HDU 5151Sit sit sit

An interval DP problem, a bestcoder of the B question, thought for a long time did not think out.

Title Description: There are a total of n chairs, n students to sit in turn, at the same time meet 3 conditions can not sit down: 1, the chair is not the most left, not the right, 2, the chair is sitting around, 3, about the chair different colors
Ask the last n people can sit down, how many different situations.

The dp[i][j] indicates the number of rows [i,j] of the interval, (to see the other person not to understand the resolution, people are in order to determine the position of the seat, that is, in the interval [i,j] the sequence of seats. When [I,j]->[1,n], the type of DP is not the number of n individuals in order to decide the position of the type of sitting )

Transfer equation: dp[i][j] = SUM (dp[i][k-1]*dp[k+1][j]* c[j-i][i-k]) where v[k-1]==v[k+1], the last one is guaranteed to be the most lawful. The reason for multiplying the number of combinations is that in the j-i individuals the person who decides to sit in the previous interval (they are ordered after the election).

Code June:

#include <iostream> #include <cstdio> #include <cstring> #include <cmath>const int N = 100+10; const INT mod = 1e9+7;typedef long long ll;ll dp[n][n];int a[n];ll c[n][n];void init () {for (int i=0;i<=100;i++) C[i]    [0] = 1; for (int i=1;i<=100;i++) for (int j=1;j<=i;j++) c[i][j] = (C[i-1][j-1] + c[i-1][j])%mod;}    ll DFS (int i,int j) {if (i>j) return 0;    if (i==j) return dp[i][j]=1;    if (dp[i][j]!=-1) return dp[i][j];    ll ans = 0;    Ans = (Dfs (i+1,j) + DFS (i,j-1))%mod;        for (int k=i+1;k<=j-1;k++) {if (a[k-1]!=a[k+1]) continue;    Ans = (ans + dfs (i,k-1) *dfs (k+1,j)%mod*c[j-i][k-i]%mod)%mod; } return Dp[i][j] = ans;}    int main () {int n;    Init ();        while (scanf ("%d", &n)!=eof) {for (int i=1;i<=n;i++) scanf ("%d", &a[i]);        Memset (Dp,-1,sizeof (DP));        ll ans = DFS (1,n);    printf ("%lld\n", ans); } return 0;}

HDU 5151Sit sit sit