hdu 5626 Clarke and points
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Clarke and points
Problem Description
The Manhattan Distance between point A(XA,YA) and B(XB,YB) is |XA - XB| + |Xb - YB|;
the coordinate of each point is generated by the followed code.
Input
long long seed;
inline long long rand(long long l, long long r) {
static long long mo=1e9+7, g=78125;
return l+((seed*=g)%=mo)%(r-l+1);
}
cin >> n >> seed;
for (int i = 0; i < n; i++)
x[i] = rand(-1000000000, 1000000000),
y[i] = rand(-1000000000, 1000000000);
long long seed;
inline long long rand(long long l, long long r) {
static long long mo=1e9+7, g=78125;
return l+((seed*=g)%=mo)%(r-l+1);
}
cin >> n >> seed;
for (int i = 0; i < n; i++)
x[i] = rand(-1000000000, 1000000000),
y[i] = rand(-1000000000, 1000000000);
Output
For each test case, print a line with an integer represented the maximum distance.
Sample Input
2
3 233
5 332
Sample Output
1557439953
1423870062
这道题原本不难但值得分析;开始一直在纠结两个变量之间的大小关系;因为这关系到去绝对值之后是否要变号的问题;但是还是看了题解。。。
题解:对于二维的变量,由于加了绝对值,则只有两种关系,要不就两个数要都大,这样ans = (Xi+Yi) - (Xj + Yj);同时注意到此时的| (Xi - Yi) - (Xj - Yj) | < | ans |;(两个正数相加肯定大于相减,这时就算计算了另一个对结果也没影响,后面亦同);若是一大一小,那么ans = (Xi - Yi) - (Xj - Yj);此时满足 |ans| > | (Xi+Yi) - (Xj + Yj) |;
这时能说ans就是取max( | (Xi+Yi) - (Xj + Yj) |,| (Xi - Yi) - (Xj - Yj) | );即维护最大最小的x+y与x - y;
ps:在线计算,节省空间;
#include<iostream> #include<cstdio> #include<cstring> #include<string.h> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include<stdlib.h> #include<time.h> #include<stack> #include<set> using namespace std; #define rep0(i,l,r) for(int i = (l);i < (r);i++) #define rep1(i,l,r) for(int i = (l);i <= (r);i++) #define rep_0(i,r,l) for(int i = (r);i > (l);i--) #define rep_1(i,r,l) for(int i = (r);i >= (l);i--) #define MS0(a) memset(a,0,sizeof(a)) #define MS1(a) memset(a,-1,sizeof(a)) typedef __int64 ll; template<typename T> void read1(T &m) { T x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} m = x*f; } template<typename T> void read2(T &a,T &b){read1(a);read1(b);} template<typename T> void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);} template<typename T> void out(T a) { if(a>9) out(a/10); putchar(a%10+‘0‘); } ll n,seed; inline ll rand(ll l, ll r) { static ll mo=1e9+7, g=78125; return l+((seed*=g)%=mo)%(r-l+1); } const ll inf = 0x3f3f3f3f3f3f3f3fLL; int main() { int T; read1(T); while(T--){ read2(n,seed); ll mn0 = inf,mn1 = -inf,mx0 = inf,mx1 = -inf,x,y,ans; rep0(i,0,n){ x = rand(-1000000000, 1000000000); y = rand(-1000000000, 1000000000); ll tmp = x + y,temp = x - y; mn0 = min(mn0,tmp);mn1 = max(mn1,tmp); mx0 = min(mx0,temp);mx1 = max(mx1,temp); } ans = max(mn1 - mn0,mx1 - mx0); out(ans);puts(""); } return 0; }
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