恩，第一题动态规划，用单调队列优化到O(n)，第二题，树形动规。
然后重点第三题，这是APIO2010中的第三题（信号覆盖/signaling）也要用到的东西，统计平面中不包含某一点的三角形的个数。
看图：
![USACO[Elite 2010 U S Open Competition/gold]解题报告 - 天之骄子 - 天之骄子的家](https://blog.boleyn.su/wp-content/uploads/2010/06/1655635813014060582.jpg "USACO[Elite 2010 U S Open Competition/gold]解题报告 - 天之骄子 - 天之骄子的家")
/*
PROG: hop
LANG: C++
ID: boleyn.2
*/
/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-6-3
DESCRIPTION:
TITLE: Cow Hopscotch [John Pardon, 2010]
CONTENT:
The cows have reverted to their childhood and
are playing a game similar to human hopscotch.
Their hopscotch game features a line of N (3 <=
N <= 250,000) squares conveniently labeled 1..N
that are chalked onto the grass.
Like any good game, this version of hopscotch
has prizes! Square i is labeled with some
integer monetary value V_i (-2,000,000,000 <=
V_i <= 2,000,000,000). The cows play the game to
see who can earn the most money.
The rules are fairly simple:
* A cow starts at square “0” (located just before square 1; it
has no monetary value).
* She then executes a potentially empty sequence of jumps toward
square N. Each square she lands on can be a maximum of K (2
<= K <= N) squares from its predecessor square (i.e., from
square 1, she can jump outbound to squares 2 or 3 if K==2).
* Whenever she wishes, the cow turns around and jumps back
towards square 0, stopping when she arrives there. In addition
to the restrictions above (including the K limit), two
additional restrictions apply:
* She is not allowed to land on any square she touched on her
outbound trip (except square 0, of course).
* Except for square 0, the squares she lands on during the
return trip must directly precede squares she landed on
during the outbound trip (though she might make some larger
leaps that skip potential return squares altogether).
She earns an amount of money equal to the sum of the monetary values
of all the squares she jumped on. Find the largest amount of cash
a cow can earn.
By way of example, consider this five box cow-hopscotch course where
K has the value 3:
Square Num: 0 1 2 3 4 5 6
+—+ +—+ +—+ +—+ +—+ +—+ +—+
|///|–| |–| |–| |–| |–| |–| |
+—+ +—+ +—+ +—+ +—+ +—+ +—+
Value: – 0 1 2 -3 4 5
One (optimal) sequence Bessie could jump (shown with respective
bracketed monetary values) is: 1[0], 3[2], 6[5], 5[4], 2[1], 0[0]
would yield a monetary total of 0+2+5+4+1+0=12.
If Bessie jumped a sequence beginning with 0, 1, 2, 3, 4, … then
she would be unable to return since she could not legally jump back
to an untouched square.
PROBLEM NAME: hop
INPUT FORMAT:
* Line 1: Two space separated integers: N and K
* Lines 2..N+1: Line i+1 contains a single integer: V_i
SAMPLE INPUT (file hop.in):
6 3
0
1
2
-3
4
5
OUTPUT FORMAT:
* Line 1: A single line with a single integer that is the maximum
amount of money a cow can earn
SAMPLE OUTPUT (file hop.out):
12
*/
#include <stdio.h>
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
const int MAXN=250000;
const long long int oo=1000000000000000000ll;
int N,K;
int V[MAXN+1];
long long int _sum[MAXN+2];
long long int _f[MAXN+2];
long long int answer;
int main()
{
freopen(“hop.in”,“r”,stdin);
freopen(“hop.out”,“w”,stdout);
scanf(“%d %d\n”,&N,&K);
for (int i=1;i<=N;i++)
scanf(“%d\n”,&V[i]);
long long int* sum=_sum+1;
long long int* f=_f+1;
for (int i=1;i<=N;i++)
if (V[i]>0) sum[i]=sum[i-1]+V[i];
else sum[i]=sum[i-1];
answer=-oo;
int head,tail;
int q[MAXN];
f[q[head=tail=0]=-1]=0;
for (int i=0;i<=N;i++)
{
long long int get=f[q[head]]-sum[q[head]+1]+sum[i-1]+V[i];
if (get>answer) answer=get;
while (q[head]+K<i) head++;
if (i!=N) f[i]=f[q[head]]-sum[q[head]+1]+sum[i-1]+V[i]+V[i+1];
while (head<=tail&&f[q[tail]]-sum[q[tail]+1]<f[i]-sum[i+1]) tail–;
q[++tail]=i;
}
cout<<answer<<endl;
fclose(stdin);
fclose(stdout);
return 0;
}
/*
PROG: slide
LANG: C++
ID: boleyn.2
*/
/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-4-23
DESCRIPTION:
TITLE: Water Slides [John Pardon, 2010]
CONTENT:
Inspired by the new water park at Machu
Picchu in Peru, Farmer John has decided
to build one for the cows. Its biggest
attraction is to be a giant water slide
of a peculiar design.
The superslide comprises E (1 <= E <=
150,000) mini slides connecting V (2 <=
V <= 50,000) small pools conveniently
labeled 1..V. Every mini slide must be
traversed in its proper direction and
may not be traversed backwards. The
cows start at pool number 1 and
traverse successive mini slides until
they end up in the pool number V, the
final pool. Every pool (except V, the
last one and 1, the first one)
includes at least one mini slide
entering it and at least one
(different) mini slide for exiting
it. Furthermore, a cow can reach the
end of the ride (pool V) from any
pool by going down a sequence of mini
slides. Finally, since this is a
slide, it is not possible to leave a
pool and then encounter that pool
again after traversing some set of
mini slides.
Each mini slide i runs from pool P_i to pool Q_i (1 <= P_i <= V; 1
<= Q_i <= V; P_i != Q_i) and has an associated fun value F_i (0 <=
F_i <= 2,000,000,000). Bessie’s total fun for any given trip down
the superslide is the sum of the fun values of all the mini slides
traversed.
Bessie naturally wants to have as much fun as possible, given the
long time that she spends in the slide’s queue waiting for the ride.
Generally, she carefully chooses which mini slide to follow out of
each pool. She is a cow, however, and no more than K (1 <= K <= 10)
times as she splashes down the slide, she loses control and follows
a random mini slide out of a pool (this can even happen on pool 1).
If Bessie chooses so as to maximize her fun in the worst case, how
much fun is she guaranteed to have for a given super-slide?
By way of example, consider a small park that has 3 pools (pool
id’s shown in brackets) and four mini slides; K has the value 1
(fun values shown outside of brackets):
[1]
/ \
5 -> / \ <- 9
/ \
[2]—3—[3]
\__5__/
She alway starts at pool 1 and ends and pool 3. If she had her way,
she’d ride direct from pool 1 to pool 2 and then on the higher-fun
mini slide (with fun value 5) to slide 3 for a total fun value of
5+5=10. But, if she loses control at pool 1, she might slide directly
from pool 1 to pool 3 for total fun 9. If she loses control at pool
2, she could reduce her total fun to just 5+3 = 8.
Bessie wants to find the most fun she can have so she strives to
choose 1->3 for a total fun of 9. If she loses control at pool 1
and ends up on mini slide 1->2, she knows she will not lose control
at pool 2 and will end up with fun 10. Thus, she knows her minimum
fun will always be at least 9.
PROBLEM NAME: slide
INPUT FORMAT:
* Line 1: Three space separated integers: V, E, and K
* Lines 2..E + 1: Line i+1 contains three space separated integers:
P_i, Q_i, and F_i
SAMPLE INPUT (file slide.in):
3 4 1
2 3 5
1 2 5
1 3 9
2 3 3
OUTPUT FORMAT:
* Line 1: A single line with a single integer that is the minimum fun
that Bessie can guarantee she can have.
SAMPLE OUTPUT (file slide.out):
9
*/
#include <stdio.h>
#include <iostream>
using namespace std;
#include <vector>
using std::vector;
const long long int oo=1000000000000000000ll;
const long long int MAXV=50000,MAXE=150000,MAXK=10;
long long int V,E,K;
long long int P[MAXE],Q[MAXE],F[MAXE];
bool calced[MAXV+1][MAXK+1];
vector<int> child[MAXV+1],length[MAXV+1];
long long int value[MAXV+1][MAXK+1];
long long int answer;
long long int min(long long int a,long long int b)
{
return a<b?a:b;
}
long long int max(long long int a,long long int b)
{
return a>b?a:b;
}
long long int dp(int node,int k)
{
if (!calced[node][k])
{
calced[node][k]=true;
for (int i=0;i<child[node].size();i++)
value[node][k]=max(value[node][k],
dp(child[node][i],k)+length[node][i]);
if (k>0)
for (int i=0;i<child[node].size();i++)
{
value[node][k]=min(value[node][k],
dp(child[node][i],k-1)+length[node][i]);
}
}
return value[node][k];
}
void print(long long int long_long_int)
{
if (long_long_int>=10)
print(long_long_int/10);
printf(“%d”,int(long_long_int%10));
}
int main()
{
freopen(“slide.in”,“r”,stdin);
freopen(“slide.out”,“w”,stdout);
scanf(“%d %d %d\n”,&V,&E,&K);
for (int i=0;i<E;i++)
scanf(“%d %d %d\n”,&P[i],&Q[i],&F[i]);
for (int i=0;i<E;i++)
{
child[P[i]].push_back(Q[i]);
length[P[i]].push_back(F[i]);
}
answer=dp(1,K);
/*
printf(“%lld\n”,answer); //for Linux/Unix
printf(“%I64d\n”,answer); //for Windows
*/
//for anywhere
print(answer);
printf(“\n”);
fclose(stdin);
fclose(stdout);
return 0;
}
/*
PROG: tricount
LANG: C++
ID: boleyn.2
*/
/*
PROGRAM: $PROGRAM
AUTHOR: Su Jiao
DATE: 2010-6-3
DESCRIPTION:
TITLE: Triangle Counting [Tom Conerly, 2010]
CONTENT:
Bessie is standing guard duty after
the big bad wolf was spotted stalking
cows over at Farmer Don’s spread.
Looking down from her guard tower in
utter boredom, she’s decided to
perform intellectual exercises in
order to keep awake.
After imagining the field as an X,Y
grid, she recorded the coordinates of
the N (1 <= N <= 100,000)
conveniently numbered 1..N cows as
X_i,Y_i (-100,000 <= X_i <= 100,000;
-100,000 <= Y_i <= 100,000; 1 <= i <=
N). She then mentally formed all possible triangles that could be
made from subsets of the entire set of cow coordinates. She counts
a triangle as ‘golden’ if it wholly contains the origin (0,0). The
origin does not fall on the line between any pair of cows. Additionally,
no cow is standing exactly on the origin.
Given the list of cow locations, calculate the number of ‘golden’
triangles that contain the origin so Bessie will know if she’s doing
a good job.
By way of example, consider 5 cows at these locations:
-5,0 0,2 11,2 -11,-6 11,-5
Below is a schematic layout of the field from Betsy’s point of view:
…………|…………
…………*……….*.
…………|…………
——-*—-+————
…………|…………
…………|…………
…………|…………
…………|…………
…………|……….*.
.*……….|…………
…………|…………
All ten triangles below can be formed from the five points above:
By inspection, 5 of them contain the origin and hence are ‘golden’.
PROBLEM NAME: tricount
INPUT FORMAT:
* Line 1: A single integer: N
* Lines 2..N+1: Each line contains two integers, the coordinates of a
single cow: X_i and Y_i
SAMPLE INPUT (file tricount.in):
5
-5 0
0 2
11 2
-11 -6
11 -5
OUTPUT FORMAT:
* Line 1: A single line with a single integer that is the count of the
number of times a triangle formed by the cow locations
contains the origin
SAMPLE OUTPUT (file tricount.out):
5
*/
#include <fstream>
using std::ifstream;
using std::ofstream;
using std::endl;
#include <utility>
using std::pair;
using std::make_pair;
#define Point pair<int,int>
#define x first
#define y second
struct Compare
{
static
inline
long long int cross(const Point& a,const Point& b)
{
return (long long int)(a.x)*(long long int)(b.y)-(long long int)(b.x)*(long long int)(a.y);
}
static
inline
bool less(const Point& a,const Point& b)
{
return cross(a,b)>0;
}
static
void sort(Point* l,Point* r)
{
Point* i=l;
Point* j=r;
Point mid=*(l+((r-l)>>1));
while (i<=j)
{
while (less(*i,mid)) i++;
while (less(mid,*j)) j–;
if (i<=j)
{
Point s=*i;
*i=*j;
*j=s;
i++;
j–;
}
}
if (l<j) sort(l,j);
if (i<r) sort(i,r);
}
};
class Application
{
ifstream cin;
ofstream cout;
static const int MAXN=100000;
long long int N;
Point p[MAXN];
public:
Application(const char* input,const char* output)
:cin(input),cout(output)
{
cin>>N;
for (int i=0;i<N;i++)
cin>>p[i].x>>p[i].y;
}
~Application()
{
cin.close();
cout.close();
}
int run()
{
Compare compare;
compare.sort(p,p+N-1);
long long int result=0;
for (int i=0,j=0;i<N;i++)
{
while (compare.less(p[i],p[(j+1)%N])) j=(j+1)%N;
int get=(j-i+N)%N;
if (get>=2) result+=((get)*(get-1))>>1;
}
cout<<(N*(N-1)*(N-2))/(3*2*1)-result<<endl;
return 0;
}
};
int main()
{
Application app(“tricount.in”,“tricount.out”);
return app.run();
}