Dima and Magic Guitar CodeForces, dimacodeforces

Dima and Magic Guitar CodeForces, dimacodeforces

Dima and Magic Guitar CodeForces-certificate e

Question:

Http://blog.csdn.net/u011026968/article/details/38716425
Http://vawait.com/2013/11/codeforces-366e/
Http://www.cnblogs.com/jianglangcaijin/archive/2013/11/25/3441319.html

For any two adjacent numbers x and y in s, it is required to find any two points in the rectangle that are equal to x and y respectively, and then calculate their Manhattan distance, this question requires that the maximum distance of all the obtained Manhattan is the maximum. It is easy to think of, it should be to make one-to-point Manhattan distance the largest, other points can be arbitrary. That is, for all the adjacent two numbers in s, find the two points in the rectangle that are equal to the two numbers and have the largest distance between Manhattan.

The distance from Manhattan is equal to the following maximum:

(Xa-xb) + (ya-yb)
(Xa-xb)-(ya-yb)
-(Xa-xb) + (ya + yb)
-(Xa-xb)-(ya-yb)

That is, the maximum values of these values:

(Xa + ya)-(xb + yb)
(Xa-ya)-(xb-yb)
(-Xa + ya)-(-xb + yb)
(-Xa-ya)-(-xb-yb)

Therefore, the maximum Manhattan distance between the points a and B is required, that is, the maximum values of these four types. The maximum values of each type are the maximum and the minimum. That is, the maximum and minimum values of xa + ta, xa-ya,-xa + ya, and-xa-ya in all the points whose values are a are recorded respectively.

(This question is not about how to handle the problem when it is impossible, nor does it have such data .)

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 int a[100110]; 6 int max1[11][4],min1[11][4]; 7 int n,m,k,s,ans; 8 int main() 9 {10     int i,j,t;11     scanf("%d%d%d%d",&n,&m,&k,&s);12     memset(min1,0x3f,sizeof(min1));13     memset(max1,140,sizeof(max1));14     for(i=1;i<=n;i++)15         for(j=1;j<=m;j++)16         {17             scanf("%d",&t);18             max1[t][0]=max(max1[t][0],i+j);19             max1[t][1]=max(max1[t][1],i-j);20             max1[t][2]=max(max1[t][2],-i+j);21             max1[t][3]=max(max1[t][3],-i-j);22             min1[t][0]=min(min1[t][0],i+j);23             min1[t][1]=min(min1[t][1],i-j);24             min1[t][2]=min(min1[t][2],-i+j);25             min1[t][3]=min(min1[t][3],-i-j);26         }27     scanf("%d",&a[1]);28     for(i=2;i<=s;i++)29     {30         scanf("%d",&a[i]);31         for(j=0;j<=3;j++)32             ans=max(ans,max(max1[a[i-1]][j]-min1[a[i]][j],max1[a[i]][j]-min1[a[i-1]][j]));33     }34     printf("%d",ans);35     return 0;36 }