04 2018 档案
摘要："ref" cpp include include include include using namespace std; typedef long long ll; int n, q, a[100005], l[100005], r[100005], sta[100005], din, blc,
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摘要：好像复杂度来说不是正解……不加谜之优化（下叙）能被loj上的加强数据卡 cpp include include include include using namespace std; int n, m, hea[100005], cnt, uu[200005], fa[100005], siz[1
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摘要：这竟然是一道分块题…… cpp include include include include using namespace std; int n, m, q, blc, qnt, fa[50005], mxa[50005], mxb[50005], sz[50005], bcj, ans[500
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摘要：大傻逼trie树，更傻逼的是我这都没独立想出来，以后要少看题解，多多思考
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摘要：CQOI 板子大赛之 pollard rho cpp include include using namespace std; typedef long long ll; ll e, N, c, jzm, p, q, r, d, n; ll gcd(ll a, ll b){ return !b?a:
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摘要：k d tree 裸题……………… cpp include include include using namespace std; typedef long long ll; int n, k, rot, nowD, din; ll dui[205]; const ll oo=9e18; stru
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摘要：题面不符？…… cpp include include include using namespace std; typedef long long ll; int n, m, nowD, rot, din, kk; const ll oo=9e18; struct Point{ int d[2],
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摘要：k d tree + 重构的思想，就能卡过luogu和bzoj啦orz cpp include include include using namespace std; int n, m, rot, nowD, opt, uu, vv, ans, rub[1000005], din, cur; co
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摘要：分治+最小割 看到题解的第一句话是这个就秒懂了，然后乱七八糟的错误、越界、RE…… cpp include include include include include include using namespace std; int n, m, uu, vv, ww, hea[855], cnt
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摘要：学了斜率优化这题就能一气呵成地做出来啦qwqqwq cpp include include using namespace std; typedef long long ll; int n, m, x[3005], s[3005], l, r, dp[3005][3005], que[3005];
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摘要："大米饼写的太棒辣qwqqwq" cpp include include using namespace std; typedef long long ll; int n, l, r, que[1000005]; ll x[1000005], p[1000005], c[1000005], dp[1
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摘要：懒得写 cpp include include using namespace std; typedef long long longlivetheelder; int n, l, ll, rr, que[50005]; longlivetheelder w[50005], dp[50005]; s
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摘要：被 bs 了姿势水平……好好学习数学QAQQAQQAQ "ref" cpp include include include using namespace std; typedef long long ll; int n, m, pri[10000005], cnt, mu[10000005], q
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摘要："ref" cpp include include include using namespace std; int nn, n, rnk[60005], tmp[60005], cnt[60005], m=128, p, saa[60005]; char s[15], ss[60005]; voi
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摘要："ref" cpp include include include include using namespace std; typedef long long ll; int n, m, a[2005], b[2005]; ll dp[2005][2005]; ll f(){ memset(dp,
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摘要：越靠近叶子越优先删掉
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摘要：做了 [JSOI2008]Blue Mary开公司 以后发现这 tm 不就是个傻逼树剖+李超线段树吗，做了以后发现我才是傻逼……树剖竟然写错了……这题是我目前写过最长的代码了qwq cpp include include using namespace std; typedef long long
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摘要："ref" 好像叫什么李超线段树？……这篇不是太通用……
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摘要：跑最大费用最大流，注意到每次 spfa 出来的 cost 一定是越来越少的，啥时小于 $0$ 了就停了吧。 cpp include include include include using namespace std; typedef long long ll; int n, a[205], b[
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摘要："ref" "ref" 一个点就是一个数对 $(x,y)$。 记状态 $f[i][1/0][1/0][1/0]$ 和 $g[i][1/0][1/0][1/0]$，其中三个 $1/0$ 取值分别代表“$x$ 在前 $i$ 位卡满 $n$（的前 $i$ 位）/小于它”、“$y$ 在前 $i$ 位卡满 $
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摘要：定义区间是内部只含有乘号的区间。 对于区间左端点是 $l \geq 2$ 的情况，左端点前头是加号的情况和前头是减号的情况的个数是相同的。因此这些区间不对答案产生贡献。 所以区间左端点必定是 $1$。当 $r=n$ 时，这个区间产生一次贡献。当 $r include include using na
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摘要：矩阵树定理+模意义下整数高斯消元 cpp include include include include include using namespace std; typedef long long ll; typedef pair pii; int n, uu, vv, ww, w[21][21]
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摘要：欲学矩阵树定理~~必先自宫~~学习 "一些行列式的姿势" 然后 "做一道例题"
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摘要："orz" cpp include include using namespace std; typedef long long ll; int n, m, k, C[105][105], u[105], r[105], g[105]; const int mod=1e9+7; int ksm(in
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摘要："too hard" cpp include include using namespace std; int n, d, m, uu, vv, hea[500005], cnt, w[500005], f[500005][22], g[500005][22]; bool vis[500005];
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摘要：trie 树思想运用到主席树上 "orz" cpp include include using namespace std; int n, m, a[200005], cnt, rot[200005], bb, xx, uu, vv; struct Node{ int l, r, sum; }nd[
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摘要：神tm st表+并查集 cpp include include include using namespace std; typedef long long ll; int n, m, a, b, c, d, fa[19][100005]; const int mod=1000000007; int
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摘要：点分治+线性基 （为了这六个字窝调了一下午一晚上QAQ cpp include include include include include using namespace std; typedef long long ll; int n, uu, vv, m, hea[20005], cnt,
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摘要："……" cpp include include include include include using namespace std; typedef pair par; typedef long long ll; int s[510005][26], n, len, cnt, idx[5100
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摘要：真的是裸背包啊…… cpp include include using namespace std; typedef long long ll; int n, k[125], c[125], sum[125]; ll m, dp[125][240795]; int main(){ cin n m;
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摘要："ref" cpp include include include include using namespace std; typedef long long ll; int ua, ub, uc, ud, n, m, a[50005], chu[50005], blc, bel[50005],
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摘要："problem" 很妙啊……发现状态转移矩阵每一行都可以由上一行平移得到，每次只算第一行然后平移，$O(n^3)$ 就变成了 $O(n^2)$。 cpp include include using namespace std; int n, k; struct Matrix{ int num[20
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摘要：枚举右端点，然后看左端点合法情况。 先预处理每个颜色 $i$ 的最大出现位置 $max_i$ 和最小出现位置 $min_i$。对于枚举右端点在一个位置 $i$，凡是 $max_k i$ 的颜色 $k$ 都是不能要的。那么要满足右端点往右都合法，就要找出一个 $j i$ 这样的最大的 $j$。那么左端
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摘要：状压 dp。参考 "there" cpp include include include include using namespace std; typedef long long ll; int T, h, w, m, n, dp[1055][1055], qx[55], qy[55], dx,
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摘要：给出的图是一个二分图（显然……吗），一个图的最大团=其补图的最大独立集，因此二分图的最大独立集就是补图的最大团。 欲使补图最大团变大，则要最大独立集变大。二分图最大独立集=点数 最小点覆盖。最小点覆盖=最大匹配。 即搞掉哪些边使得最大匹配变小。即二分图的必经边。 二分图的必经边的判断：流量为 $1$
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摘要："there" cpp include include using namespace std; int n, m, mod, nxt[25], too[25][15]; char ss[25]; struct Matrix{ int num[25][25]; Matrix operator (co
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摘要：所有的下标从 $0$ 开始。 考虑枚举 $C$ （第一个加上负的等于第二个加上其绝对值）和第二个手链的偏移量 $p$。答案就是 $$\sum_{i=0}^{n 1}(x_i+C y_{(i+p) \bmod n})^2$$ 复制一遍 $y$ 数组就能去掉取模了，再展开就是 $$\sum_{i=0}^
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摘要：鬼畜线段树…… "Orz Capella" cpp include include using namespace std; int n, m, uu, vv, opt; double xx[100005], yy[100005], ss, tt, faq[4]; struct SGT{ doubl
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摘要："there" cpp include include using namespace std; int n, m, dfn[100005], idx, hea[100005], cnt, uu, vv, siz[100005], fa[100005][19]; int dep[100005], v
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摘要：裸dinic就跑过去了，哪用得着平面图最小割=最短路…… cpp include include include include using namespace std; int n, m, w, ss, tt, hea[1000005], cnt, lev[1000005], cur[100000
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摘要：狗题卡我精度…… "sol" cpp include include include include include include using namespace std; int n; struct Point{ double x, y; }pt[100005], O; double r; co
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摘要：lct裸题 cpp include include using namespace std; typedef long long ll; int n, m, uu, vv, val[100005], tagm[100005], taga[100005], sum[100005], siz[10000
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摘要：参考 "there" 和 "there" "题单" cpp include include using namespace std; int n, m, val[300005], ch[300005][2], sum[300005], fa[300005], uu, vv, opt; int rev
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摘要：prufer序列及其推论。参考 "there" cpp include include include include using namespace std; typedef long long ll; int n, d[155], cnt[155]; ll ans=1; vector vec[3
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摘要："there" cpp include include include include include include using namespace std; int n, m, aa[300005], bb[300005], lim=1, tmpcnt, rev[1100005]; const
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摘要："这里" cpp include include include include include using namespace std; typedef long long ll; int T, n, lim, c[100005], rev[270005], tmpcnt, a[100005];
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摘要：参考资料 "picks" "miskcoo" "menci" "胡小兔" "unname" "自为风月马前卒" 上面是FFT的，学完了就来看 "NTT" 吧 "原根" 例题：luogu3803 fft优化后模板 cpp include include include using namespace
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摘要：```cpp include include include using namespace std; typedef long long ll; int n, m, uu, vv, hea[100005], cnt, siz[100005], dfn[100005], loo[100005], i
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摘要：参考 "这里" cpp include include include using namespace std; typedef long long ll; int n, m, ise[100005], fa[100005][19], dep[100005], uu, vv, ww, hea[100
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摘要：拆点。 cpp include include include include using namespace std; int n, m, hea[2005], cnt, ss, tt, uu, vv, maxFlow, ans, lev[2005]; const int oo=0x3f3f3f3
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摘要：二分图的（不）可行边。 cpp include include include include include using namespace std; int n, m, T, ss, tt, hea[20005], cnt, uu[100005], vv[100005], cur[20005],
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摘要：最小路径可重点覆盖。先传递闭包，然后拆点，$n $最大匹配，看算法竞赛进阶指南。 cpp include include include using namespace std; int n, m, uu, vv, mat[205], ans; bool w[205][205], vis[205];
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摘要：```cpp include include include include include using namespace std; int n, m, tu, tv, v, xu[55], yu[55], xv[55], yv[55], hea[2705], cnt, cur[2705]; in
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摘要：day0 动身去太原。太原五中虽然挺小的但是很好看啊qwq（进门口一个“通天堂”（逃 试机。似乎看到了__stdcall！！然而没敢去认orz。linux选手似乎是9个。准考证（一整张A4纸）发下来了。不用建子目录？lemon？ 回酒店，吃饭。吃完饭还搞了一份鸡排（每到一个地方，就要吃那个地方的鸡排
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