HGOI 20200723
最后一天夏令营了（仅仅是提高组结束，后面还有冲省队。完蛋）
A：棋盘
送分，人均100
对于每行每列
如果数量 \(\ge k\) 那么肯定不行
如果数量 \(=0\) 直接答案加 \(1\)
否则对 \(a[l\)$r][j-k+1$\(j]\) 的这个矩形进行一个加操作，对旁边两个减
重复的加
然后求一遍前缀和
就过了，\(O(n^2)\)
#include <bits/stdc++.h>
using namespace std ;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int) a.size()
#define pb push_back
#define ll long long
#define fi first
#define se second
const int N = 2010;
int n, k, ans, Ans ;
char mp[N][N] ;
int res[N][N] ;
void add(int X1, int Y1, int X2, int Y2) {
++res[X1][Y1] ; --res[X1][Y2 + 1] ;
--res[X2 + 1][Y1] ; ++res[X2 + 1][Y2 + 1] ;
}
signed main() {
scanf("%d%d", &n, &k) ;
rep(i, 1, n) scanf("%s", mp[i] + 1) ;
rep(i, 1, n) {
int lef = 1, rig = n, label = 0 ;
rep(j, 1, n)
if (mp[i][j] == 'B') {
lef = max(lef, j - k + 1) ;
rig = min(rig, j) ;
label = true ;
}
if (!label) {
ans++ ;
continue ;
}
if (lef <= rig) add(max(i - k + 1 , 1), lef, i, rig);
}
rep(j, 1, n) {
int lef = 1, rig = n, label = 0 ;
rep(i, 1, n) {
if (mp[i][j] == 'B') {
lef = max(lef, i - k + 1) ;
rig = min(rig, i) ;
label = true ;
}
}
if (!label) {
ans++ ;
continue ;
}
if (lef <= rig) add(lef, max(1, j - k + 1), rig, j) ;
}
rep(i, 1, n)
rep(j, 1, n)
res[i][j] = res[i][j - 1] + res[i - 1][j] - res[i - 1][j - 1] + res[i][j] ;
rep(i, 1, n)
rep(j, 1, n)
Ans = max(Ans, ans + res[i][j]) ;
printf("%d\n" , Ans);
return 0;
}
B：序列
T2
一眼dp
\(dp[i] = dp[j]+ calc(j+1,i)\)
\(calc(l,r)\) 表示 \(l\)~\(r\) 的极差
查询可以用ST表O(1)解决
可得60分
80分：发现决策单调，记录一个 \(p\) 即可
莫名挂到70但告诉我只挂了2个点。无语
70分现场代码：
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int)a.size()
#define pb push_back
#define mp make_pair
#define ll long long
#define fi first
#define se second
#define Log2(n) (int) (log((double) (n)) / log(2.0))
const int mod = 1073741824 ;
const int N = 500020 ;
int n ;
int a[N], Max[N][25], Min[N][25] ;
ll dp[N] ;
namespace READ {
int x, y, z, m ;
int b[N], p[N], l[N], r[N] ;
void read() {
scanf("%d%d%d%d%d%d", &x, &y, &z, &b[1], &b[2], &m) ;
rep(i, 1, m) scanf("%d%d%d", &p[i], &l[i], &r[i]) ;
rep(i, 3, n) b[i] = (1ll * x * b[i - 1] + 1ll * y * b[i - 2] + z) % mod ;
p[0] = 0 ;
rep(i, 1, n) {
rep(j, p[i - 1] + 1, p[i]) a[j] = (b[j] % (r[i] - l[i] + 1) + l[i]) ;
}
}
}
void rmq() {
rep(i, 1, n) Max[i][0] = a[i] ;
rep(j, 1, Log2(n + 1))
rep(i, 1, n + 1 - (1 << j))
Max[i][j] = max(Max[i][j - 1], Max[i + (1 << (j - 1))][j - 1]) ;
rep(i, 1, n) Min[i][0] = a[i] ;
rep(j, 1, Log2(n + 1))
rep(i, 1, n + 1 - (1 << j))
Min[i][j] = min(Min[i][j - 1], Min[i + (1 << (j - 1))][j - 1]) ;
}
int ask(int l, int r) {
if (l > r) return 0 ;
int k = Log2(r - l + 1) ;
return max(Max[l][k], Max[r - (1 << k) + 1][k]) - min(Min[l][k], Min[r - (1 << k) + 1][k]) ;
}
signed main() {
scanf("%d", &n) ;
READ::read() ;
rep(i, 1, n) printf("%d ", a[i]) ; cout << endl ;
rmq() ;
if (n <= 5000) {
dp[0] = 0 ;
rep(i, 1, n) {
dp[i] = 0 ;
rep(j, 0, i - 1) dp[i] = max(dp[i], dp[j] + (ll) ask(j + 1, i)) ;
}
} else {
dp[0] = 0 ;
int p = 0 ;
rep(i, 1, n) {
dp[i] = 0 ;
rep(j, p, i - 1)
if (dp[i] < dp[j] + (ll) ask(j + 1, i)){
dp[i] = dp[j] + (ll) ask(j + 1, i) ;
p = j - 1 ;
}
}
}
printf("%lld\n", dp[n]) ;
return 0 ;
}
100分：
瓶颈在 \(ST\) 表的预处理上，显然没有 \(O(n),O(1)\) 的数据结构，因此只能换思路
发现一个区间，如果存在一个极值，那么断开可能变大
如果随便断开答案不可能变大
因此只需对极值的情况考虑，而且只需要记录前一个 \(dp\) 的值就行了，空间时间两边都优化
#include <bits/stdc++.h>
using namespace std ;
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
#define siz(a) (int) a.size()
#define pb push_back
#define mp make_pair
#define ll long long
#define fi first
#define se second
const int mod = 1073741824 ;
const int iinf = 0x3f3f3f3f ;
const ll linf = 0x3f3f3f3f3f3f3f3f ;
const int N = 10000020 ;
int n ;
int a[N] ;
ll ans[2] ;
namespace READ {
int x, y, z, m ;
int b[N], p[N], l[N], r[N] ;
void read() {
scanf("%d%d%d%d%d%d", &x, &y, &z, &b[1], &b[2], &m) ;
rep(i, 1, m) scanf("%d%d%d", &p[i], &l[i], &r[i]) ;
rep(i, 3, n) b[i] = (1ll * x * b[i - 1] + 1ll * y * b[i - 2] + z) % mod ;
p[0] = 0 ;
rep(i, 1, n) {
rep(j, p[i - 1] + 1, p[i]) a[j] = (b[j] % (r[i] - l[i] + 1) + l[i]) ;
}
}
}
signed main() {
scanf("%d", &n) ;
READ::read() ;
// rep(i, 1, n) printf("%d ", a[i]) ; cout << endl ;
ans[0] = -linf, ans[1] = 0 ;
int Max = 0, Min = iinf ;
int p = n ;
rep(i, 1, n) {
if (i > 1 && i < n && ((a[i - 1] < a[i] && a[i] >= a[i + 1])
|| (a[i - 1] > a[i] && a[i] <= a[i + 1]))) {
ll nxt[2] = {-linf, -linf} ;
nxt[0] = max(nxt[0], ans[0] + max(Max, a[p]) - min(Min, a[p])) ;
if (p != i - 1) nxt[0] = max(nxt[0], ans[1] + Max - Min) ;
else nxt[0] = max(nxt[0], ans[1]) ;
nxt[1] = max(nxt[1], ans[0] + max(Max, max(a[p], a[i])) - min(Min, min(a[p], a[i]))) ;
nxt[1] = max(nxt[1], ans[1] + max(Max, a[i]) - min(Min, a[i])) ;
ans[0] = nxt[0], ans[1] = nxt[1] ;
Max = 0, Min = iinf, p = i ;
} else {
Max = max(Max, a[i]) ;
Min = min(Min, a[i]) ;
}
// cout << Max << " " << Min << " " << p << " " << ans[0] << " " << ans[1] << endl ;
}
printf("%lld\n", max(ans[0] + max(a[p], Max) - min(a[p], Min), ans[1] + Max - Min)) ;
}
C：游戏
这个题太难了，打个暴力都贼难
暴力：
如果u和v不连通或u与x不连通或v与x不连通那么必定必能经过
这种情况特殊处理
然后发现 \((u,v)\) 方案数 \(=u\) 到 \(x\) 的 \(u\) 合法点数 * \(x\) 到 \(v\) 的 \(v\) 合法点数
合法难求，求不合法的
也就是在路径上有环的
\(Tarjan\) 缩点求出哪些点有环
计算即可
这样的算法在没有修改的情况下是很快的
但是有了修改就GG
然后之后的就一脸懵了
搬题解
std :
#include <bits/stdc++.h>
using namespace std;
int read() {
char c = getchar();
while (c != '-' && !isdigit(c)) c = getchar();
int neg = 0;
if (c == '-')
neg = 1, c = getchar();
int num = 0;
while (isdigit(c)) num = num * 10 + c - '0', c = getchar();
return neg ? -num : num;
}
int head[200001], ver[5000001], nxt[5000001], sz;
void addedge(int u, int v) { ver[++sz] = v, nxt[sz] = head[u], head[u] = sz; }
int backup[200001], backupsz;
int cnt, rt;
int dfn[200001], low[200001], now;
int stk[200001], top;
int vis[200001], scc[200001], size[200001], num;
void tarjan(int x) {
dfn[x] = low[x] = ++now;
stk[++top] = x;
vis[x] = 1;
for (int i = head[x]; i; i = nxt[i])
if (!vis[ver[i]])
tarjan(ver[i]), low[x] = min(low[x], low[ver[i]]);
else if (vis[ver[i]] == 1)
low[x] = min(low[x], dfn[ver[i]]);
if (dfn[x] == low[x]) {
++num;
while (stk[top] != x) {
vis[stk[top]] = 2;
scc[stk[top--]] = num;
}
vis[x] = 2, scc[x] = num, --top;
}
}
vector<int> e1[200001], e2[200001];
int first[100001], maxlen;
int x[100001], l[100001], r[100001], last;
int single[200001], t1[200001], t2[200001];
int n, x0;
void link(int x0, int l0, int r0, int len) {
if (l0 > r0)
return;
if (r0 - l0 + 1 < len) {
link(x0, l0, r0, len >> 1);
return;
}
int l1 = (l0 - 1) / len + 1;
if (len * (l1 - 1) + 1 == l0)
addedge(x0, first[len] + l1), link(x0, len * l1 + 1, r0, len);
else
link(x0, l0, len * l1, len), link(x0, len * l1 + 1, r0, len);
return;
}
long long check(int mid) {
memcpy(backup, head, sizeof(backup));
backupsz = sz;
for (int i = last + 1; i <= mid; i++) link(x[i], l[i], r[i], maxlen);
memset(vis, 0, sizeof(vis)), memset(size, 0, sizeof(size)), now = num = 0;
for (int i = 1; i <= cnt; i++)
if (!vis[i])
tarjan(i);
for (int i = 1; i <= num; i++) single[i] = 0;
for (int i = 1; i <= cnt; i++) e1[i].clear(), e2[i].clear();
for (int i = 1; i <= n; i++) single[scc[i]] = 1, ++size[scc[i]];
for (int i = 1; i <= cnt; i++)
for (int j = head[i]; j; j = nxt[j]) {
int u = scc[i], v = scc[ver[j]];
if (u != v)
e1[u].push_back(v), e2[v].push_back(u);
if (i <= n && ver[j] <= n && u == v)
single[u] = 0;
}
memset(t1, 0, sizeof(t1)), memset(t2, 0, sizeof(t2));
t1[scc[x0]] = t2[scc[x0]] = 1;
int cnt1 = 0, cnt2 = 0, c1 = n, c2 = n;
for (int i = 1; i <= num; i++)
if (t2[i]) {
c2 -= size[i];
for (int j = 0; j < e2[i].size(); j++) t2[e2[i][j]] = 1;
}
for (int i = 1; i <= num; i++)
if (single[i] || !size[i])
t2[i] = 0;
for (int i = 1; i <= num; i++)
if (t2[i]) {
cnt2 += size[i];
for (int j = 0; j < e2[i].size(); j++) t2[e2[i][j]] = 1;
}
for (int i = num; i >= 1; i--)
if (t1[i]) {
c1 -= size[i];
for (int j = 0; j < e1[i].size(); j++) t1[e1[i][j]] = 1;
}
for (int i = num; i >= 1; i--)
if (single[i] || !size[i])
t1[i] = 0;
for (int i = num; i >= 1; i--)
if (t1[i]) {
cnt1 += size[i];
for (int j = 0; j < e1[i].size(); j++) t1[e1[i][j]] = 1;
}
return 1ll * (n - cnt1) * (n - cnt2) + 1ll * cnt1 * c2 + 1ll * cnt2 * c1;
}
void reset() {
memcpy(head, backup, sizeof(head));
sz = backupsz;
}
int main() {
int m, q;
long long k;
cin >> n >> m >> x0 >> q >> k;
for (int i = 1; i <= m; i++) {
int u, v;
u = read(), v = read();
addedge(u, v);
}
cnt = n, maxlen = 1;
for (int i = 2; i <= n; i <<= 1) {
first[i] = cnt, maxlen = i;
for (int j = 1; j + i - 1 <= n; j += i) {
++cnt;
for (int k = j; k < j + i; k++) addedge(cnt, k);
}
}
for (int i = 1; i <= q; i++) x[i] = read(), l[i] = read(), r[i] = read();
int l0 = 0, r0 = q + 1;
while (l0 < r0) {
int mid = (l0 + r0 + 1) / 2;
if (check(mid - 1) >= k)
l0 = mid, last = mid - 1;
else
r0 = mid - 1, reset();
}
cout << l0 << endl;
}
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