USACO 2016 US Open Contest, Platinum Problem 2. Bull in a China Shop

原题下载:

USACO2016-OPEN-P2

答案:
Here's my solution to this problem annotated with what each section of code aims to accomplish.

#include
#include
#include
#include

#include
#include
#include
#include

using namespace std;

typedef vector pat;

int POLYMOD[2] = {
975919579,
975979579,
};

int POLYMUL[2] = {
382737283,
382878283,
};

#define MAXCOL 1010
#define MAXROW 510

#define HASHES 2
#define MAXPOW (MAXROW * MAXCOL)

int POWTAB[HASHES][MAXPOW];

void init_tab() {
if (POWTAB[0][0]) {
return;
}
for (int i = 0; i < HASHES; i++) {
for (int j = POWTAB[i][0] = 1; j < MAXPOW; j++) {
POWTAB[i][j] = (1ll * POWTAB[i][j - 1] * POLYMUL[i]) % POLYMOD[i];
}
}
}

/* Tracks a polynomial hash over a 2D array. */
struct xhash {
xhash() {
init_tab();
memset(H, 0, sizeof(H));
}

xhash(const pat& p) {
init_tab();
memset(H, 0, sizeof(H));

/* Calculate the hash of the given input matrix. We linearize the array by
* setting a[r * MAXCOL + c] = p[i][j] and then apply a standard polynomial
* hash. */
for (int i = 0; i < p.size(); i++) {
for (int j = 0; j < p[0].size(); j++) {
if (p[i][j] == '.') {
continue;
}

/* We set v this way to ensure that v_1 - v_2 could never represent a
* valid character. This is important for ensuring the integrity of
* subtracting two hashes. */
int v = 26 + (p[i][j] - 'a');
for (int k = 0; k < HASHES; k++) {
H[k] = (H[k] + 1ll * POWTAB[k][i * MAXCOL + j] * v) %
POLYMOD[k];
}
}
}
}

void offset(int r, int c) {
/* Offsetting the matrix by (r, c) translates into offsetting the
* linearized array by r * MAXCOL + c. Therefore we multiply each hash
* by x^(r * MAXCOL + c). */
for (int i = 0; i < HASHES; i++) {
H[i] = (1ll * H[i] * POWTAB[i][r * MAXCOL + c]) % POLYMOD[i];
}
}

/* Compute the difference of hashes. This gives you a hash of what would
* remain in *this if you got rid of everything present in x. In the case
* that x isn't actually a subset of *this the hash should just represent
* garbage and won't get matched. */
xhash operator-(const xhash& x) const {
xhash nh;
for (int i = 0; i < HASHES; i++) {
nh.H[i] = H[i] - x.H[i];
if (nh.H[i] < 0) {
nh.H[i] += POLYMOD[i];
}
}
return nh;
}

bool operator==(const xhash& x) const {
return !memcmp(H, x.H, sizeof(H));
}

bool operator<(const xhash& x) const {
return memcmp(H, x.H, sizeof(H)) < 0;
}

int H[HASHES];
};

/* For use in C++'s unordered_map. */
struct xhash_downhash {
int operator()(const xhash& h) const {
return h.H[0];
}
};

/* Vertically flips the pattern. */
void vflip(pat& p) {
int R = p.size();
for (int i = 0; i < R - i - 1; i++) {
p[i].swap(p[R - i - 1]);
}
}

/* Rotates the pattern 90 degrees. */
void rotate(pat& p) {
int R = p.size();
int C = p[0].size();
pat op(C, string(R, '.'));
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) { op[C - j - 1][i] = p[i][j]; } } p = op; } /* Read in a pattern and canonicalize it. */ pat read_pat() { int R, C; cin >> R >> C;

pat res(R);
for (int i = 0; i < R; i++) { cin >> res[i];
}

/* Remove unneeded padding from the sides. */
int mnr = R, mxr = 0;
int mnc = C, mxc = 0;
for (int i = 0; i < R; i++) {
for (int j = 0; j < C; j++) {
if (res[i][j] != '.') {
mnr = min(mnr, i);
mxr = max(mxr, i + 1);
mnc = min(mnc, j);
mxc = max(mxc, j + 1);
}
}
}

pat nres;
for (int i = mnr; i < mxr; i++) {
nres.push_back(res[i].substr(mnc, mxc - mnc));
}

/* Try all orientations and take the lexicographically least one. This
* ensures that all equivalant piece representations are actually equal. */
res = nres;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 4; j++) {
if (nres < res) {
res = nres;
}
rotate(nres);
}
vflip(nres);
}
return res;
}

typedef vector<vector > sum_table;

sum_table target_sums;
vector piece_sums;

/* Calculate the number of non-empty entries after every position in the
* linearized array. This gets used in find_offset later. */
sum_table calc_sums(const pat& p) {
int N = p.size();
int M = p[0].size();
sum_table sums(N, vector(M));

int lst = 0;
for (int i = N - 1; i >= 0; i--) {
for (int j = M - 1; j >= 0; j--) {
if (p[i][j] != '.') {
lst++;
}
sums[i][j] = lst;
}
}
return sums;
}

/* Efficiently find the last non-zero position in base after subtracting
* (possibly) several other matricies. */
pair<int, int> find_offset(const sum_table& base,
const vector& subtract_inds = {},
const vector<pair<int, int> >& offsets = {}) {
/* Binary search over the linearized array. */
int N = base.size();
int M = base[0].size();
int lo = 0;
int hi = N * M - 1;
while (lo < hi) {
int md = (lo + hi + 1) / 2;
int r = md / M;
int c = md % M;

/* Calculate how many non-zero entries there are after (r, c) in the
* linearized array. */
int val = base[r][c];
for (int i = 0; i < subtract_inds.size(); i++) { int j = subtract_inds[i]; int er = r - offsets[i].first; int ec = c - offsets[i].second; int PN = piece_sums[j].size(); int PM = piece_sums[j][0].size(); if (ec >= PM) {
er++;
ec = 0;
}
if (er >= PN) {
/* Do nothing. */
} else if (er < 0) { val -= piece_sums[j][0][0]; } else { val -= piece_sums[j][er][max(0, ec)]; } } /* Search right if there are more non-zero entries, otherwise search left. */ if (val) { lo = md; } else { hi = md - 1; } } return make_pair(lo / M, lo % M); } int main() { freopen("bcs.in", "r", stdin); freopen("bcs.out", "w", stdout); int K; cin >> K;
pat target = read_pat();

int R = target.size();
int C = target[0].size();
xhash target_hash = target;
target_sums = calc_sums(target);

/* Read in the pieces into a map after canonicalization. Keep track of
* how many times an equivalant piece occurs and deduplicate. */
map<pat, int> piece_map;
for (int i = 0; i < K; i++) {
piece_map[read_pat()]++;
}

/* Build data structures based on the deduplicated pieces in each of
* their orientations. */
vector piece_counts;
vector<pair<int, int> > piece_offsets;
vector piece_index;
vector piece_hashes;

unordered_map<xhash, int, xhash_downhash> the_hash;
for (auto it : piece_map) {
pat p = it.first;
int index = piece_counts.size();
piece_counts.push_back(it.second);
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 4; k++) { if (p.size() > R || p[0].size() > C) {
rotate(p);
continue;
}

xhash h(p);
piece_sums.push_back(calc_sums(p));
piece_offsets.push_back(find_offset(piece_sums.back()));
piece_hashes.push_back(h);
piece_index.push_back(index);

h.offset(R - piece_offsets.back().first,
C - piece_offsets.back().second);

/* Deduplication ensures that no two pieces should have the same hash.
*/
the_hash[h] = index;
rotate(p);
}
vflip(p);
}
}

set<tuple<int, int, int> > sols;
pair<int, int> base_1 = find_offset(target_sums);
for (int i = 0; i < piece_hashes.size(); i++) {
/* Find the offset so that the last item in piece_hashes[i] covers the last
* item in the target. */
pair<int, int> off_1 = base_1;
off_1.first -= piece_offsets[i].first;
off_1.second -= piece_offsets[i].second;
if (off_1.first < 0 || off_1.second < 0) {
continue;
}

xhash hash_1 = piece_hashes[i];
hash_1.offset(off_1.first, off_1.second);

pair<int, int> base_2 = find_offset(target_sums, {i}, {off_1});
for (int j = 0; j < piece_hashes.size(); j++) {
/* Find the offset so that the last uncovered item in * piece_hashes[j]
* covers the last still uncovered item in the target. */
pair<int, int> off_2 = base_2;
off_2.first -= piece_offsets[j].first;
off_2.second -= piece_offsets[j].second;
if (off_2.first < 0 || off_2.second < 0) {
continue;
}

xhash hash_2 = piece_hashes[j];
hash_2.offset(off_2.first, off_2.second);

/* Canonicalize the position of the last still uncovered item of the
* target so we can look up if we have a matching hash. */
pair<int, int> off_3 = find_offset(target_sums, {i, j}, {off_1, off_2});
if (off_3.first < 0 || off_3.second < 0) { continue; } xhash hash_remains = target_hash - hash_1 - hash_2; hash_remains.offset(R - off_3.first, C - off_3.second); /* Check if we have a match, if we do insert it into the solutions list. */ auto it = the_hash.find(hash_remains); if (it != the_hash.end()) { int L[3]; L[0] = piece_index[i]; L[1] = piece_index[j]; L[2] = it->second;
sort(L, L + 3);
sols.insert(make_tuple(L[0], L[1], L[2]));
}
}
}

/* Convert the solutions list into an actual result on the input array. This
* takes into account the pieces that were deduplicated in the beginning. */
int result = 0;
for (auto sol : sols) {
int c0 = piece_counts[get<0>(sol)];
int c1 = piece_counts[get<1>(sol)];
int c2 = piece_counts[get<2>(sol)];
if (get<0>(sol) == get<2>(sol)) {
result += c0 * (c0 - 1) * (c0 - 2) / 6;
} else if (get<0>(sol) == get<1>(sol)) {
result += c0 * (c0 - 1) / 2 * c2;
} else if (get<1>(sol) == get<2>(sol)) {
result += c0 * c1 * (c1 - 1) / 2;
} else {
result += c0 * c1 * c2;
}
}
cout << result << endl;

return 0;
}

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