AtCoder Regular Contest 81
C.Make a Rectangle
给出一堆木棍的长度
从中选4根,询问在能围成矩形的情况下,矩形的最大面积
开个map统计一下就行
分正方形和矩形分别统计即可
复杂度$O(n \log n)$
#include <map> #include <cmath> #include <cstdio> #include <cstring> #include <iostream> using namespace std; extern inline char gc() { static char RR[23456], *S = RR + 23333, *T = RR + 23333; if(S == T) fread(RR, 1, 23333, stdin), S = RR; return *S ++; } inline int read() { int p = 0, w = 1; char c = gc(); while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); } while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc(); return p * w; } #define ri register int #define ll long long ll ans; int n, mx; map <int, int> num; int main() { n = read(); for(ri i = 1; i <= n; i ++) num[read()] ++; for(map<int, int> :: iterator it = num.begin(); it != num.end(); it ++) { int p1 = it -> first, p2 = it -> second; if(p2 >= 4) ans = max(ans, 1ll * p1 * p1); if(p2 >= 2) ans = max(ans, 1ll * p1 * mx), mx = p1; } printf("%lld\n", ans); return 0; }
D.Coloring Dominoes
给出一个$2 * m$的网格,被$1 *2$的骨牌满覆盖
有3种颜色,给骨牌染色,要求相领的骨牌颜色不相同,询问方案数
分4种情况讨论即可
自己是($1 * 2$或者$2 *2$),前置状态是($1 * 2$或者$2 * 2$)
复杂度$O(n)$,$n = 52$...
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define ri register int #define sid 105 #define mod 1000000007 int n; char S[3][sid]; int main() { cin >> n >> S[1] + 1 >> S[2] + 1; ri i = 1, ans; if(S[1][i] == S[2][i]) i = 1, ans = 3; else i = 2, ans = 6; for(i = i + 1; i <= n; i ++) { int p1 = S[1][i] == S[2][i]; int p2 = S[1][i - 1] == S[2][i - 1]; if(!p1) i ++; if(p1 && p2) ans = (1ll * ans * 2) % mod; if(!p1 && !p2) ans = (1ll * ans * 3) % mod; if(!p1 && p2) ans = (1ll * ans * 2) % mod; } printf("%d\n", ans); return 0; }
E.Don't Be a Subsequence
询问最短的,字典序最小的,不是串$S$的子序列的字符串
考虑序列自动机
那么,所有不能被$S$所识别的序列一定是在识别了一段前置之后多了一个字母
考虑虚拟一个节点$n + 1$,能转移到它说明不是串$S$的子序列
之后就是一个拓扑图$dp$和字典序的事了
复杂度$O(26 * n)$
#include <cstdio> #include <cstring> #include <iostream> using namespace std; #define ri register int #define sid 200050 char s[sid]; int nxt[sid][26], f[sid]; int main() { scanf("%s", s + 1); int n = strlen(s + 1); for(ri i = 0; i < 26; i ++) nxt[n][i] = n + 1; for(ri i = n - 1; i >= 0; i --) { for(ri j = 0; j < 26; j ++) nxt[i][j] = nxt[i + 1][j]; nxt[i][s[i + 1] - 'a'] = i + 1; } for(ri i = n; i >= 0; i --) { f[i] = 1e9; for(ri j = 0; j < 26; j ++) f[i] = min(f[i], 1 + f[nxt[i][j]]); } int o = 0; while(o != n + 1) { for(ri j = 0; j < 26; j ++) if(f[o] == f[nxt[o][j]] + 1) { o = nxt[o][j]; printf("%c", j + 'a'); break; } } return 0; }
F. Flip and Rectangles
可以任意地翻转一些行或者一些列
询问最大子矩阵
鬼畜的性质:一个矩阵中,如果所有的$2 *2$子矩形都有偶数个黑点,那么它就能被翻转出来
证明网上一大堆
然后令$v[i][j]$表示以$(i, j)$为左上角存不存在$2 *2$偶数黑点矩阵
问题转化为求最大子矩阵
注意特判一行,一列
注意计算答案时,求出来的最大子矩阵的长和宽是原题中最大子矩阵长和宽 - 1的结果
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define ri register int #define sid 2005 char s[sid]; int n, m, ans; bool v[sid][sid], c[sid][sid]; short h[sid][sid], l[sid][sid], r[sid][sid]; short L[sid][sid], R[sid][sid]; int main() { scanf("%d%d", &n, &m); for(ri i = 1; i <= n; i ++) { scanf("%s", s + 1); for(ri j = 1; j <= m; j ++) c[i][j] = (s[j] == '#') ? 1 : 0; } for(ri i = 1; i < n; i ++) for(ri j = 1; j < m; j ++) v[i][j] = c[i][j] ^ c[i + 1][j] ^ c[i][j + 1] ^ c[i + 1][j + 1] ^ 1; for(ri i = 1; i <= n; i ++) { int lst = 0; for(ri j = 1; j <= m; j ++) if(v[i][j]) l[i][j] = lst; else L[i][j] = 0, lst = j; lst = m + 1; for(ri j = m; j >= 1; j --) if(v[i][j]) r[i][j] = lst; else R[i][j] = m + 1, lst = j; } for(ri i = 0; i <= m + 1; i ++) R[0][i] = m + 1; ans = max(n, m); for(ri i = 1; i <= n; i ++) for(ri j = 1; j <= m; j ++) if(v[i][j]) { h[i][j] = h[i - 1][j] + 1; L[i][j] = max((short)(l[i][j] + 1), L[i - 1][j]); R[i][j] = min((short)(r[i][j] - 1), R[i - 1][j]); ans = max(ans, (R[i][j] - L[i][j] + 2) * (h[i][j] + 1)); } printf("%d\n", ans); return 0; }
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