【ACM】HDU 6639 Faraway 2019杭电多校第六场1006（枚举）

最新推荐文章于 2019-08-29 10:43:00 发布

wannatolive

最新推荐文章于 2019-08-29 10:43:00 发布

阅读量455

点赞数

CC 4.0 BY-SA版权

文章标签： ACM HDU

本文链接：https://blog.youkuaiyun.com/qq_41279172/article/details/98878030

本文详细解析了HDU 6639题目的算法解决方案，该题目要求根据士兵们提供的信息，计算可能的目标位置数量。通过使用模运算和坐标遍历的方法，文章提供了一段C++代码实现，展示了如何高效地解决这一复杂问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6639

Faraway

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 268 Accepted Submission(s): 137

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 (xe,ye), 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 (|xi−xe|+|yi−ye|)modki=ti.

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≤109) in the first line, denoting the number of soldiers and the upper bound of xe,ye.

For the next n lines, each line contains four integers xi,yi,ki,ti(0≤xi,yi≤m,2≤ki≤5,0≤ti<ki), 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 5

1 2 4 2

3 1 2 1

2 5

1 2 4 2

1 2 4 3

Sample Output

题目大意：

有n个形如 |x-a|+|y-b| mod k =t 的等式，给出a,b,k,t,求同时满足所有的等式的x,y有多少种

思路：

代码：

#include <bits/stdc++.h>
using namespace std;
int a[15],b[15],x[15],y[15],xe[15],ye[15];
int check(int xx,int yy,int n)
{
    for(int i=1;i<=n;i++)
    {
        if((abs(xx-xe[i])+abs(yy-ye[i]))%a[i]!=b[i])
            return 0;
    }
    return 1;
}
int cal(int l,int r)
{
    r-=l;
    if(r<0)
        return 0;
    else
        return r/60+1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        long long ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d%d%d%d",&xe[i],&ye[i],&a[i],&b[i]);
        x[0]=0,y[0]=0;//边界
        x[n+1]=m+1,y[n+1]=m+1;
        for(int i=1;i<=n;i++)
            x[i]=xe[i],y[i]=ye[i];
        sort(x,x+n+2);
        sort(y,y+n+2);
//        for(int i=0;i<=n+1;i++)
//            printf("%d\n",x[i]);
        for(int i=0;i<=n;i++)
        {
            for(int j=0;j<=n;j++)
            {
                for(int k=0;k<60;k++)
                {
                    for(int h=0;h<60;h++)
                    {
                        int xx=x[i]+k,yy=y[j]+h;
                        if(check(xx,yy,n))
                            ans+=(long long)cal(xx,x[i+1]-1)*cal(yy,y[j+1]-1);
                    }
                }
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}