POJ 3034 Whac-a-Mole(DP)

题目链接 

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。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int dp[12][40][41];
bool o[12][41][41];
int n,ans,d;
int gcd(int a,int b)
{
    return b == 0?a:gcd(b,a%b);
}
int fun(int t,int x,int y,int tx,int ty,int ax,int ay)
{
    int temp = 0;
    while(x != ax||y != ay)//开始这里写成&&
    {
        if(o[t+1][x][y])
        temp ++;
        x += tx;
        y += ty;
    }
    return temp+o[t+1][x][y];
}
void judge(int t,int x,int y)
{
    int i,j,g;
    for(i = x-d;i <= x+d;i ++)
    {
        for(j = y-d;j <= y+d;j ++)
        {
           if((i-x)*(i-x)+(j-y)*(j-y) > d*d) continue;
           if(i < 0||j < 0) continue;
           if(i > n+10||j > n+10) continue;
           g = gcd(abs(i-x),abs(j-y));
           if(g == 0) g = 1;
           int tx,ty;
           tx = (i-x)/g;
           ty = (j-y)/g;
           dp[t+1][i][j] = max(dp[t+1][i][j],dp[t][x][y]+fun(t,x,y,tx,ty,i,j));
           ans = max(dp[t+1][i][j],ans);
        }
    }
}
int main()
{
    int m,i,j,k,y,x,t;
    while(scanf("%d%d%d",&n,&d,&m)!=EOF)
    {
        if(n == 0&&d == 0&&m == 0) break;
        memset(dp,0,sizeof(dp));
        memset(o,0,sizeof(o));
        for(i = 1; i <= m; i ++)
        {
            scanf("%d%d%d",&x,&y,&t);
            o[t][x+5][y+5] = 1;
        }
        ans = 0;
        for(i = 0; i <= 10; i ++)
        {
            for(j = 0; j < n+10; j ++)
            {
                for(k = 0; k < n+10; k ++)
                {
                    judge(i,j,k);
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}