Codeforces 30D King's Problem? Simulation

First, sort n points, find the K after sorting, and then discuss the situation.

When k = n + 1, it is obviously a small value of k-> 1-> n | k-> n-> 1, because the sum of the two sides of the triangle is greater than the third side.

When 1 <= k & k <= n:

1, k-> 1-> n + 1-> k + 1-> n | k-> n + 1-> k-1-> 1, the corresponding step is ignored when k + 1 | K-1 does not exist.

2, k-> 1-> n + 1-> n-> k + 1 | k-> n + 1-> 1-> k-1, the corresponding step is ignored when k + 1 | K-1 does not exist.

3. This is a wonderful strategy. At the beginning, I intuitively thought that this situation would not exist, but five WA people taught me how to behave.

K-> I-> n + 1-> I-1-> 1 (I <k) | k-> I-> 1-> n + 1-> I + 1-> n (I> k ).

It can be understood that, at the expense of abs (site [k]-site [I]), the total value is minimized by changing the starting point, and the IQ is urgent.

# Include <algorithm> # include <iostream> # include <cstring> # include <cstdlib> # include <cstdio> # include <queue> # include <cmath> # include <stack> # include <map> # include <ctime> # include <iomanip> # pragma comment (linker, "/STACK: 1024000000"); # define EPS (1e-6) # define _ LL long # define ULL unsigned long # define LL _ int64 # define INF 0x3f3f3f3f # define Mod 1000000007/** I/O Accelerator Interface .. **/# Define g (c = getchar () # define d isdigit (g) # define p x = x * 10 + c-'0' # define n x = x * 10 + '0'-c # define pp l/= 10, p # define nn l/= 10, ntemplate <class T> inline T & RD (T & x) {char c; while (! D); x = c-'0'; while (d) p; return x;} template <class T> inline T & RDD (T & x) {char c; while (g, c! = '-'&&! Isdigit (c); if (c = '-') {x = '0'-g; while (d) n ;} else {x = c-'0'; while (d) p;} return x;} inline double & RF (double & x) // scanf ("% lf ", & x); {char c; while (g, c! = '-' & C! = '.'&&! Isdigit (c); if (c = '-') if (g = '. ') {x = 0; double l = 1; while (d) nn; x * = l;} else {x = '0'-c; while (d) n; if (c = '. ') {double l = 1; while (d) nn; x * = l ;}} else if (c = '. ') {x = 0; double l = 1; while (d) pp; x * = l;} else {x = c-'0'; while (d) p; if (c = '. ') {double l = 1; while (d) pp; x * = l ;}} return x ;} # undef nn # undef pp # undef n # undef p # undef d # undef gusing namespace std; LL num [100010]; double Cal (LL x, LL x0, LL y0 ){ Return sqrt (x-x0) * (x-x0) + y0 * y0);} int main () {int n, I; LL x, y, k; scanf ("% d % I64d", & n, & k); for (I = 1; I <= n; ++ I) scanf ("% I64d ", & num [I]); scanf ("% I64d % I64d", & x, & y); if (k = n + 1) {sort (num + 1, num + n + 1); double anw = num [n]-num [1] + min (Cal (num [n], x, y ), cal (num [1], x, y); for (I = 2; I <n; ++ I) anw = min (anw, min (num [n]-num [I], num [I]-num [1]) + num [n]-num [1] + Cal (num [I], x, y); printf ("%. 10lf \ n ", anw ); Return 0;} k = num [k]; sort (num + 1, num + n + 1); for (I = 1; I <= n & num [I]! = K; ++ I); k = I; double Min = 1000000000; double tmp; tmp = num [k]-num [1] + num [n]-num [1]; for (I = 2; I <= n; ++ I) min = min (Min, tmp + Cal (num [i-1], x, y) + Cal (num [I], x, y) -(num [I]-num [i-1]); tmp = num [n]-num [k] + num [n]-num [1]; for (I = 2; I <= n; ++ I) Min = min (Min, tmp + Cal (num [i-1], x, y) + Cal (num [I], x, y)-(num [I]-num [i-1]); tmp = num [n]-num [1]; if (k-1) {Min = min (Min, tmp + Cal (num [n], x, y) + Cal (num [k-1], x, y)-(num [k]-num [k-1]); Min = min (Min, tmp + Cal (num [n], x, y) + Cal (num [1], x, y)-(num [k]-num [k-1]);} else {Min = min (Min, tmp + Cal (num [n], x, y);} if (k + 1 <= n) {Min = min (Min, tmp + Cal (num [1], x, y) + Cal (num [k + 1], x, y) -(num [k + 1]-num [k]); Min = min (Min, tmp + Cal (num [1], x, y) + Cal (num [n], x, y)-(num [k + 1]-num [k]);} else {Min = min (Min, tmp + Cal (num [1], x, y);} Min = min (Min, n Um [n]-num [1] + num [k]-num [1] + Cal (num [n], x, y); Min = min (Min, num [n]-num [1] + num [n]-num [k] + Cal (num [1], x, y )); for (I = k + 1; I <= n; ++ I) {if (I! = N) {Min = min (Min, num [I]-num [k] + num [I]-num [1] + Cal (num [1], x, y) + Cal (num [I + 1], x, y) + num [n]-num [I + 1]);} else {Min = min (Min, num [I]-num [k] + num [I]-num [1] + Cal (num [1], x, y) + Cal (num [I + 1], x, y) ;}for (I = k-1; I> = 1; -- I) {if (I! = 1) {Min = min (Min, num [k]-num [I] + num [n]-num [I] + Cal (num [n], x, y) + Cal (num [i-1], x, y) + num [i-1]-num [1]);} else {Min = min (Min, num [k]-num [I] + num [n]-num [I] + Cal (num [n], x, y) + Cal (num [i-1], x, y) ;}} printf ("%. 10lf \ n ", Min); return 0 ;}

Codeforces 30D King's Problem? Simulation