Mice and Holes CodeForces, holescodeforces

Mice and Holes CodeForces, holescodeforces

Mice and Holes CodeForces-797F

There are n rats and m holes on a digital axis. The coordinates of the mouse are x [1],..., x [n], the coordinate of the hole is p [1],..., p [m], each hole can accommodate c [1],..., c [m]: Ask the minimum value of the sum of the distances that every mouse runs after hiding in the hole.

Analysis:

If the sequence of holes and mice is not given, a set operation is required, which is enumeration.

Therefore, the holes and mice are sorted in the order of coordinates, and then the State is defined: ans [I] [j], which indicates the minimum distance and size of the first j mice.

This is because if a mouse runs overlapping routes, the answer is definitely not the best. The first sort after this definition of the state, it implies that only the first I-1 holes put the first k mouse, then the first I hole put the k + 1 TO THE j mouse, routes do not overlap.

(I cannot think of the following parts)

However, you may not be able to think about it in practice. You must make a bold attempt (for example, try sorting, and it is very likely that the status can be defined by sorting ).

So ans [I] [j] = min (ans [I-1] [j] + 0, ans [I-1] [J-1] + {j ~ J point to I distance and}, ans [I-1] [J-2] + {J-1 ~ The Point-to-I distance of j and}..., ans [I-1] [j-c [I] + {j-c [I] + 1 ~ The distance from j to I and })

Look at ans [I] [J-1] = min (ans [I-1] [J-1] + 0, ans [I-1] [J-2] + {J-1 ~ J-1 point to I distance and},..., ans [I-1] [j-c [I]-1] + {j-c [I] ~ J-1 point to I distance and })

Taking a closer look is like taking the minimum value in a "Sliding Window", but the form cannot be accelerated, it is still too slow.

So let's take a look:

Ans [I] [j] = min (ans [I-1] [j] + 0, ans [I-1] [J-1] + 1 to j point to I distance and-1 to (J-1) point to I distance and, ans [I-1] [J-2] + 1 to j point to I distance and-1 to (J-2) point to I distance and ,..., ans [I-1] [j-c [I] + 1 to j point to I distance and-1 to (j-c [I]) point to I distance and)
Ans [I] [J-1] = min (ans [I-1] [J-1] + 0, ans [I-1] [J-2] + 1 to J-1 point to I distance and-1 to J-2 point to I distance and ,..., ans [I-1] [j-c [I]-1] + 1 to J-1 point to I distance and-1 to j-c [I]-1 point to I distance and)

That is, ans [I] [j] = the point-to-I distance from 1 to j and the point-to-point I distance from + min {ans [I-1] [j-k]-1 to j-k and} (0 <= k <= c [I])

Ans [I] [J-1] = 1 to J-1 point to I distance and + min {ans [I-1] [j-k-1]-1 to j-k-1 point to point I distance and} (0 <= k <= c [I]-1)

That is, ans [I] [j] = 1 to j point to I distance and + min (ans [I-1] [j]-1 to j point to I distance and ,..., ans [I-1] [j-c [I]-1 to j-c [I] Point to I distance and)

Ans [I] [J-1] = 1 to J-1 point to I distance and + min (ans [I-1] [J-1]-1 to J-1 location to I distance and ,..., ans [I-1] [j-c [I]-1]-1 to j-c [I]-1 point to I distance and)

In this way, we can maintain min {ans [I-1] [j-...]-...} with a monotonous queue every time I changes.

Also, ans [I] [...] is only related to ans [I-1] [...] and can save space with a rolling array.

1 # include <cstdio> 2 # include <cstring> 3 # include <algorithm> 4 using namespace std; 5 typedef long LL; // I forgot to open LL 6 struct Hole 7 {8 LL p, c; 9 bool operator <(const Hole & B) const10 {11 return p <B. p; 12} 13} h [5010]; 14 LL n, m; 15 LL x [5010]; 16 LL ans [2] [5010]; 17 // an error was written as ans [5010] [2] 18 LL q [5010], l, r, i1; 19 LL s1 [5010]. // indicates the total number of mice in the first I hole: 20 LL s2 [5010]; // s2 [j] indicates 1 ~ J's mouse to I distance and 21 int main () 22 {23 LL I, j; 24 scanf ("% I64d % I64d", & n, & m ); 25 for (I = 1; I <= n; I ++) 26 scanf ("% I64d", & x [I]); 27 for (I = 1; I <= m; I ++) 28 scanf ("% I64d % I64d", & h [I]. p, & h [I]. c); 29 sort (x + 1, x + n + 1); 30 sort (h + 1, h + m + 1); 31 for (I = 1; I <= m; I ++) 32 s1 [I] = s1 [I-1] + h [I]. c; 33 memset (ans, 0x3f, sizeof (ans); 34 ans [0] [0] = 0; // The ans [0] [..] set 035 if (n> s1 [m]) 36 {37 printf ("-1"); 38 return 0; 39} 40 for (I = 1; I <= m; I ++) 41 {42 i1 ^ = 1; 4 3 l = r = 0; 44 for (j = 1; j <= min (s1 [I], n); j ++) 45 s2 [j] = s2 [J-1] + abs (x [j]-h [I]. p); 46 // q [r ++] = 0; 47 for (j = 0; j <= min (s1 [I], n); j ++) // initially treated as 148 {49 if (r> l & q [l] <j-h [I]. c) l ++; // The j-c [I] ~ J, that is, remove the one before j-c [I, add j50 while (r> l & ans [i1 ^ 1] [q [r-1]-s2 [q [r-1]> = ans [i1 ^ 1] [j]-s2 [j]) 51 r --; 52 q [r ++] = j; 53 ans [i1] [j] = ans [i1 ^ 1] [q [l]-s2 [q [l] + s2 [j]; 54} 55} 56 printf ("% I64d", ans [m % 2] [n]); 57 return 0; 58}