POJ 3034 Whac-a-Mole(DP)
理解了题意之后，这个题感觉状态转移还是挺好想的，实现起来确实有点繁琐，代码能力还有待加强，经过很长时间才发现bug。注意坐标可能是负的。
1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 int dp[12][40][41]; 8 bool o[12][41][41]; 9 int n,ans,d; 10 int gcd(int a,int b) 11 { 12 return b == 0?a:gcd(b,a%b); 13 } 14 int fun(int t,int x,int y,int tx,int ty,int ax,int ay) 15 { 16 int temp = 0; 17 while(x != ax||y != ay)//开始这里写成&& 18 { 19 if(o[t+1][x][y]) 20 temp ++; 21 x += tx; 22 y += ty; 23 } 24 return temp+o[t+1][x][y]; 25 } 26 void judge(int t,int x,int y) 27 { 28 int i,j,g; 29 for(i = x-d;i <= x+d;i ++) 30 { 31 for(j = y-d;j <= y+d;j ++) 32 { 33 if((i-x)*(i-x)+(j-y)*(j-y) > d*d) continue; 34 if(i < 0||j < 0) continue; 35 if(i > n+10||j > n+10) continue; 36 g = gcd(abs(i-x),abs(j-y)); 37 if(g == 0) g = 1; 38 int tx,ty; 39 tx = (i-x)/g; 40 ty = (j-y)/g; 41 dp[t+1][i][j] = max(dp[t+1][i][j],dp[t][x][y]+fun(t,x,y,tx,ty,i,j)); 42 ans = max(dp[t+1][i][j],ans); 43 } 44 } 45 } 46 int main() 47 { 48 int m,i,j,k,y,x,t; 49 while(scanf("%d%d%d",&n,&d,&m)!=EOF) 50 { 51 if(n == 0&&d == 0&&m == 0) break; 52 memset(dp,0,sizeof(dp)); 53 memset(o,0,sizeof(o)); 54 for(i = 1; i <= m; i ++) 55 { 56 scanf("%d%d%d",&x,&y,&t); 57 o[t][x+5][y+5] = 1; 58 } 59 ans = 0; 60 for(i = 0; i <= 10; i ++) 61 { 62 for(j = 0; j < n+10; j ++) 63 { 64 for(k = 0; k < n+10; k ++) 65 { 66 judge(i,j,k); 67 } 68 } 69 } 70 printf("%d\n",ans); 71 } 72 return 0; 73 }
