Poj3034 -- Whac-a-Mole (dp)

Poj3034 -- Whac-a-Mole (dp)

 

The n * n Square, the hammer moves d at most each time, and the hamster appears at (x, y) time at the t time to maintain a unit of time, two rats won't appear at the same time. The hammer can hit the mouse and ask how many rats can be smashed at most. (The initial hammer can be anywhere)

Dp [t] [I] [j]: the maximum number of hamster that a hammer can hit at the position (I, j) at the t moment.

State transition equation: Because the hammer moves d at most, the points (tx, ty) from (x-d, y-d) to (x + d, y + d) are enumerated ), this is the first point that (x, y) moves to (tx, ty), and then calculates the point in the range of d. Count the number of hit hamster

Note: you are willing to move the nodes outside the boundary.

 

20 5 41 0 10 1 10 5 21 6 20 0 0

Result 4

 

So moving all the points (5, 5) can avoid negative coordinates.

 

#include 
 
  #include 
  
   #include using namespace std ;int dp[12][41][41] ;int Map[12][41][41] ;void f(int n,int t,int x,int y,int d) {    int i , j , k , p , q , num ;    for(i = max(0,x-d) ; i <= min(n-1,x+d) ; i++) {        for(j = max(0,y-d) ; j <= min(n-1,y+d) ; j++) {            if( i == x && j == y ) continue ;            p = i - x ;            q = j - y ;            k = num = 0 ;            while( x+k*p >= 0 && x+k*p < n && y+k*q >= 0 && y+k*q < n && k*k*(q*q+p*p) <= d*d ) {                if( Map[t][x+k*p][y+k*q] ) num++ ;                dp[t][x][y] = max(dp[t][x][y],dp[t-1][x+k*p][y+k*q]+num) ;                k++ ;            }        }    }    return ;}int main() {    int n , d , m ;    int x , y , t ;    int i , j , max_t , ans ;    while( scanf(%d %d %d, &n, &d, &m) && n+d+m != 0 ) {        memset(dp,0,sizeof(dp)) ;        memset(Map,0,sizeof(Map)) ;        max_t = ans = 0 ;        while( m-- ) {            scanf(%d %d %d, &x, &y, &t) ;            Map[t][x+5][y+5] = 1 ;            max_t = max(max_t,t) ;        }        n += 12 ;        for(t = 1 ; t <= max_t ; t++) {            for(i = 0 ; i < n ; i++) {                for(j = 0 ; j < n ; j++) {                    f(n,t,i,j,d) ;                    ans = max(ans,dp[t][i][j]) ;                }            }        }        printf(%d, ans) ;    }    return 0 ;}