[HDU6639]Faraway

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 524288/524288 K (Java/Others)

分数：2100，本来是个傻逼题但是怪自己太蠢了没想明白。
Problem Description

n soldiers are dispatched to somewhere in Byteland. These soldiers are going to set off now, but the target location is not so clear.

Assume the target location is at $(x_e,y_e)$, it is clear that xe,ye are both non-negative integers within [0,m]. For the i-th soldier, the only thing he knows is that $(|x_i−x_e|+|y_i−y_e|)$ $mod$ $k_i=t_i$.

To find the correct target location, these soldiers are working on the information they have now. Please write a program to figure out the number of possible target locations.

Input
The first line of the input contains an integer $T(1≤T≤10)$, denoting the number of test cases.

In each test case, there are two integers $n,m(1≤n≤10,1≤m≤10^9)$ in the first line, denoting the number of soldiers and the upper bound of $x_e,y_e$.

For the next $n$ lines, each line contains four integers $x_i,y_i,k_i,t_i(0≤x_i,y_i≤m,2≤k_i≤5,0≤t_i<k_i)$, denoting what each soldier knows.

Output
For each test case, print a single line containing an integer, denoting the number of possible target locations.

Sample Input

2
2 5
1 2 4 2
3 1 2 1
2 5
1 2 4 2
1 2 4 3

Sample Output

10
0

题意：
给定 $n$个方程组成的方程组，形如， $(|x_i−x_e|+|y_i−y_e|)$ $mod$ $k_i=t_i$
求 $(x_e,y_e)$ $(0<=x_e,y_e<=m)$的方案数。

题解：
由于是绝对值，所以每个绝对值讲解空间切成了四份，最多是 $n^2$份，那么我们将 $x_i$和 $y_i$排序之后枚举解空间，对于每个解空间求解的方案数。对于每个解空间，我们枚举 $x_e,y_e$对60(2,3,4,5的最小公倍数)取模的结果，然后带入方程组验证。如果可行，那么求这个解在当前解空间内有多少个不同的方案，注意解空间可能会出现边界相交的情况，所以我们需要将区间变为前闭后开的防止重复计算。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll lcm=60LL,ans;
int n;
ll m,x[14],y[14],k[14],t[14];
ll ax[14],ay[14];
void calc(ll Lx,ll Rx,ll Ly,ll Ry){
    for(ll dx=0;dx<lcm;dx++){
        for(ll dy=0;dy<lcm;dy++){
            bool flag=1;
            ll xe=dx+Lx,ye=Ly+dy;
            for(int i=1;i<=n-1;i++){
                if((abs(x[i]-xe)+abs(y[i]-ye))%k[i]!=t[i]){
                    flag=0;
                    break;
                }
            }
            if(!flag)continue;
            ll cax=Rx-xe-1,cay=Ry-ye-1;
            if(cax<0)cax=0;
            else cax=cax/lcm+1;
            if(cay<0)cay=0;
            else cay=cay/lcm+1;
            ans+=cax*cay;
        }
    }
}
int w33ha(){
    scanf("%d%lld",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lld%lld%lld%lld",&x[i],&y[i],&k[i],&t[i]);
        ax[i]=x[i];
        ay[i]=y[i];
    }
    ans=0;
    n++;
    ax[0]=0;ay[0]=0;
    ax[n]=m+1;ay[n]=m+1;
    sort(ax+1,ax+n+1);
    sort(ay+1,ay+n+1);
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(ax[i]>=ax[i+1])continue;
            if(ay[j]>=ay[j+1])continue;
            calc(ax[i],ax[i+1],ay[j],ay[j+1]);
        }
    }
    printf("%lld\n",ans);
    return 0;
}
int main(){
    int T;scanf("%d",&T);
    while(T--)w33ha();
    return 0;
}