POJ 3034 Whac-a-Mole(三维dp+处理小技巧)

题目感觉不错，以后还得再做做。

题意：打地鼠，n*n的棋盘，每秒特定的格子上会有地鼠出没。给一个锤子，比如锤子前一次在(x1, y1)，现在移动到

(x2, y2)。则它可以打掉(x1, y1) -> (x2, y2)连线上的所有地鼠。锤子每次最多移动的距离为d。求最后最多能打到

多少个地鼠。

主要的思路就是用该点求出与该点相邻的点的状态。

dp[t][x][y]表示时间为t是，锤子在(x, y)坐标的最大值。枚举x,y，求t
+ 1时间到(xx, yy)的最优值， dis(x, y, xx,

 yy) <= d。在计算(x, y) -> (xx, yy)的时候不能一个单位一个单位的加，否则会超时。dx表示x -> xx的偏移量，

dy表示y -> yy的偏移量。

处理小技巧：求出dx, dy的最大公约数tp。 dx /= tp, dy /= tp。这样从(x, y) 到(xx, yy)时没次x
+= dx, y += dy;

Whac-a-Mole

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3179		Accepted: 961

Description



While visiting a traveling fun fair you suddenly have an urge to break the high score in the Whac-a-Mole game. The goal of the Whac-a-Mole game is to… well… whack moles. With a hammer. To make the job easier you have first consulted the fortune teller and
now you know the exact appearance patterns of the moles.

The moles appear out of holes occupying the n2 integer points (x, y) satisfying 0 ≤ x, y < n in a two-dimensional coordinate system. At each time step, some moles will appear and then disappear
again before the next time step. After the moles appear but before they disappear, you are able to move your hammer in a straight line to any position (x2, y2) that is at distance at most d from your current position
(x1, y1). For simplicity, we assume that you can only move your hammer to a point having integer coordinates. A mole is whacked if the center of the hole it appears out of is located on the line between (x1, y1)
and (x2, y2) (including the two endpoints). Every mole whacked earns you a point. When the game starts, before the first time step, you are able to place your hammer anywhere you see fit.

Input

The input consists of several test cases. Each test case starts with a line containing three integers n, d and m, where n and d are as described above, and m is the total number of moles that will appear
(1 ≤ n ≤ 20, 1 ≤ d ≤ 5, and 1 ≤ m ≤ 1000). Then follow m lines, each containing three integers x, y and t giving the position and time of the appearance of a mole (0 ≤ x, y < n and
1 ≤ t ≤ 10). No two moles will appear at the same place at the same time.

The input is ended with a test case where n = d = m = 0. This case should not be processed.

Output

For each test case output a single line containing a single integer, the maximum possible score achievable.

Sample Input

4 2 6
0 0 1
3 1 3
0 1 2
0 2 2
1 0 2
2 0 2
5 4 3
0 0 1
1 2 1
2 4 1
0 0 0

Sample Output

4
2

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-4
#define M 1000100
//#define LL __int64
#define LL long long
#define INF 0x7ffffff
#define PI 3.1415926535898

const int maxn = 63;

using namespace std;
int N;
int n, d, m;
struct node
{
int x, y;
double d;
} f[1010];
int dp[maxn][maxn][maxn];
int mp[maxn][maxn][maxn];

bool cmp(node a, node b)
{
return a.d < b.d;
}
void init()
{
N = 0;
for(int i = -5; i <= 5; i++)
{
for(int j = -5; j <= 5; j++)
{
if(i*i + j*j <= 25)
{
f
.x = i;
f
.y = j;
f[N++].d = sqrt((i*i*1.0)+(j*j*1.0));
}
}
}
sort(f, f+N, cmp);
}

int Sum(int t, int x, int y, int xx, int yy, int dx, int dy)
{
int cnt = 0;
while(1)
{
cnt += mp[t][xx][yy];
if(x == xx && y == yy)
break;
xx += dx;
yy += dy;
}
return cnt;
}

bool judge(int x, int y)
{
if(x < 0 || x >= n+2*d+1 || y < 0 || y >= n+2*d+1)
return false;
return true;
}

int main()
{
init();
while(cin >>n>>d>>m)
{
if(!n && !d && !m)
break;
int x1, y1, t;
int max_t = 0;
memset(dp, 0, sizeof(dp));
memset(mp, 0, sizeof(mp));
for(int i = 0; i < m; i++)
{
cin >>x1>>y1>>t;
mp[t][x1+d][y1+d] = 1;
max_t = max(max_t, t);
}
max_t++;
for(int tt = 1; tt < max_t; tt++)
{
for(int x = 0; x < 2*d+n+1; x++)
{
for(int y = 0; y < 2*d+n+1; y++)
{
for(int i = 0; i < N && double(d) >= f[i].d; i++)
{
int dx = f[i].x;
int dy = f[i].y;
int xx = x+dx;
int yy = y+dy;
if(!judge(xx, yy))
continue;
int ans;
if(i == 0)//(0,0)这种情况
ans = mp[tt][x][y];
else
{
int s = __gcd(abs(dx), abs(dy));
dx /= s;
dy /= s;
ans = Sum(tt, xx, yy, x, y, dx, dy);
}
dp[tt+1][xx][yy] = max(dp[tt][x][y]+ans, dp[tt+1][xx][yy]);
}
}
}
}
int Max = 0;
for(int i = d; i < d+n; i++)
for(int j = d; j < d+n; j++)
if(Max < dp[max_t][i][j])
Max = dp[max_t][i][j];
cout<<Max<<endl;
}
return 0;
}