[Bzoj 1037] [ZJOI2008] Birthday party (DP)

Title Link: http://www.lydsy.com:808/JudgeOnline/problem.php?id=1037

The topic is too god, to kneel ....

This question completely unexpectedly DP's idea, on-line of the solution to turn over to not a few reliable, eventually finally found a reliable point of the key to understand ...

First, let's simplify the subject by giving you n 0,m 1 to arrange them so that the difference between the number of 0 and 1 in any successive sequence is less than or equal to K.

Then we use f[a][b][c][d] to indicate that there is currently a 0,b of 1, 0:1 maximum number of K, 1:0 of the maximum number of scenarios of t

DP process, each added a new number, we can either add 0, also can add 1.

Plus 0 is the premise that there are 0 of the remaining numbers, and the current 0:1 is less than K

Plus 1 is the premise that there are 0 of the remaining numbers, and the current 1:0 is less than K

Finally, we will count the answers and add up the number of all legal schemes for more than 0:1 I, 1:0 J.

This is the whole process of the problem.

#include <iostream> #include <stdio.h> #include <stdlib.h> #include <string.h> #include < algorithm> #define MAXN 155#define maxm 22#define MOD 12345678using namespace Std;int F[MAXN][MAXN][MAXM][MAXM];    F[a][b][c][d]=a a 0,b of 1, 0:1 max of K, 1:0 maximum number of T scheme number int N,m,k,ans;int main () {scanf ("%d%d%d", &n,&m,&k);    F[0][0][0][0]=1; for (int i=0;i<=n;i++) for (int j=0;j<=m;j++) for (int. k=0;k<=k;k++) for (int t=0;t& lt;=k;t++) {if (F[i][j][k][t]) {if (I<n&amp ; &k<k)//New addition of a 0 F[i+1][j][k+1][max (t-1,0)]= (F[i+1][j][k+1][max (t-1,0)]+f[i][j][k][t])%mo                        D if (j<m&&t<k)//New Add a 1 F[i][j+1][max (k-1,0)][t+1]= (F[i][j+1][max (k-1,0)][t+1]+f[i] [j] K [T])                    %mod;    }} int ans=0; for (int i=0;i<=k;i++)//enumeration 0:1 up to many I for (intj=0;j<=k;j++)//enumeration 1:0 up to a maximum of J ans= (Ans+f[n][m][i][j])%mod;    printf ("%d\n", ans); return 0;}



[Bzoj 1037] [ZJOI2008] Birthday party (DP)