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// Smart Carts | |
// Solution by Jacob Plachta | |
using namespace std; | |
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); | |
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); | |
} | |
int N; | |
int nxt[2][LIM], prv[2][LIM]; | |
vector<int> seq[2][2]; | |
int L[2][2]; | |
PR pos[2][LIM]; | |
int F[2], subS[2]; | |
vector<int> CS; | |
int csInd, sumC, baseAns; | |
int ProcessQuery0(int* K) | |
{ | |
int i, ans = 0; | |
// check whether either initial line is too long to be valid | |
Fox(i, 2) | |
if (L[0][i] > K[i]) | |
return(-1); | |
// count initially-satisfied carts | |
Fox(i, N) | |
ans += nxt[0][i] == nxt[1][i]; | |
return(ans); | |
} | |
void PrecomputeForQuery1(int C) | |
{ | |
int i, j; | |
// determine which carts are free | |
Fox(i, 2) | |
F[i] = min(C - L[0][1 - i], L[0][i]); | |
auto IsFree = [&](int i, int e) { // if e=1, include non-free, accessible latches | |
PR p = pos[0][i]; | |
return(p.y > L[0][p.x] - F[p.x] - e); | |
}; | |
// compute number of potentially-satisfiable carts | |
baseAns = 0; | |
Fox(i, N) | |
baseAns += nxt[0][i] == nxt[1][i] || // initially satisfied? | |
(IsFree(i, 0) && IsFree(nxt[1][i], 1)); // free, and target accessible? | |
// split free, potentially-satisfiable carts into chains | |
// and compute achievable subset sums of chain lengths | |
sumC = 0; | |
Fill(subS, 0); | |
bitset<LIM> BC = 1; | |
Fox(i, 2) | |
{ | |
auto& s = seq[1][i]; | |
int c = 0, f = -1; | |
Fox(j, Sz(s)) | |
{ | |
if (!IsFree(s[j], 0)) | |
continue; | |
c++; | |
// chain forced to belong to a certain line? | |
if (c == 1 && j && IsFree(s[j - 1], 1)) | |
f = pos[0][s[j - 1]].x; | |
// chain ends with this cart? | |
if (j + 1 == Sz(s) || !IsFree(s[j + 1], 0)) | |
{ | |
if (f < 0) | |
sumC += c, BC |= (BC << c); | |
else | |
subS[f] += c; | |
c = 0, f = -1; | |
} | |
} | |
} | |
// store results | |
CS.clear(); | |
Fox(i, sumC + 1) | |
if (BC[i]) | |
CS.pb(i); | |
csInd = 0; | |
} | |
int ProcessQuery1(int* K) | |
{ | |
// compute space for free carts in each line | |
int i, S[2]; | |
Fox(i, 2) | |
S[i] = K[i] - (L[0][i] - F[i]) - subS[i]; | |
// check whether all free cart chains may be packed into the lines | |
if (min(S[0], S[1]) < 0) | |
return(baseAns - 1); | |
while (csInd + 1 < Sz(CS) && CS[csInd + 1] <= S[0]) | |
csInd++; | |
if (sumC - CS[csInd] <= S[1]) | |
return(baseAns); | |
return(baseAns - 1); // 1 cart must remain unsatisfied due to chain breakage | |
} | |
int ProcessQuery2(int* K) | |
{ | |
int i; | |
// check whether either target line is too long to be fully satisfied | |
Fox(i, 2) | |
if (L[1][i] > K[i]) | |
return(N - 1); | |
return(N); | |
} | |
int ProcessQuery(int d, int c, int x, int y) | |
{ | |
int i, K[2] = { min(c, x), min(c, y) }; | |
// check whether either initial line is too long to be valid | |
Fox(i, 2) | |
if (L[0][i] > c) | |
return(-1); | |
// check whether there are too many carts to form valid final lines | |
if (N > K[0] + K[1]) | |
return(-1); | |
return d == 0 | |
? ProcessQuery0(K) | |
: d == 1 | |
? ProcessQuery1(K) | |
: ProcessQuery2(K); | |
} | |
void ProcessCase() | |
{ | |
int i, j, z; | |
// init | |
Fill(prv, -1); | |
// input | |
Read(N); | |
Fox(i, N) | |
{ | |
Fox(z, 2) | |
{ | |
Read(nxt[z][i]), nxt[z][i]--; | |
prv[z][nxt[z][i]] = i; | |
} | |
} | |
// split into lines (initial/target) | |
Fox(z, 2) | |
{ | |
Fox(i, 2) | |
{ | |
seq[z][i].clear(); | |
j = N + i; | |
while (j >= 0) | |
{ | |
pos[z][j] = mp(i, Sz(seq[z][i])); | |
seq[z][i].pb(j); | |
j = prv[z][j]; | |
} | |
L[z][i] = Sz(seq[z][i]) - 1; | |
} | |
} | |
// process queries | |
int d, c, x, y; | |
LL G[LIM] = { 0 }; | |
Fox(d, min(3, N + 1)) | |
{ | |
Fox(c, N + 1) | |
{ | |
if (d == 1) | |
PrecomputeForQuery1(c); | |
Fox(x, N + 1) | |
{ | |
Fox(y, N + 1) | |
G[ProcessQuery(d, c, x, y) + 1] += d == 2 ? N - 1 : 1; | |
} | |
} | |
} | |
Fox(i, N + 2) | |
printf(" %lld", G[i]); | |
printf("\n"); | |
} | |
int main() | |
{ | |
int T, t; | |
Read(T); | |
Fox1(t, T) | |
{ | |
printf("Case #%d:", t); | |
ProcessCase(); | |
} | |
return(0); | |
} |