HDU 5151 sit sit sit zone DP

Topic Links:

http://acm.hdu.edu.cn/showproblem.php?pid=5151

Exercises

There are n chairs, numbered 1 to N.

Now there are n students, numbered 1 to N, starting from the first classmate to choose the seat to sit, and this classmate can not sit at the same time meet the following three conditions of the Chair.

1. There are adjacent seats around

2, around the adjacent seats are empty.

3, the left and right sides of the seat color is different.

Ask how many kinds of sitting methods are in total.

Exercises

DP[I][J] Indicates the number of classes of chairs numbered I to J.

Now the first person to start the seat chooses the T-seat, then Dp[i][j]+=dp[i][t-1]*dp[t+1][j]*c[j-i][t-i], where the combination number C represents the number of combinations that arrange the t-i individual on the left.

Code:

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5 using namespacestd;6 7typedefLong LongLL;8 Const intMoD = 1e9 +7;9 Const intMAXN =111;Ten  One intN; A intARR[MAXN]; - LL DP[MAXN][MAXN],C[MAXN][MAXN]; -  the voidPre () { -c[0][0] =1; -      for(inti =1; i < MAXN; i++) { -c[i][0] =1; +          for(intj =1; J <= I; J + +) { -C[I][J] = (c[i-1][j-1] + c[i-1][J])%MoD; +         } A     } at } -  -LL DFS (intIintj) { -     if(i = = j)returnDP[I][J] =1; -     if(i > J)return 0; -     if(Dp[i][j]! =-1)returnDp[i][j]; inLL &ret = dp[i][j] =0; -ret= (Ret+dfs (i +1, j) +dfs (I, J-1))%MoD; to      for(intMID = i +1; Mid <= J-1; mid++) { +         if(Arr[mid-1] ^ Arr[mid +1])Continue; -ret = (Ret+dfs (i, Mid-1) *dfs (Mid +1, j)%mod*c[j-i][mid-i])%MoD; the     } *     returnret; $ }Panax Notoginseng  - voidinit () { theMemset (DP,-1,sizeof(DP)); + } A  the intMain () { + pre (); -      while(SCANF ("%d", &n) = =1&&N) { $ init (); $          for(inti =1; I <= N; i++) scanf ("%d", arr +i); -LL Ans=dfs (1, n); -printf"%lld\n", ans); the     } -     return 0;Wuyi}

HDU 5151 sit sit sit zone dp