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Posted by Sd.Invol. 则如果i能走到x，$f[i][j]$要么从$f[x][j] cost$转移，表示不用机会，或者从$ max(f[x][j-1] cost)$转移，从中选最大值。直接一位一位DP复杂度是$O(9 nnm)$的，肯定不行，考虑到位最多只有$2 n$种，可以把相同的放在一块考虑，先DP出$g[i][j]$表示i位数字加起来得到j的有多少种，再dp就是$O(9 nn2 n)$了。至于这个$k$该如何去找，可以考虑从$[0,p)$枚举$x$每一位上的取值，这样取所有$x equiv i mod p$的位置，将指数的和加入答案，再找下一位。所以答案为$ Pi {i=0} {N-1}(K (K-1)i)$。 Posted by Sd.Invol. N的矩阵，并且满足当i j时,dij=0 当i=j时，dij等于vi的度数。 The Matrix Revolutions(ICPC Regionals 2014 : Asia - Shanghai Problem A). Posted by Sd.Invol. With tags Data Structure. 所以我们可以...

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Posted by Sd.Invol. 则如果i能走到x，$f[i][j]$要么从$f[x][j] cost$转移，表示不用机会，或者从$ max(f[x][j-1] cost)$转移，从中选最大值。直接一位一位DP复杂度是$O(9 nnm)$的，肯定不行，考虑到位最多只有$2 n$种，可以把相同的放在一块考虑，先DP出$g[i][j]$表示i位数字加起来得到j的有多少种，再dp就是$O(9 nn2 n)$了。至于这个$k$该如何去找，可以考虑从$[0,p)$枚举$x$每一位上的取值，这样取所有$x equiv i mod p$的位置，将指数的和加入答案，再找下一位。所以答案为$ Pi {i=0} {N-1}(K (K-1)i)$。 Posted by Sd.Invol. N的矩阵，并且满足当i j时,dij=0 当i=j时，dij等于vi的度数。 The Matrix Revolutions(ICPC Regionals 2014 : Asia - Shanghai Problem A). Posted by Sd.Invol. With tags Data Structure. 所以我们可以...

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7 如果机器人在某一方向有墙挡着，返回 1，因为往那个方向再怎么走都不会死

8 然后现在考虑某一个方向，如果走x步就会死，那显然步数序列中就不能有x个这个方向

9 四个方向的x 1加起来答案为n m 2

10 medium

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archive,messages,tags,with tags topcoder,editorial,easy,如果机器人在某一方向有墙挡着，返回 1，因为往那个方向再怎么走都不会死,然后现在考虑某一个方向，如果走x步就会死，那显然步数序列中就不能有x个这个方向,四个方向的x 1加起来答案为n m 2,medium,考虑1所在置换群子群的大小，答案为$ 1 n $的lcm,转移带环，用dijkstra即可,直接dp，看方案是否存在且唯一,利用期望线性性，枚举一段字符串考虑它们全是某种字符的概率,hard

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旅 | sd-invol.github.io Reviews

https://sd-invol.github.io

Posted by Sd.Invol. 则如果i能走到x，$f[i][j]$要么从$f[x][j] cost$转移，表示不用机会，或者从$ max(f[x][j-1] cost)$转移，从中选最大值。直接一位一位DP复杂度是$O(9 nnm)$的，肯定不行，考虑到位最多只有$2 n$种，可以把相同的放在一块考虑，先DP出$g[i][j]$表示i位数字加起来得到j的有多少种，再dp就是$O(9 nn2 n)$了。至于这个$k$该如何去找，可以考虑从$[0,p)$枚举$x$每一位上的取值，这样取所有$x equiv i mod p$的位置，将指数的和加入答案，再找下一位。所以答案为$ Pi {i=0} {N-1}(K (K-1)i)$。 Posted by Sd.Invol. N的矩阵，并且满足当i j时,dij=0 当i=j时，dij等于vi的度数。 The Matrix Revolutions(ICPC Regionals 2014 : Asia - Shanghai Problem A). Posted by Sd.Invol. With tags Data Structure. 所以我们可以...

INTERNAL PAGES

sd-invol.github.io

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TC SRM641~650

http://sd-invol.github.io/2015/01/10/Topcoder-SRM-641-650

TC SRM641 650 - 旅. 经过仔细的简化后，可以成为$O(n time m)$个点，$O({150} 2)$条边的网络，费用流就可以轻松跑了. Min(f[i] , g[i]) 1. 点$(x,y)$以点$(A,B)$做中心对称，会到$(2A-x,2B-y)$。点$(x,y)$以点$(A,B)$做中心对称，再以点$(C,D)$做中心对称，会到$(2(C-A) x,2(D-B) y)$。其实就是禁止了 TS 的出现，即如果有一条$u$到$v$代价为$w$的边，还需要加一条从$v$到$u$代价为无穷大的边，表示割掉这条边的代价为正无穷，从而禁止了 TS 的出现。非常有趣的题，首先枚举最大值取了多少个，一个要点是取最多的物品和最少的物品数量不会超过30，所以最大值取的数量应该是在${N over K} pm 50$左右。

2

北航第十届程序设计竞赛现场决赛题解

http://sd-invol.github.io/2014/12/22/BCPC2014-Onsite

F[0][8] = 1.0. F[i][j] = f[i - 1][j - k[i] * p[i] / 100.0. 8 leq j - k[i] 35$). F[i][j] = f[i - 1][j] * (100 - p[i]) / 100.0. 8 leq j 35$). 1 起点是(0,0);. 2 转折点是管道活动空间的端点,即(Xi, Li) 或 (Xi, Hi);. Dp[i] = min{ dp[j] dist(i, j) 0 = j i & (i, j)可以无阻碍连线}. 剩下就是判断( i, j )是否可以无阻碍相连。对于当前扫描到的点j，我们求出i和j 的斜率k，若k在Li，Ri之间，则 i，j 可以直接相连。如题目中给的初始的立方体就是由(0,0,0)点作为原点以及v1:(L,0,0), v2:(0,L,0)和v3:(0,0,L)这三个方向的向量决定的。 A,b,c)=(0,0,0) a*v1 b*v2 c*v3. A’,b’,c’)=p0 a*v1’ b*v2’ c*v3’. A 0-x a i-x-1 . a n-x-n.

3

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http://sd-invol.github.io/page3

HihoCoder Challenge 3 全图传送. Hdu 4789 ICPC Ranking.

4

hdu 4789 ICPC Ranking

http://sd-invol.github.io/2014/10/26/Chengdu-2013-I

Hdu 4789 ICPC Ranking - 旅. Hdu 4789 ICPC Ranking.

5

hihoCoder Challenge 3 全图传送

http://sd-invol.github.io/2014/10/29/hihoCoder-Challenge-3C

HihoCoder Challenge 3 全图传送 - 旅. HihoCoder Challenge 3 全图传送.

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Posted by Sd.Invol. 则如果i能走到x，$f[i][j]$要么从$f[x][j] cost$转移，表示不用机会，或者从$ max(f[x][j-1] cost)$转移，从中选最大值。直接一位一位DP复杂度是$O(9 nnm)$的，肯定不行，考虑到位最多只有$2 n$种，可以把相同的放在一块考虑，先DP出$g[i][j]$表示i位数字加起来得到j的有多少种，再dp就是$O(9 nn2 n)$了。至于这个$k$该如何去找，可以考虑从$[0,p)$枚举$x$每一位上的取值，这样取所有$x equiv i mod p$的位置，将指数的和加入答案，再找下一位。所以答案为$ Pi {i=0} {N-1}(K (K-1)i)$。 Posted by Sd.Invol. N的矩阵，并且满足当i j时,dij=0 当i=j时，dij等于vi的度数。 The Matrix Revolutions(ICPC Regionals 2014 : Asia - Shanghai Problem A). Posted by Sd.Invol. With tags Data Structure. 所以我们可以...

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