给我唐完了,Rank
A. 王国边缘
好像简单倍增就做完了。
由于昨天 T5 在我脑海中留下了挥之不去的印象,今天一上来看到这题就发现是一个内向基环树森林。然后被硬控硬控硬控,最后一个小点加一点优化就能过没调出来,挂 30pts,菜菜菜菜菜。
注意到倍增理论复杂度是 \(\mathcal{O(n\log n)}\) 的,而我们基环树玩家的理论复杂度可以做到完全线性,所以一下午直接调出了 8.2k 基环树做法。
中途插曲还不少,包括但不限于判完完整环之后暴力查询过了,轻松卡成 \(\mathcal{O(n^2)}\);然后在环上加了前缀和优化,不过链上还是 \(\mathcal{O(n)}\) 跑,又过了;尝试交题解,被告知要贴代码,遂加入离线 dfs 优化,至此达成 \(\mathcal{O(n)}\)。
点击查看代码
#include<bits/stdc++.h>
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{
char ch = getchar(); lx x = 0, f = 1;
for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
const int Ratio = 0;
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
ll n, m, q;
string s;
int las[N], fx[N], dfn[N], dt;
int hh[N], to[N], ne[N], cnt;
bool huan[N], vis[N], yz[N];
int bushu[N], daxiao[N], jieru[N];
ll len[N], w[N], dep[N], dis[N];
int newid[N], tot, st[N];
ll sumid[N];
ll ans[N];
namespace Wtask
{
short main()
{
m %= mod;
fo(i, 1, q)
{
ll qd = qr, k = qr;
qd %= mod;
k %= mod;
printf("%lld\n", (qd + k * m % mod) % mod);
}
return Ratio;
}
}
int h1[N], t1[N], n1[N], tnc;
ll w1[N];
vector<pii> que[N];
namespace Wlixian
{
int deep[N];
ll shendu[N];
inline void Wdfs(int u, int sd)
{
for(pii qq : que[u])
{
int x = qq.fi, y = qq.se;
ans[y] = (ans[y] + (shendu[u] - shendu[deep[sd - x]] + mod) % mod) % mod;
}
for(int i = h1[u]; i != -1; i = n1[i])
{
int v = t1[i];
if(huan[v]) continue;
deep[sd + 1] = v;
shendu[v] = (shendu[u] + w1[i]) % mod;
Wdfs(v, sd + 1);
}
}
inline void Wwork()
{
fo(i, 1, n) if(huan[i]) deep[1] = i, Wdfs(i, 1);
}
}
namespace Wisadel
{
int siz[N];
inline int Wfind(int x)
{
if(x == fx[x]) return x;
return fx[x] = Wfind(fx[x]);
}
inline void Wmerge(int u, int v)
{
int _ = Wfind(u), __ = Wfind(v);
if(siz[_] < siz[__]) fx[_] = __, siz[__] += siz[_];
else fx[__] = _, siz[_] += siz[__];
}
inline void Wadd(int u, int v, ll val)
{
to[++cnt] = v;
w[cnt] = val;
ne[cnt] = hh[u];
hh[u] = cnt;
}
inline void Wda(int u, int v, ll val)
{
t1[++tnc] = v;
w1[tnc] = val;
n1[tnc] = h1[u];
h1[u] = tnc;
}
int tag = 1e9;
inline void Wsearch(int u, int id, int tar)
{
newid[u] = id;
for(int i = hh[u]; i != -1; i = ne[i])
{
int v = to[i];
if(v == tar) return ;
sumid[id + 1] = sumid[id] + w[i];
Wsearch(v, ++tot, tar);
}
}
inline void Wdfs(int u)
{
yz[u] = 1;
vis[u] = 1;
dfn[u] = ++dt;
for(int i = hh[u]; i != -1; i = ne[i])
{
int v = to[i];
if(vis[v])
{
tag = dfn[v];
huan[u] = 1;
sumid[++tot] = 0;
st[Wfind(u)] = tot;
Wsearch(v, ++tot, v);
len[Wfind(u)] = (dep[u] - dep[v] + w[i] + mod) % mod;
}
else if(yz[v]) continue;
else dep[v] = (dep[u] + w[i]) % mod, Wdfs(v);
}
vis[u] = 0;
if(dfn[u] >= tag) jieru[u] = u, huan[u] = 1, daxiao[Wfind(u)]++;
if(dfn[u] == tag) tag = 1e9;
}
inline void Wdfss(int u)
{
yz[u] = 1;
for(int i = hh[u]; i != -1; i = ne[i])
{
int v = to[i];
if(huan[v]) dis[u] = w[i], bushu[u] = 1, jieru[u] = v;
else Wdfss(v), dis[u] = dis[v] + w[i], bushu[u] = bushu[v] + 1, jieru[u] = jieru[v];
}
}
short main()
{
freopen("kingdom.in", "r", stdin), freopen("kingdom.out", "w", stdout);
n = qr, m = qr, q = qr;
fo(i, 1, n) fx[i] = i, siz[i] = 1;
memset(h1, -1, sizeof h1);
memset(hh, -1, sizeof hh);
cin >> s;
s = " " + s;
int zaisuan = -1;
fo(i, 1, n) if(s[i] == '1'){zaisuan = i - 1; break;}
if(zaisuan == -1)
{
fo(i, 1, q)
{
ll qd = qr, k = qr;
qd %= mod;
k %= mod;
printf("%lld\n", (qd + k) % mod);
}
return Ratio;
}
int ok = 1;
fo(i, 1, n) if(s[i] != '1'){ok = 0; break;}
if(ok) return Wtask::main();
fo(i, zaisuan + 1, n) las[i] = s[i] == '1' ? i : las[i - 1];
fo(i, 1, zaisuan) las[i] = las[n] - n;
fo(i, 1, n)
{
if(i + m <= n)
{
ll dis;
if(las[i + m] <= i)
{
dis = 1;
Wmerge(i + 1, i);
Wadd(i, i + 1, dis);
Wda(i + 1, i, dis);
}
else
{
dis = las[i + m] - i;
Wmerge(las[i + m], i);
Wadd(i, las[i + m], dis);
Wda(las[i + m], i, dis);
}
}
else if(m < n)
{
int to = (m - (n - i)) % n;
int zto = to;
if(las[to] <= 0)
{
zto = las[to] + n;
if(zto <= i)
{
ll dis = 1;
zto = i + 1;
if(zto > n) zto = 1;
Wmerge(zto, i);
Wadd(i, zto, dis);
Wda(zto, i, dis);
}
else
{
ll dis = zto - i;
Wmerge(zto, i);
Wadd(i, zto, dis);
Wda(zto, i, dis);
}
}
else
{
ll dis = las[to] - i + n;
Wmerge(las[to], i);
Wadd(i, las[to], dis);
Wda(las[to], i, dis);
}
}
else
{
ll to = (m - (n - i)) % n;
ll tim = (m - (n - i)) / n;
if(!to) to += n, tim--;
int zto = las[to];
if(zto <= 0) zto += n, tim--;
ll dis = (n - i + zto + tim * n % mod) % mod;
Wmerge(zto, i);
Wadd(i, zto, dis);
Wda(zto, i, dis);
}
}
fo(i, 1, n) if(!yz[i]) Wdfs(i);
fill(yz + 1, yz + 1 + n, 0);
fo(i, 1, n) if(!yz[i] && !huan[i]) Wdfss(i);
fo(i, 1, q)
{
ll qidian = qr, cishu = qr, res = 0;
if(cishu == 0)
{
ans[i] = qidian % mod;
continue;
}
int dxqidian = qidian % n;
if(!dxqidian) dxqidian = n;
int belong = Wfind(dxqidian);
res = qidian % mod;
if(cishu < bushu[dxqidian]) que[dxqidian].P_B(M_P(cishu, i));
else
{
cishu -= bushu[dxqidian];
ll timesl = cishu / daxiao[belong];
cishu -= timesl * daxiao[belong];
timesl %= mod;
res = (res + timesl * len[belong] % mod) % mod;
res = (res + dis[dxqidian]) % mod;
if(cishu)
{
if(newid[jieru[dxqidian]] - st[belong] + cishu <= daxiao[belong]) res = ((res + sumid[newid[jieru[dxqidian]] + cishu]) % mod - sumid[newid[jieru[dxqidian]]] + mod) % mod;
else
{
int zcc = newid[jieru[dxqidian]] + cishu - daxiao[belong];
res = (res + ((sumid[zcc] - sumid[newid[jieru[dxqidian]]] + mod) % mod + len[belong]) % mod) % mod;
}
}
}
ans[i] = res;
}
Wlixian::Wwork();
fo(i, 1, q) printf("%lld\n", ans[i]);
return Ratio;
}
}
signed main(){return Wisadel::main();}
// 佳墙坂诶迦币等渔塞
B. 买东西题
反悔贪心板子。
其实做完两个特殊性质就把正解启发得差不多了,结合一下就过了。
考虑物品按 \(a_i\) 升序,券按 \(w_i\) 升序,双指针扫,这样能够保证当前券可被之前的所有物品使用,方便反悔操作。
发现 \(a_i-b_i\) 实质上就是优惠价给的满减值,我们维护一个堆存放当前可选的满减值,每次操作只有取或不取。若从堆中取了满减,那么将 \(a_i-b_i\) 加入堆。这个操作还挺好理解的,如果 \(a_i-b_i\) 在之后某次抉择中成了最优决策,那么将 \(i\) 用的券给当前物品用,然后 \(i\) 直接用优惠价买一定是优的。
然后就做完了,复杂度 \(\mathcal{O(n\log n)}\)。
点击查看代码
#include<bits/stdc++.h>
#define fo(x, y, z) for(int (x) = (y); (x) <= (z); (x)++)
#define fu(x, y, z) for(int (x) = (y); (x) >= (z); (x)--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define lx ll
inline lx qr()
{
char ch = getchar(); lx x = 0, f = 1;
for(; ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
for(; ch >= '0' && ch <= '9'; ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ 48);
return x * f;
}
#undef lx
#define qr qr()
#define pii pair<int, int>
#define ppp pair<pii, pii>
#define fi first
#define se second
#define M_P(x, y) make_pair(x, y)
#define P_B(x) push_back(x)
const int Ratio = 0;
const int N = 1e6 + 5;
const int mod = 1e9 + 7;
int n, m;
struct good{int a, b;} g[N];
struct rmm{int w, v;} c[N];
priority_queue<int, vector<int>, less<int> > q;
namespace Wisadel
{
short main()
{
freopen("buy.in", "r", stdin), freopen("buy.out", "w", stdout);
n = qr, m = qr;
fo(i, 1, n) g[i].a = qr, g[i].b = qr;
fo(i, 1, m) c[i].w = qr, c[i].v = qr;
sort(g + 1, g + 1 + n, [](good A, good B){return A.a < B.a;});
sort(c + 1, c + 1 + m, [](rmm A, rmm B){return A.w < B.w;});
int j = 1; ll ans = 0;
fo(i, 1, n)
{
while(j <= m && c[j].w <= g[i].a) q.push(c[j++].v);
if(q.empty() || g[i].a - g[i].b >= q.top()) ans += g[i].b;
else
{
ans += g[i].a - q.top(); q.pop();
q.push(g[i].a - g[i].b);
}
}
printf("%lld\n", ans);
return Ratio;
}
}
signed main(){return Wisadel::main();}
// 佳墙坂诶迦币等渔塞
C. IMAWANOKIWA (Construction ver.)
大粪掏题。
D. 魔法少女们
魔法题。
末
为什么只写了两个题就发,你是不是想摆烂?
因为 T1 打了一个下午,T3 看起来就要改很长时间,怕来不及记录就先发了。
谁问你了?
?
开场直接被隔壁发现是原题,甚至三天前的比赛直接搬。然后因为没人做过所以正常进行,输麻了。
状态?感觉不是,可能主要因为 T1 打了太久被影响了心态。其实 T1 当时已经打了正解的 \(\frac{3}{4}\) 左右了,考虑当时只剩一个半小时就没继续看,结果错误是弱智取模,输麻了。
T2 感觉也不该是想不出来的,不过丁真
T3 T4 压根不想看,感觉 T4 讲的优化是那种学了一辈子用不上几次的那种,粉兔讲题更是直言我卡你那个了就不会再卡这个,有点玄学。
完结撒花~
没有