3892: [Usaco2014 Dec]Marathon
3892: [Usaco2014 Dec]Marathon
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 169 Solved: 100
[Submit][Status][Discuss]
Description
Unhappy with the poor health of his cows, Farmer John enrolls them in an assortment of different physical fitness activities. His prize cow Bessie is enrolled in a running class, where she is eventually expected to run a marathon through the downtown area of the city near Farmer John's farm! The marathon course consists of N checkpoints (3 <= N <= 500) to be visited in sequence, where checkpoint 1 is the starting location and checkpoint N is the finish. Bessie is supposed to visit all of these checkpoints one by one, but being the lazy cow she is, she decides that she will skip up to K checkpoints (K < N) in order to shorten her total journey. She cannot skip checkpoints 1 or N, however, since that would be too noticeable. Please help Bessie find the minimum distance that she has to run if she can skip up to K checkpoints. Since the course is set in a downtown area with a grid of streets, the distance between two checkpoints at locations (x1, y1) and (x2, y2) is given by |x1-x2| + |y1-y2|.
Input
Output
Sample Input
0 0
8 3
1 1
10 -5
2 2
Sample Output
HINT
Source
题解:一道神奇的DP,b[i,j]代表从1到i,共计跳跃了j次不难得出递推式
\(b[i,j]=min(b[i-k-1,j-k]+dis(i-k-1,i))\)
,然后\(O\left(N{M}^{2} \right)\)瞎搞万事
/**************************************************************
Problem:
User: HansBug
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/ var
i,j,k,l,m,n:longint;
a:array[..,..] of longint;
b:array[..,..] of longint;
function dis(x,y:longint):longint;inline;
begin
exit(abs(a[x,]-a[y,])+abs(a[x,]-a[y,]));
end;
function min(x,y:longint):longint;inline;
begin
if x<y then min:=x else min:=y;
end;
begin
readln(n,m);
for i:= to n do readln(a[i,],a[i,]);
fillchar(b,sizeof(b),-);
b[,]:=;
for i:= to n do
begin
for j:= to min(i-,m) do
begin
b[i,j]:=maxlongint;
for k:= to j do
if b[i-k-,j-k]<>- then
begin
b[i,j]:=min(b[i,j],b[i-k-,j-k]+dis(i-k-,i));
end;
end;
end;
writeln(b[n,min(n-,m)]);
end.
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