题目链接 Problem J
这道题思路还是很直观的,但是有两个难点:
1、题目中说$1<=NM<=10^{6}$,但没具体说明$N$和$M$的值,也就是可能出现:
$N = 1, M = 1000000$ 这样的数据。
2、对每个查询的分类讨论。
#include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(int i(a); i <= (b); ++i) #define dec(i,a,b) for(int i(a); i >= (b); --i) #define for_edge(i,x) for(int i = H[x]; i; i = X[i]) const int N = 1000000 + 10; const int Q = 1000 + 10; struct node{ int num, x, y, id; } c[N], f[N]; int ret[Q], nc[Q], nd[Q], ne[Q], ccc[Q]; int ct[N]; int n, m, q; int cnt; char st[N]; int T; int v[N], t[N]; int l, r; int color; int col[N]; int x, y; struct nnnn{ int x, y; } nf[6]; inline bool check(int nu, int pos){ if (nu < 1 || nu > cnt) return false; if (v[nu]) return false; int xx1 = c[nu].x, yy1 = c[nu].y, xx2 = c[pos].x, yy2 = c[pos].y; if (xx1 < 1 || xx1 > n || yy1 < 1 || yy1 > m) return false; if (abs(xx1 - xx2) + abs(yy1 - yy2) != 1) return false; if (c[nu].num != c[pos].num) return false; return true; } inline bool cc(int nu, int pos){ if (nu < 1 || nu > cnt) return false; int xx1 = c[nu].x, yy1 = c[nu].y, xx2 = c[pos].x, yy2 = c[pos].y; if (xx1 < 1 || xx1 > n || yy1 < 1 || yy1 > m) return false; if (abs(xx1 - xx2) + abs(yy1 - yy2) != 1) return false; return true; } inline int pos(int x, int y){ return (x - 1) * m + y;} inline void work(int nu){ t[nu] = color; f[++r].x = c[nu].x, f[r].y = c[nu].y, f[r].id = nu; v[nu] = 1; } int main(){ scanf("%d", &T); while (T--){ scanf("%d%d%d", &n, &m, &q); cnt = 0; rep(i, 1, n){ scanf("%s", st + 1); rep(j, 1, m) ++cnt, c[cnt].num = (int)st[j] - 48, c[cnt].x = i, c[cnt].y = j; } color = 0; rep(i, 1, cnt) v[i] = 0, t[i] = 0; rep(i, 1, cnt) if (v[i] == 0){ ++color; t[i] = color; ct[color] = c[i].num; f[1].x = c[i].x, f[1].y = c[i].y, f[1].id = i; for (l = r = 1; l <= r; ++l){ int now = f[l].id; if (check(now - 1, now)) work(now - 1); if (check(now + 1, now)) work(now + 1); if (check(now - m, now)) work(now - m); if (check(now + m, now)) work(now + m); } } int maxcol = 0; rep(i, 1, cnt) col[i] = 0; int ta = 0; rep(i, 1, n){ rep(j, 1, m){ ++ta; ++col[t[ta]]; maxcol = max(maxcol, t[ta]); } } while (q--){ scanf("%d%d", &x, &y); int df = 0, np = pos(x, y); if (cc(np - 1, np)) ret[++df] = np - 1; if (cc(np + 1, np)) ret[++df] = np + 1; if (cc(np - m, np)) ret[++df] = np - m; if (cc(np + m, np)) ret[++df] = np + m; rep(i, 1, df) nc[i] = t[ret[i]]; sort(nc + 1, nc + df + 1); nd[1] = nc[1]; int dx = 1; rep(i, 2, df) if (nc[i] != nc[i - 1]) nd[++dx] = nc[i]; bool fl = false; rep(i, 1, dx) if (nd[i] == t[np]){ fl = true; break;} if (!fl){ rep(i, 1, dx) ne[i] = ct[nd[i]]; rep(i, 0, 9) ccc[i] = 0; rep(i, 1, dx) ccc[ne[i]] += col[nd[i]]; int lk = 0; rep(i, 0, 9) lk = max(lk, ccc[i]); printf("%d ", lk + 1); } else { int spj = 0; rep(i, 1, dx) if (nd[i] == t[np]) spj += col[nd[i]]; int dd = 0; rep(i, 1, dx) if (nd[i] != t[np]) nf[++dd].x = ct[nd[i]], nf[dd].y = nd[i]; rep(i, 0, 9) ccc[i] = 0; rep(i, 1, dd) ccc[nf[i].x] += col[nf[i].y]; int lk = 0; rep(i, 0, 9) lk = max(lk, ccc[i]); printf("%d ", max(lk + 1, spj)); } } } return 0; }