洛谷 3613 - 睡覺困難綜合征
阿新 • • 發佈：2019-01-04
long 得到 father {} 睡覺 a10 from spa air
該題是[NOI2014]起床困難綜合征的樹上加修改版
先說說《起床困難綜合征》，由於不同位運算之間不滿足交換律，故必須按順序執行操作
考慮位運算套路 —— 拆位，對於未知的初始值，它的每一位也是未知的，所以可以用兩個變量 $a_0, a_1$ 來當作初始值當前位為 $0$ 或 $1$ 時最終可以得到的值，最後貪心即可
附上部分代碼：
1 int a1 = 0, a2 = - 1; 2 for (int i = 1; i <= N; i ++) { 3 scanf ("%s", opt + 1); 4 int p = getnum ();5 if (opt[1] == ‘A‘) 6 a1 &= p, a2 &= p; 7 else if (opt[1] == ‘O‘) 8 a1 |= p, a2 |= p; 9 else if (opt[1] == ‘X‘) 10 a1 ^= p, a2 ^= p; 11 } 12 int ps = 0, ans = 0; 13 for (int j = 30; j >= 1; j --) { 14 if (a1 & (1 << (j - 1))) 15ans += (1 << (j - 1)); 16 else if ((a2 & (1 << (j - 1))) && ps + (1 << (j - 1)) <= M) 17 ans += (1 << (j - 1)), ps += (1 << (j - 1)); 18 }
那麽現在將操作擴展到樹上，使用 $LCT$ 來維護其代表的鏈的 $a_0, a_1$ ，通過找規律可以發現：
設新的 $a_0, a_1$ 分別為 $f_0, f_1$ ，舊的分別為 $x.a_0, x.a_1$ 及 $y.a_0, y.a_1$ （其中 $x, y$ 為從左往右），現在要合並 $x, y$
對於 $f_0$ 很好想到要將 $x.a_0 \ opt \ y.a_0$ 與 $x.a_0 \ opt \ y.a_1$ 合並即可得解，公式即為 $(x.a_0 \& y.a_1) | (~ x.a_0 \& y.a_0)$ ，其中， $(x.a_0 \& y.a_t)$ 為通解，第二個的取反即為消去 $x.a_0$ 中為 $1$ 的位對第二個操作的影響
$pushup$ 寫完就是 $LCT$ 常規操作了，求答案也是貪心即可
註意：由於有 $makeroot$ 操作，又求答案順序一定是從左往右，所以還需要維護一下從右往左的答案，用於在 $reverse$ 的時候交換
代碼
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 5 using namespace std; 6 7 typedef long long LL; 8 typedef unsigned long long uLL; 9 10 const int MAXN = 1e05 + 10; 11 12 // {&, |, ^} -> {1, 2, 3} 13 int opt[MAXN]= {0}; 14 LL value[MAXN]= {0}; 15 16 struct optionSt { 17 LL a0, a1; 18 19 optionSt () {} 20 optionSt (LL fa0, LL fa1) : 21 a0 (fa0), a1 (fa1) {} 22 23 } ; 24 optionSt merge (optionSt x, optionSt y) { 25 LL f0 = ((x.a0 & y.a1) | (~ x.a0 & y.a0)); 26 LL f1 = ((x.a1 & y.a1) | (~ x.a1 & y.a0)); 27 return optionSt (f0, f1); 28 } 29 /*pair<LL, LL> merge (pair<LL, LL> p1, pair<LL, LL> p2) { 30 LL a00 = p1.first, a10 = p1.second; 31 LL a01 = p2.first, a11 = p2.second; 32 LL a0 = ((a00 & a11) | (~ a00 & a01)); 33 LL a1 = ((a10 & a11) | (~ a10 & a01)); 34 return make_pair (a0, a1); 35 }*/ 36 optionSt calc (int op, LL val) { 37 switch (op) { 38 case 1: 39 return optionSt (0ll, val); 40 case 2: 41 return optionSt (val, - 1ll); 42 case 3: 43 return optionSt (val, ~ val); 44 } 45 } 46 47 int father[MAXN]= {0}; 48 int son[MAXN][2]= {0}; 49 optionSt tval[MAXN], froml[MAXN], fromr[MAXN]; 50 int revtag[MAXN]= {0}; 51 52 int isroot (int p) { 53 return son[father[p]][0] != p && son[father[p]][1] != p; 54 } 55 int sonbel (int p) { 56 return son[father[p]][1] == p; 57 } 58 void reverse (int p) { 59 if (! p) 60 return ; 61 swap (son[p][0], son[p][1]); 62 swap (froml[p], fromr[p]); // 註意要更改左右順序 63 revtag[p] ^= 1; 64 } 65 void pushup (int p) { 66 froml[p] = fromr[p] = tval[p]; 67 if (son[p][0]) { 68 froml[p] = merge (froml[son[p][0]], froml[p]); 69 fromr[p] = merge (fromr[p], fromr[son[p][0]]); 70 } 71 if (son[p][1]) { 72 froml[p] = merge (froml[p], froml[son[p][1]]); 73 fromr[p] = merge (fromr[son[p][1]], fromr[p]); 74 } 75 } 76 void pushdown (int p) { 77 if (revtag[p]) { 78 reverse (son[p][0]), reverse (son[p][1]); 79 revtag[p] = 0; 80 } 81 } 82 void rotate (int p) { 83 int fa = father[p], anc = father[fa]; 84 int s = sonbel (p); 85 son[fa][s] = son[p][s ^ 1]; 86 if (son[fa][s]) 87 father[son[fa][s]] = fa; 88 if (! isroot (fa)) 89 son[anc][sonbel (fa)] = p; 90 father[p] = anc; 91 son[p][s ^ 1] = fa, father[fa] = p; 92 pushup (fa), pushup (p); 93 } 94 int Stack[MAXN]; 95 int top = 0; 96 void splay (int p) { 97 top = 0, Stack[++ top] = p; 98 for (int nd = p; ! isroot (nd); nd = father[nd]) 99 Stack[++ top] = father[nd]; 100 while (top > 0) 101 pushdown (Stack[top]), top --; 102 for (int fa = father[p]; ! isroot (p); rotate (p), fa = father[p]) 103 if (! isroot (fa)) 104 sonbel (p) == sonbel (fa) ? rotate (fa) : rotate (p); 105 pushup (p); 106 } 107 void Access (int p) { 108 for (int tp = 0; p; tp = p, p = father[p]) 109 splay (p), son[p][1] = tp, pushup (p); 110 } 111 void Makeroot (int p) { 112 Access (p), splay (p), reverse (p); 113 } 114 void Split (int x, int y) { 115 Makeroot (x); 116 Access (y), splay (y); 117 } 118 void link (int x, int y) { 119 Makeroot (x); 120 father[x] = y; 121 } 122 123 int N, M, K; 124 125 uLL Query (int x, int y, LL lim) { 126 Split (x, y); 127 uLL ans = 0, ps = 0; 128 LL a0 = froml[y].a0, a1 = froml[y].a1; 129 for (int j = K; j >= 1; j --) { 130 if (a0 & (1ll << (j - 1))) 131 ans += (1ll << (j - 1)); 132 else if ((a1 & (1ll << (j - 1))) && ps + (1ll << (j - 1)) <= lim) 133 ans += (1ll << (j - 1)), ps += (1ll << (j - 1)); 134 } 135 return ans; 136 } 137 138 int getint () { 139 int num = 0; 140 char ch = getchar (); 141 142 while (! isdigit (ch)) 143 ch = getchar (); 144 while (isdigit (ch)) 145 num = (num << 3) + (num << 1) + ch - ‘0‘, ch = getchar (); 146 147 return num; 148 } 149 LL getLL () { 150 LL num = 0; 151 char ch = getchar (); 152 153 while (! isdigit (ch)) 154 ch = getchar (); 155 while (isdigit (ch)) 156 num = (num << 3) + (num << 1) + ch - ‘0‘, ch = getchar (); 157 158 return num; 159 } 160 uLL getuLL () { 161 uLL num = 0; 162 char ch = getchar (); 163 164 while (! isdigit (ch)) 165 ch = getchar (); 166 while (isdigit (ch)) 167 num = (num << 3) + (num << 1) + ch - ‘0‘, ch = getchar (); 168 169 return num; 170 } 171 void work (uLL x) { 172 if (x >= 10) 173 work (x / 10); 174 putchar (x % 10 + ‘0‘); 175 } 176 void println (uLL x) { 177 work (x), puts (""); 178 } 179 180 int main () { 181 N = getint (), M = getint (), K = getint (); 182 for (int i = 1; i <= N; i ++) { 183 opt[i] = getint (), value[i] = getLL (); 184 tval[i] = calc (opt[i], value[i]); 185 } 186 for (int i = 1; i < N; i ++) { 187 int u = getint (), v = getint (); 188 link (u, v); 189 } 190 for (int Case = 1; Case <= M; Case ++) { 191 int op = getint (); 192 if (op == 1) { 193 int x = getint (), y = getint (); 194 uLL lim = getLL (); 195 println (Query (x, y, lim)); 196 } 197 else if (op == 2) { 198 int p = getint (), nopt = getint (); 199 LL delta = getLL (); 200 Access (p), splay (p); 201 opt[p] = nopt, value[p] = delta; 202 tval[p] = calc (opt[p], value[p]); 203 pushup (p); 204 } 205 } 206 207 return 0; 208 } 209 210 /* 211 5 5 3 212 1 7 213 2 6 214 3 7 215 3 6 216 3 1 217 1 2 218 2 3 219 3 4 220 1 5 221 1 1 4 7 222 1 1 3 5 223 2 1 1 3 224 2 3 3 3 225 1 1 3 2 226 */ 227 228 /* 229 2 2 2 230 2 2 231 2 2 232 1 2 233 2 2 2 2 234 1 2 2 2 235 */
洛谷 3613 - 睡覺困難綜合征