Solution -「SDOI 2016」「洛谷 P4076」墙上的句子
\(\mathcal{Description}\)
Link.
（概括得说不清话了还是去看原题吧 qwq。
\(\mathcal{Solution}\)
首先剔除回文串——它们一定对答案产生 \(1\) 的贡献。我们称一个句子是“正序”的，当且仅当句子的所有单词同时满足自己的字典序不小于翻转后的字典序；“逆序”则当且仅当句子的所有单词同时满足自己的字典序严格大于翻转后的字典序。从这条显眼的性质入手：
此外观察者发现，对每一行（列）来说，按照确定后的阅读顺序读出的所有单词同时满足“自己的字典序不小于翻转后的字典序”，或同时满足“自己的字典序不大于翻转后的字典序”。
也就是说任何一个句子都是“正序”或“逆序”的，而正序句子和正序句子同时选择不会对答案额外贡献，逆序句子亦然。所以答案一定是贡献自正逆序句子之间。考虑最小割黑白染色模型：
在一张无向图中，钦定一些点为黑色，一些点为白色，为其余点染色，使得连接黑白两色的边数尽量少。
接下来就自然而然了：
- 源点 \(S\)，汇点 \(T\)，行 \(r_{1..n}\) 列 \(c_{1..m}\) 共 \(n+m\) 个点（为它们染色），每个正序单词对应 \(w_s,w_r\)，表示其正序阅读和逆序阅读两种情况；
- \(S\) 连向被钦定正序（注意需要用句子正序方向和阅读方向综合判断）的行/列结点，流量 \(+\infty\)（不可割）；
- 能正序读出 \(w\) 的所有行/列结点连向 \(w_s\)，流量 \(+\infty\)（一旦有一个句子被正序读，\(w_s\) 就在字典中，不可割）；
- \(w_s\) 连向 \(w_r\)，流量为 \(1\)（可以被割，对答案贡献 \(1\)）；
- \(w_r\) 连向所有能逆序读出 \(w\) 的行/列结点，流量 \(+\infty\)（同理）。
- \(T\) 同理 \(S\)。
当然，单词去重。建图后跑最小割即可。复杂度 \(\mathcal O(\operatorname{Dinic}(n^2,n^3))\)（\(n,m\) 同阶）。
\(\mathcal{Code}\)
/* Clearink */
#include <map>
#include <set>
#include <queue>
#include <cstdio>
#include <vector>
#define rep( i, l, r ) for ( int i = l, repEnd##i = r; i <= repEnd##i; ++i )
#define per( i, r, l ) for ( int i = r, repEnd##i = l; i >= repEnd##i; --i )
typedef unsigned long long ULL;
const int MAXN = 80, INF = 0x3f3f3f3f;
const ULL BASE = 127;
int n, m, node, S, T;
int rdir[MAXN + 5], cdir[MAXN + 5];
int rcor[MAXN + 5], ccor[MAXN + 5];
char table[MAXN + 5][MAXN + 5];
std::set<ULL> parl;
std::map<ULL, int> virn;
inline int imin ( const int a, const int b ) { return a < b ? a : b; }
struct MaxFlowGraph {
static const int MAXND = 1e6, MAXEG = 1e6;
int ecnt, head[MAXND + 5], S, T, bound, curh[MAXND + 5], d[MAXND + 5];
struct Edge { int to, flow, nxt; } graph[MAXEG * 2 + 5];
inline MaxFlowGraph (): ecnt ( 1 ) {}
inline void clear () {
ecnt = 1;
for ( int i = 0; i <= bound; ++i ) head[i] = 0;
}
inline void restore () {
for ( int i = 2; i <= ecnt; i += 2 ) {
graph[i].flow += graph[i ^ 1].flow;
graph[i ^ 1].flow = 0;
}
}
inline void link ( const int s, const int t, const int f ) {
graph[++ecnt].to = t, graph[ecnt].flow = f, graph[ecnt].nxt = head[s];
head[s] = ecnt;
}
inline Edge& operator [] ( const int k ) { return graph[k]; }
inline void operator () ( const int s, const int t, const int f ) {
#ifdef RYBY
printf ( "%d %d ", s, t );
if ( f == INF ) puts ( "INF" );
else printf ( "%d\n", f );
#endif
link ( s, t, f ), link ( t, s, 0 );
}
inline bool bfs () {
static std::queue<int> que;
for ( int i = 0; i <= bound; ++i ) d[i] = -1;
d[S] = 0, que.push ( S );
while ( !que.empty () ) {
int u = que.front (); que.pop ();
for ( int i = head[u], v; i; i = graph[i].nxt ) {
if ( graph[i].flow && !~d[v = graph[i].to] ) {
d[v] = d[u] + 1;
que.push ( v );
}
}
}
return ~d[T];
}
inline int dfs ( const int u, const int iflow ) {
if ( u == T ) return iflow;
int ret = 0;
for ( int& i = curh[u], v; i; i = graph[i].nxt ) {
if ( graph[i].flow && d[v = graph[i].to] == d[u] + 1 ) {
int oflow = dfs ( v, imin ( iflow - ret, graph[i].flow ) );
ret += oflow, graph[i].flow -= oflow, graph[i ^ 1].flow += oflow;
if ( ret == iflow ) break;
}
}
if ( !ret ) d[u] = -1;
return ret;
}
inline int calc ( const int s, const int t ) {
S = s, T = t;
int ret = 0;
for ( ; bfs (); ret += dfs ( S, INF ) ) {
for ( int i = 0; i <= bound; ++i ) curh[i] = head[i];
}
return ret;
}
} graph;
struct Word {
std::vector<char> str;
inline int order () const {
if ( str.empty () ) return 0;
for ( int l = 0, r = str.size () - 1; ~r; ++l, --r ) {
if ( str[l] < str[r] ) return 1;
if ( str[r] < str[l] ) return -1;
}
return 0;
}
inline ULL hash () const {
if ( str.empty () ) return 0;
ULL ret = 0;
rep ( i, 0, str.size () - 1 ) ret = ret * BASE + ( str[i] - 'A' + 1 );
return ret;
}
inline ULL rhash () const {
if ( str.empty () ) return 0;
ULL ret = 0;
per ( i, str.size () - 1, 0 ) ret = ret * BASE + ( str[i] - 'A' + 1 );
return ret;
}
};
inline int virid ( const Word& tmp, const bool r ) {
ULL h = r ? tmp.rhash () : tmp.hash ();
if ( !virn.count ( h ) ) {
virn[h] = ++node;
graph ( node, node + 1, 1 );
return ++node - !r;
}
return virn[h] + r;
}
inline void initCorDir () {
static Word tmp;
rep ( i, 1, n ) table[i][0] = table[i][m + 1] = '_';
rep ( i, 1, m ) table[0][i] = table[n + 1][i] = '_';
rep ( i, 1, n ) {
for ( int l = 1, r; l <= m; l = r + 1 ) {
tmp.str.clear ();
for ( r = l; table[i][r] != '_'; tmp.str.push_back ( table[i][r++] ) );
if ( l == r ) continue;
int type = tmp.order ();
if ( !type ) parl.insert ( tmp.hash () );
else {
rcor[i] = type;
int id;
if ( rcor[i] == 1 ) {
graph ( i, id = virid ( tmp, 0 ), INF );
graph ( id + 1, i, INF );
} else {
graph ( id = virid ( tmp, 1 ), i, INF );
graph ( i, id - 1, INF );
}
}
}
}
rep ( i, 1, m ) {
for ( int l = 1, r; l <= n; l = r + 1 ) {
tmp.str.clear ();
for ( r = l; table[r][i] != '_'; tmp.str.push_back ( table[r++][i] ) );
if ( l == r ) continue;
int type = tmp.order ();
if ( !type ) parl.insert ( tmp.hash () );
else {
ccor[i] = type;
int id;
if ( ccor[i] == 1 ) {
graph ( i + n, id = virid ( tmp, 0 ), INF );
graph ( id + 1, i + n, INF );
} else {
graph ( id = virid ( tmp, 1 ), i + n, INF );
graph ( i, id - 1, INF );
}
}
}
}
}
inline void clear () {
/*
* node, rcor, ccor, graph are cleared.
* */
virn.clear (), parl.clear ();
}
int main () {
// freopen ( "wall.in", "r", stdin );
// freopen ( "wall.out", "w", stdout );
int cas;
for ( scanf ( "%d", &cas ); cas--; ) {
clear ();
scanf ( "%d %d", &n, &m );
S = n + m + 1, T = node = S + 1;
rep ( i, 1, n ) scanf ( "%d", &rdir[i] ), rcor[i] = 0;
rep ( i, 1, m ) scanf ( "%d", &cdir[i] ), ccor[i] = 0;
rep ( i, 1, n ) scanf ( "%s", table[i] + 1 );
initCorDir ();
rep ( i, 1, n ) if ( rdir[i] ) {
if ( rcor[i] * rdir[i] >= 0 ) graph ( S, i, INF );
else graph ( i, T, INF );
}
rep ( i, 1, m ) if ( cdir[i] ) {
if ( ccor[i] * cdir[i] >= 0 ) graph ( S, i + n, INF );
else graph ( i + n, T, INF );
}
graph.bound = node;
int flw = graph.calc ( S, T );
printf ( "%d\n", flw * 2 + ( int ) parl.size () );
graph.clear ();
}
return 0;
}
