Free pie HDU - 1176 basis dp

/ * Problem is there is a state transition equation,
As long as it is not a problem to launch the equation, the first
For the first gameboy for him the next second only
0-10 is moved in the eleven positions,
1 to 9 and for which he can move for the nine positions (assuming his present position x) to x + 1, or x-1, or x;
0 and two positions 10 can move only two positions,
Can dp [t] [x], t seconds see pie t-1 to obtain the next position (x + 1, x, x-1) when the second
Because the required maximum value of the entire process, thus seeking dp is accumulated and,
Therefore, the state transition equation is dp [i] [j] + = max (dp [i + 1] [j-1], max (dp [i + 1] [j], dp [i + 1] [j + 1 ]));
*/
#include<cstring>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<stdio.h>
using namespace std;
int dp[150000][20];
int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        int t,x,tt=0;
        memset(dp,0,sizeof(dp));
        for(int i=0; i<n; i++)
        {
            Scanf ( " % D% D " , X &, & T);
             // T is the time
             // X + is the position plus. 1. 1 
            DP [T] [X + . 1 ] ++ ;
            tt=max(tt,t);
        }
        int RES = 0 ;
         // only beginning position is determined, the time from the rear forward 
        for ( int I = TT- . 1 ; I> = 0 ; i-- )
             // Wang Qianmian pushed in this position pie number received, this position is received and coupled to the one second 
            for ( int J = 12 is ; J> = 0 ; J, )
                DP [I] [J] + = max (DP [I + . 1 ] [J- . 1 ], max (DP [I + . 1 ] [J], DP [I + . 1 ] [J + . 1 ]));
         // initial position 5 position, 
        COUT << DP [ 0 ] [ . 6 ] << endl;
    }
    return 0;
}