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// Finshakes
// Solution by Jacob Plachta
#define DEBUG 0
#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;
#define LL long long
#define LD long double
#define PR pair<int,int>
#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x*x); }
string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); }
const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);
#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif
bool Read(int &x)
{
char c,r=0,n=0;
x=0;
for(;;)
{
c=GETCHAR();
if ((c<0) && (!r))
return(0);
if ((c=='-') && (!r))
n=1;
else
if ((c>='0') && (c<='9'))
x=x*10+c-'0',r=1;
else
if (r)
break;
}
if (n)
x=-x;
return(1);
}
#define MAXN 500
#define MAXM 51
int N,M,W;
int H[MAXN];
int L[MAXM],R[MAXM];
vector<int> ch[MAXM];
int dyn[MAXM][MAXM][MAXM][MAXM],tmp[MAXM][MAXM][MAXM];
// dyn[i][a][m][b] = max. finshakes in i's subtree, such that:
// a = number of fish at left edge of i's interval
// m = max. number of fish in any pool within i's interval
// b = number of fish at right edge of i's interval
void rec(int i)
{
int j,c,a,a2,b,b2,m,m2,m3,d,d2,d3;
bool o,e1,e2;
// iterate over children
dyn[i][0][0][0]=0;
Fox(j,Sz(ch[i]))
{
// recurse to child
rec(c=ch[i][j]);
if (!j)
{
memcpy(dyn[i],dyn[c],sizeof(dyn[i]));
continue;
}
// does this child overlap with the previous child?
o=R[ch[i][j-1]]==L[c];
// incorporate child into DP values
Fill(tmp,-1);
Fox(a,M+1)
Fox(m,M+1)
Fox(b,M+1)
if ((d=dyn[i][a][m][b])>=0)
Fox(a2,M+1)
Fox(m2,M+1)
Fox(b2,M+1)
if ((d2=dyn[c][a2][m2][b2])>=0)
{
m3=max(m,m2);
d3=d+d2;
if (o)
{
Max(m3,b+a2);
d3+=b*a2;
}
Max(tmp[a][m3][b2],d3);
}
memcpy(dyn[i],tmp,sizeof(dyn[i]));
}
// do the children extend to this interval's edges?
if (Sz(ch[i]))
{
e1=L[ch[i][0]]==L[i];
e2=R[ch[i][Sz(ch[i])-1]]==R[i];
Fill(tmp,-1);
Fox(a,M+1)
Fox(m,M+1)
Fox(b,M+1)
Max(tmp[e1 ? a : 0][m][e2 ? b : 0],dyn[i][a][m][b]);
memcpy(dyn[i],tmp,sizeof(dyn[i]));
}
// incorporate this fish
if (i==M)
return;
Fill(tmp,-1);
Fox(a,M+1)
Fox(m,M+1)
Fox(b,M+1)
if ((d=dyn[i][a][m][b])>=0)
if (L[i]==R[i])
{
// add onto only pool in interval
Max(tmp[a+1][m+1][b+1],d+m);
}
else
{
// add onto left
Max(tmp[a+1][max(m,a+1)][b],d+a);
// add onto max
Max(tmp[a][m+1][b],d+m);
// add onto right
Max(tmp[a][max(m,b+1)][b+1],d+b);
}
memcpy(dyn[i],tmp,sizeof(dyn[i]));
}
int main()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int i,j,p,a,m,b,ans;
PR F[MAXM];
set< pair<PR,int> > S;
set< pair<PR,int> >::iterator I,J;
// testcase loop
Read(T);
Fox1(t,T)
{
// input, and compute fish intervals
Read(N),Read(M),Read(W);
Fox(i,N)
Read(H[i]),H[i]=W-H[i];
Fox(i,M)
{
Read(p),Read(j),p--;
a=b=p;
while ((a>0) && (H[a]<j))
a--;
L[i]=a;
while ((b<N-1) && (H[b]<j))
b++;
R[i]=b;
}
// sort fish by interval sizes
Fox(i,M)
F[i]=mp(R[i]-L[i],i);
sort(F,F+M);
// form tree out of fish
Fox(i,M+1)
ch[i].clear();
S.clear();
Fox(i,M)
{
j=F[i].y;
I=S.lower_bound(mp(mp(L[j],0),0));
while ((I!=S.end()) && (I->x.y<=R[j]))
{
ch[j].pb(I->y);
J=I,I++;
S.erase(J);
}
S.insert(mp(mp(L[j],R[j]),j));
}
Foxen(I,S)
ch[M].pb(I->y);
// DP
Fill(dyn,-1);
rec(M);
// get answer
ans=0;
Fox(a,M+1)
Fox(m,M+1)
Fox(b,M+1)
Max(ans,dyn[M][a][m][b]);
// output
printf("Case #%d: %d\n",t,ans);
}
return(0);
} |