省选前的任务计划
作者：互联网
因为太菜了，NOIP2020考挂了
要靠省选翻盘
可是现在压力大的一批
都没什么动力
就写一点计划 来监督自己
1.WC2021
2.string、ds尽可能熟练掌握
SAM+线段树合并
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read() {
int out = 0; bool flag = false;
register char cc = getchar();
while (cc < '0' || cc > '9') {
if (cc == '-') flag = true;
cc = getchar();
}
while (cc >= '0' && cc <= '9') {
out = (out << 3) + (out << 1) + (cc ^ 48);
cc = getchar();
}
return flag ? -out : out;
}
inline void write(ll x, char ch) {
if (x < 0) putchar('-'), x = -x;
if (x == 0) putchar('0');
else {
int num = 0; char cc[22];
while (x) cc[++num] = x % 10 + 48, x /= 10;
while (num) putchar(cc[num--]);
}
putchar(ch);
}
const int L = 5e5 + 10, N = 1e6 + 10;
int rt[N], ls[N * 40], rs[N * 40], sum[N * 40], tot, Left, Right;
void add(int &u, int l, int r, int pos) {
if (!u) u = ++tot;
if (l == r) sum[u]++;
else {
int mid = (l + r) >> 1;
if (pos <= mid) add(ls[u], l, mid, pos);
else add(rs[u], mid + 1, r, pos);
sum[u] = sum[ls[u]] + sum[rs[u]];
}
}
void merge(int &u, int x, int y) {
if (!x || !y) u = (x | y);
else {
u = ++tot;
merge(ls[u], ls[x], ls[y]), merge(rs[u], rs[x], rs[y]);
sum[u] = sum[ls[u]] + sum[rs[u]];
}
}
int find(int u, int l, int r) {
if (!u || !sum[u]) return -1;
if (l == r) return l;
int mid = (l + r) >> 1, pos = -1;
if (mid < Right) pos = find(rs[u], mid + 1, r);
if (pos == -1 && Left <= mid) pos = find(ls[u], l, mid);
return pos;
}
int n, m, tax[L], ord[N], ed[L]; char s[L], t[L]; ll ans;
struct sam {
int trans[N][26], fa[N], len[N], las, nodeid;
void init() {
for (int c = 0; c < 26; c++) trans[0][c] = 1;
for (int u = 1; u <= nodeid; u++)
for (int c = 0; c < 26; c++) trans[u][c] = 0;
las = nodeid = 1;
}
void ins(int c) {
int p = las, np = ++nodeid;
len[np] = len[p] + 1;
for ( ; p && !trans[p][c]; p = fa[p]) trans[p][c] = np;
if (!p) fa[np] = 1;
else {
int q = trans[p][c];
if (len[q] == len[p] + 1) fa[np] = q;
else {
int nq = ++nodeid; len[nq] = len[p] + 1;
for (int i = 0; i < 26; i++) trans[nq][i] = trans[q][i];
fa[nq] = fa[q], fa[q] = fa[np] = nq;
for ( ; p && trans[p][c] == q; p = fa[p]) trans[p][c] = nq;
}
}
las = np;
}
void build() {
for (int i = 1; i <= nodeid; i++) tax[len[i]]++;
for (int i = 1; i <= n; i++) tax[i] += tax[i - 1];
for (int i = 1; i <= nodeid; i++) ord[tax[len[i]]--] = i;
for (int i = nodeid; i; i--) {
int now = ord[i];
if (now != 1) merge(rt[fa[now]], rt[fa[now]], rt[now]);
}
}
} ss, st;
int main() {
scanf("%s", s + 1), n = strlen(s + 1);
ss.init();
for (int i = 1; i <= n; i++) ss.ins(s[i] - 'a'), ed[i] = ss.las;
for (int i = 1; i <= n; i++) add(rt[ed[i]], 1, n, i);
ss.build();
for (int T = read(); T; T--) {
scanf("%s", t + 1), m = strlen(t + 1);
Left = read(), Right = read(), ans = 0;
st.init();
int now = 1, l = 0;
for (int i = 1; i <= m; i++) {
int c = t[i] - 'a'; st.ins(c);
while (now && !ss.trans[now][c]) now = ss.fa[now], l = ss.len[now];
now = ss.trans[now][c], l += (now != 1);
//cout << now << " " << l << " ----------------------------- \n";
int pos = -1;
while (now > 1) {
pos = find(rt[now], 1, n);
//cout << pos << " \n";
if (pos == -1 || pos - ss.len[ss.fa[now]] < Left) now = ss.fa[now], l = ss.len[now];
else break;
}
//cout << pos << "," << now << "," << l << " ===================\n";
int mat = (pos == -1 ? 0 : min(l, pos - Left + 1));
ans += max(0, st.len[st.las] - max(mat, st.len[st.fa[st.las]]));
}
write(ans, '\n');
}
return 0;
}
View Code
splay+hash（自然溢出）
#include <bits/stdc++.h>
#define check(x) (tr[fa[x]][1] == x)
using namespace std;
inline int read() {
int out = 0;
bool flag = false;
register char cc = getchar();
while (cc < '0' || cc > '9') {
if (cc == '-') flag = true;
cc = getchar();
}
while (cc >= '0' && cc <= '9') {
out = (out << 3) + (out << 1) + (cc ^ 48);
cc = getchar();
}
return flag ? -out : out;
}
inline void write(int x, char ch) {
if (x < 0) putchar('-'), x = -x;
if (x == 0) putchar('0');
else {
int num = 0; char cc[22];
while (x) cc[++num] = x % 10 + 48, x /= 10;
while (num) putchar(cc[num--]);
}
putchar(ch);
}
const int N = 100010;
int n, q, rt, nodeid, tr[N][2], sze[N], fa[N];
unsigned int arr[N], val[N], p[N];
char ch;
void updata(int u) {
sze[u] = sze[tr[u][0]] + sze[tr[u][1]] + 1;
val[u] = val[tr[u][0]] * p[sze[tr[u][1]] + 1] + arr[u] * p[sze[tr[u][1]]] + val[tr[u][1]];
}
void rotate(int x) {
int y = fa[x], z = fa[y], opt1 = check(x), opt2 = check(y);
fa[x] = z, tr[z][opt2] = x;
fa[tr[x][opt1 ^ 1]] = y, tr[y][opt1] = tr[x][opt1 ^ 1];
fa[y] = x, tr[x][opt1 ^ 1] = y;
updata(y), updata(x);
}
void splay(int x, int goal) {
updata(x);
while (fa[x] != goal) {
int y = fa[x], z = fa[y];
if (z != goal) rotate(check(x) == check(y) ? y : x);
rotate(x);
}
if (!goal) rt = x;
}
int find(int k) {
if (k == 0) return 0;
int u = rt;
while (true) {
if (k > sze[tr[u][0]] + 1) k -= sze[tr[u][0]] + 1, u = tr[u][1];
else if (k <= sze[tr[u][0]]) u = tr[u][0];
else return u;
}
}
void ins(int k, int c) {
int u = find(k), par = u, v = tr[par][1]; while (v) par = v, v = tr[par][0];
v = ++nodeid, sze[v] = 1, arr[v] = c;
fa[v] = par, tr[par][par == u ? 1 : 0] = v, splay(v, 0);
}
void change(int k, int c) {
int u = find(k);
arr[u] = c, splay(u, 0);
}
unsigned int query(int l, int r) {
if (l == 1 && r == n) return val[rt];
if (l == 1) return splay(find(r + 1), 0), val[tr[rt][0]];
if (r == n) return splay(find(l - 1), 0), val[tr[rt][1]];
int x = find(l - 1), y = find(r + 1);
splay(x, 0), splay(y, x);
return val[tr[y][0]];
}
int main() {
p[0] = 1; for (int i = 1; i < N; i++) p[i] = p[i - 1] * 29u;
ch = getchar(); while (ch < 'a' || ch > 'z') ch = getchar();
while (ch >= 'a' && ch <= 'z') ins(n++, ch - 'a'), ch = getchar();
for (q = read(); q; q--) {
ch = getchar(); while (ch != 'Q' && ch != 'R' && ch != 'I') ch = getchar();
if (ch == 'Q') {
int x = read(), y = read();
int l = 0, r = n - max(x, y), ans = -1;
while (l <= r) {
int mid = (l + r) >> 1;
if (query(x, x + mid) == query(y, y + mid)) l = mid + 1, ans = mid;
else r = mid - 1;
}
write(ans + 1, '\n');
}
else if (ch == 'R') {
int x = read(); ch = getchar(); while (ch < 'a' || ch > 'z') ch = getchar();
change(x, ch - 'a');
}
else {
int x = read(); ch = getchar(); while (ch < 'a' || ch > 'z') ch = getchar();
ins(x, ch - 'a'); n++;
}
}
return 0;
}
View Code
3.DP、各种比赛、知识点补充、trick的积累
4.数学
5.图
计划待补充
标签：ch,cc,while,mid,省选前,int,任务,计划,getchar 来源： https://www.cnblogs.com/Urushibara-Ruka/p/14380880.html