 1 && knd[j] == 0) {
if (a[i] % a[j] == 0 && judge(a[i] / a[j]) || a[j] % a[i] == 0 && judge(a[j] / a[i])) addedge(i, j, min(b[i], b[j]), 1ll * c[i] * c[j]);
}
s = n + 1, t = n + 2;
rep(i, 1, n) if (knd[i]) addedge(s, i, b[i], 0); else addedge(i, t, b[i], 0);
while (spfa()) if (!mcf()) break;
cout << ans << endl;
}
pair
游戏
树链剖分，考虑树链剖分时，每一次的修改均是连续的一段，我们对每一段维护标记$ax + b$x为当前点到这一段起点的距离。
那么假如有新的标记$Ax + B$，我们讨论$Ax + B <= ax + b$的情况，如果影响范围在mid的一边，就把新的标记向一边推，如果新的标记覆盖了多余一半，我们将新标记放在当前节点上，将原来的标记向不足mid的部分下推。
可以说是把标记永久化了。
复杂度为$O(mlog^{3}n)$，树链剖分两个log，标记一次最多下推log层。
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (int i = a; i <= b; i++)
3 #define drep(i, a, b) for (int i = a; i >= b; i--)
4 #define REP(i, a, b) for (int i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define xx first
8 #define yy second
9 using namespace std;
10 typedef long long ll;
11 typedef pair<int, int> pii;
12 typedef unsigned long long ull;
13 const int inf = 0x3f3f3f3f;
14 const ll INF = 0x3f3f3f3f3f3f3f3fll;
15 template <typename T> void Max(T& a, T b) { if (b > a) a = b; }
16 //*********************************
17
18 const int maxn = 100005;
19
20 struct Ed {
21 int u, v, w, nx; Ed() {}
22 Ed(int _u, int _v, int _w, int _nx) :
23 u(_u), v(_v), w(_w), nx(_nx) {}
24 } E[maxn << 1];
25 int G[maxn], edtot;
26
27 void addedge(int u, int v, int w) {
28 E[++edtot] = Ed(u, v, w, G[u]);
29 G[u] = edtot;
30 E[++edtot] = Ed(v, u, w, G[v]);
31 G[v] = edtot;
32 }
33
34 int pre[maxn], dep[maxn], son[maxn], sz[maxn];
35 ll dis[maxn];
36 int f[maxn][18];
37 int dfs(int x, int fa) {
38 pre[x] = fa, dep[x] = dep[fa] + 1; f[x][0] = fa;
39 for (int i = 1; i <= 17; i++) f[x][i] = f[f[x][i - 1]][i - 1];
40 son[x] = 0, sz[x] = 1;
41 for (int i = G[x]; i; i = E[i].nx) if (E[i].v != fa) {
42 dis[E[i].v] = dis[x] + E[i].w;
43 sz[x] += dfs(E[i].v, x);
44 if (sz[E[i].v] > sz[son[x]]) son[x] = E[i].v;
45 }
46 return sz[x];
47 }
48 int pos[maxn], inv[maxn], top[maxn], ndtot;
49 void repos(int x, int tp) {
50 top[x] = tp; pos[x] = ++ndtot; inv[ndtot] = x;
51 if (son[x]) repos(son[x], tp);
52 for (int i = G[x]; i; i = E[i].nx) if (E[i].v != son[x] && E[i].v != pre[x]) repos(E[i].v, E[i].v);
53 }
54
55 int LCA(int l, int r) {
56 if (dep[l] < dep[r]) swap(l, r);
57 int t = dep[l] - dep[r];
58 drep(i, 17, 0) if (t >> i & 1) l = f[l][i];
59 if (l == r) return l;
60 drep(i, 17, 0) if (f[l][i] != f[r][i]) l = f[l][i], r = f[r][i];
61 return f[l][0];
62 }
63
64 ll Min[maxn << 2];
65 ll A[maxn << 2], B[maxn << 2], Flag[maxn << 2];
66 void pushup(int o, int l, int r) {
67 if (Flag[o]) Min[o] = min(B[o], A[o] * (dis[inv[r]] - dis[inv[l]]) + B[o]);
68 if (l != r) Min[o] = min(Min[o], min(Min[o << 1], Min[o << 1 | 1]));
69 }
70 void giveFlag(int o, int l, int r, ll a, ll b) {
71 if (!Flag[o]) { A[o] = a, B[o] = b; Flag[o] = 1; }
72 else {
73 int mid = l + r >> 1; ll len = dis[inv[mid + 1]] - dis[inv[l]];
74 if (A[o] == a || l == r) B[o] = min(B[o], b);
75 else if (a > A[o] && B[o] < b);
76 else if (a > A[o] && B[o] >= b) {
77 ll d = (B[o] - b) / (a - A[o]);
78 if (d < len) giveFlag(o << 1, l, mid, a, b);
79 else {
80 swap(A[o], a), swap(B[o], b);
81 giveFlag(o << 1 | 1, mid + 1, r, a, b + a * len);
82 }
83 }
84 else if (a < A[o] && B[o] > b) swap(A[o], a), swap(B[o], b);
85 else if (a < A[o] && B[o] <= b) {
86 ll d = (B[o] - b - 1) / (a - A[o]) + 1;
87 if (d >= len) giveFlag(o << 1 | 1, mid + 1, r, a, b + a * len);
88 else {
89 swap(A[o], a), swap(B[o], b);
90 giveFlag(o << 1, l, mid, a, b);
91 }
92 }
93 }
94
95 pushup(o, l, r);
96 }
97
98 void modify(int o, int l, int r, int ql, int qr, ll a, ll b) {
99 if (ql <= l && r <= qr) {
100 giveFlag(o, l, r, a, b + a * (dis[inv[l]] - dis[inv[ql]]));
101 return;
102 }
103
104 int mid = l + r >> 1;
105 if (ql <= mid) modify(o << 1, l, mid, ql, qr, a, b);
106 if (qr > mid) modify(o << 1 | 1, mid + 1, r, ql, qr, a, b);
107 pushup(o, l, r);
108 }
109
110 int n;
111 void modify2(int tar, int x, ll a, ll b) {
112 while (top[tar] != top[x]) {
113 int F = top[x];
114 modify(1, 1, n, pos[F], pos[x], a, b + a * (dis[F] - dis[tar]));
115 x = pre[F];
116 }
117
118 modify(1, 1, n, pos[tar], pos[x], a, b);
119 }
120
121 void modify(int l, int r, ll a, ll b) {
122 int lca = LCA(l, r);
123 modify2(lca, l, -a, b + a * (dis[l] - dis[lca]));
124 modify2(lca, r, a, b + a * (dis[l] - dis[lca]));
125 }
126
127 ll query(int o, int l, int r, int ql, int qr) {
128 ll ret = INF;
129 if (Flag[o]) {
130 int L = max(l, ql), R = min(r, qr);
131 ret = min(A[o] * (dis[inv[L]] - dis[inv[l]]) + B[o], A[o] * (dis[inv[R]] - dis[inv[l]]) + B[o]);
132 }
133
134 if (ql <= l && r <= qr) return min(ret, Min[o]);
135 else {
136 int mid = l + r >> 1;
137 if (ql <= mid) ret = min(ret, query(o << 1, l, mid, ql, qr));
138 if (qr > mid) ret = min(ret, query(o << 1 | 1, mid + 1, r, ql, qr));
139 }
140 return ret;
141 }
142
143 ll solve(int l, int r) {
144 ll ret = INF;
145 while (top[l] != top[r]) {
146 if (dep[top[l]] < dep[top[r]]) swap(l, r);
147 ret = min(ret, query(1, 1, n, pos[top[l]], pos[l]));
148 l = pre[top[l]];
149 }
150 if (dep[l] < dep[r]) swap(l, r);
151 ret = min(ret, query(1, 1, n, pos[r], pos[l]));
152
153 return ret;
154 }
155
156 int main() {
157 /*
158 freopen("game.in", "r", stdin);
159 freopen("game.out", "w", stdout);
160 */
161 int m; scanf("%d%d", &n, &m);
162 REP(i, 1, n) {
163 int u, v, w; scanf("%d%d%d", &u, &v, &w);
164 addedge(u, v, w);
165 }
166
167 dfs(1, 1), repos(1, 1);
168
169 REP(i, 0, maxn << 2) Min[i] = 123456789123456789ll;
170
171 while (m--) {
172 int op; scanf("%d", &op);
173 if (op == 1) {
174 int u, v, a, b; scanf("%d%d%d%d", &u, &v, &a, &b);
175 modify(u, v, a, b);
176 }
177 else {
178 int s, t; scanf("%d%d", &s, &t);
179 printf("%lld\n", solve(s, t));
180 }
181 }
182
183 return 0;
184 }
game
生成魔咒
后缀数组，将字符串反过来，维护一个set，每次加入的字符串找到位置，计算贡献。
#include <bits/stdc++.h>
#define rep(i, a, b) for (register int i = a; i <= b; i++)
#define drep(i, a, b) for (register int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define pb push_back
#define mp make_pair
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
//********************************
const int maxn = 100005;
int a[maxn], hsh[maxn], cd;
int sa[maxn], t[maxn], t2[maxn], c[maxn], n;
void build_sa(int m) {
int *x = t, *y = t2;
rep(i, 0, m) c[i] = 0;
rep(i, 1, n) c[x[i] = a[i]]++;
rep(i, 1, m) c[i] += c[i - 1];
drep(i, n, 1) sa[c[x[i]]--] = i;
for (int k = 1; k <= n; k <<= 1) {
int p = 0;
rep(i, n - k + 1, n) y[++p] = i;
rep(i, 1, n) if (sa[i] > k) y[++p] = sa[i] - k;
rep(i, 0, m) c[i] = 0;
rep(i, 1, n) c[x[y[i]]]++;
rep(i, 1, m) c[i] += c[i - 1];
drep(i, n, 1) sa[c[x[y[i]]]--] = y[i];
swap(x, y);
p = 1; x[sa[1]] = 0;
rep(i, 2, n)
x[sa[i]] = y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k] ? p - 1 : p++;
if (p >= n) break;
m = p;
}
}
int rnk[maxn], height[maxn];
void build_height() {
int k(0);
rep(i, 1, n) rnk[sa[i]] = i;
rep(i, 1, n) {
if (k) k--;
int j = sa[rnk[i] - 1];
while (a[j + k] == a[i + k]) k++;
height[rnk[i]] = k;
}
}
int Min[maxn][20], Log[maxn];
void build_st() {
Log[1] = 0;
rep(i, 2, n) {
Log[i] = Log[i - 1];
if (1 << Log[i] + 1 == i) Log[i]++;
}
drep(i, n, 1) {
Min[i][0] = height[i];
for (int j = 1; i + (1 << j) - 1 <= n; j++)
Min[i][j] = min(Min[i][j - 1], Min[i + (1 << j - 1)][j - 1]);
}
}
int query(int l, int r) {
int len = r - l + 1, k = Log[len];
return min(Min[l][k], Min[r - (1 << k) + 1][k]);
}
int main() {
scanf("%d", &n);
rep(i, 1, n) scanf("%d", a + i), hsh[i] = a[i]; cd = n;
sort(hsh + 1, hsh + 1 + n), cd = unique(hsh + 1, hsh + 1 + cd) - hsh - 1;
rep(i, 1, n) a[i] = lower_bound(hsh + 1, hsh + 1 + cd, a[i]) - hsh;
rep(i, 1, n / 2) swap(a[i], a[n - i + 1]);
a[++n] = 0;
a[n + 1] = inf;
build_sa(cd);
build_height();
build_st();
ll ans(0);
set <int> S; S.insert(-inf), S.insert(inf);
drep(i, n - 1, 1) {
set<int> :: iterator H = S.lower_bound(rnk[i]), P = S.upper_bound(rnk[i]);
int A, B;
if (H != S.begin()) A = *--H; else A = -inf;
B = *P;
if (A != -inf) ans += query(min(rnk[i], A) + 1, max(rnk[i], A));
if (B != inf) ans += query(min(rnk[i], B) + 1, max(rnk[i], B));
if (A != -inf && B != inf) ans -= query(min(A, B) + 1, max(A, B));
printf("%lld\n", 1ll * (n - i) * (n - i - 1) / 2 + n - i - ans);
S.insert(rnk[i]);
}
return 0;
}
incantation
排列计数
组合数*错排数
$f_{n} = (n - 1)^{2}f_{n - 2} + (n - 1)(n - 2)f_{n - 3}$
1 #include <bits/stdc++.h>
2 #define rep(i, a, b) for (long long i = a; i <= b; i++)
3 #define drep(i, a, b) for (long long i = a; i >= b; i--)
4 #define REP(i, a, b) for (long long i = a; i < b; i++)
5 #define pb push_back
6 #define mp make_pair
7 #define xx first
8 #define yy second
9 using namespace std;
10 typedef long long ll;
11 typedef pair<int, int> pii;
12 const int inf = 0x3f3f3f3f;
13 const ll INF = 0x3f3f3f3f3f3f3f3fll;
14 template <typename T> void Max(T& a, T b) { if (b > a) a = b; }
15 //*********************************
16
17 const int maxn = 1000005;
18 const int mod = 1e9 + 7;
19
20 ll fac[maxn];
21 ll f[maxn];
22 void prepare() {
23 f[0] = 1, f[1] = 0, f[2] = 1;
24 rep(n, 3, 1000000) f[n] = ((n - 1) * (n - 1) % mod * f[n - 2] % mod + (n - 1) * (n - 2) % mod * f[n - 3] % mod) % mod;
25
26 fac[0] = 1;
27 rep(n, 1, 1000000) fac[n] = fac[n - 1] * n % mod;
28 }
29
30 ll POW(ll base, int num) {
31 ll ret = 1;
32 while (num) {
33 if (num & 1) ret = ret * base % mod;
34 base = base * base % mod;
35 num >>= 1;
36 }
37 return ret;
38 }
39 ll C(int n, int m) {
40 ll ret = fac[n];
41 ret = ret * POW(fac[n - m], mod - 2) % mod;
42 ret = ret * POW(fac[m], mod - 2) % mod;
43 return ret;
44 }
45 int main() {
46 prepare();
47 int T; scanf("%d", &T);
48 while (T--) {
49 int n, m; scanf("%d%d", &n, &m);
50 if (n < m) puts("0");
51 else printf("%lld\n", C(n, m) * f[n - m] % mod);
52 }
53 return 0;
54 }
permutation
征途
$s^{2} = \frac{\sum\limits_{i = 1}^{m}(x_{i} - \overline{x})^{2}}{m}$，将$(x_{i} - \overline{x})^{2}$展开，会发现我们其实要最小化$\sum\limits_{i = 1}^{m}x_{i}^{2}$
设$f_{i,j}$代表dp到第i位(包括i)，是第j段的最小平方和。
那么$f_{i,j} = min(f_{k,j - 1}) + (s_{i}-s_{k})^{2}$， s为前缀和。
固定j以后发现只与k有关，斜率优化即可。
#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define pb push_back
#define mp make_pair
#define xx first
#define yy second
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
template <typename T> void Max(T& a, T b) { if (b > a) a = b; }
//*********************************
const int maxn = 3005;
ll f[maxn], s[maxn], a[maxn];
inline ll p(ll num) { return num * num; }
struct HH {
ll x, y;
};
double k(HH A, HH B) { return (double)(A.y - B.y) / (A.x - B.x); }
int main() {
int n, m; scanf("%d%d", &n, &m);
rep(i, 1, n) scanf("%lld", a + i), s[i] = s[i - 1] + a[i];
rep(i, 1, n) f[i] = p(s[i]);
rep(j, 2, m) {
static HH que[maxn]; int qh = 1, qt(0);
rep(i, 1, n) {
HH t = (HH) {s[i], f[i] + p(s[i])};
while (qt > qh && k(que[qt], t) < k(que[qt - 1], que[qt])) qt--;
que[++qt] = t;
while (qt > qh && k(que[qh], que[qh + 1]) < 2 * s[i]) qh++;
f[i] = que[qh].y - p(que[qh].x) + p(s[i] - que[qh].x);
}
}
cout << m * f[n] - p(s[n]) << endl;
return 0;
}
journey
时间： 2024-10-07 06:54:10