template<classT,classS> inlineboolchkmin(T& a, S b) {returna > b ? a = b,true:false;}intn;intf[N][N], a[N][N];intpos[N];boolin[N];structBit {inta[N];voidinit() { memset(a,0,sizeof(a)); }voidmodify(intx,intv) {for(inti = x; i < N; i += i & -i) a[...
}usingio :: gi;usingio :: putc;usingio :: print;intmain(void){freopen("aztec.in","r", stdin),freopen("aztec.out","w", stdout);inttestCase =1;gi(testCase);while(testCase--) {intN, M, mx =0;gi(N),gi(M); std::vector <int>A(N +1),E(N +1,0);for(inti =1; i ...
template<classT,classS> inlineboolchkmin(T& a, S b) {returna > b ? a = b,true:false;}intn;intf[N][N], a[N][N];intpos[N];boolin[N];structBit {inta[N];voidinit() { memset(a,0,sizeof(a)); }voidmodify(intx,intv) {for(inti = x; i < N; i += i & -i) a[...
Solution Since the number of different possible masks were just 7070 in the previous problem, we had been able to use dynamic programming for checking all possible xors. But what if the constraint was much bigger, say 105.105. That is when we can use Part 22 of this technique, which, in...
CodeForces 1521E Nastia and a Beautiful Matrix 题解 CF1521E链接这道题有多解,下面给出一种无需二分的做法。Hint 1 先假设已经求出了 sizesize。 观察题目所给条件,是否可以把网格分成两部分,使得每部分里互不影响,只需要考虑两部分之间的冲突?
Codeforces 1085G(1086E) Beautiful Matrix dpdp+树状数组 题意定义一个n∗nn∗n的矩阵是beautifulbeautiful的,需要满足以下三个条件:1.每一行是一个排列。2.上下相邻的两个元素的值不同。再定义两个矩阵的字典序大的矩阵大(从左往右从上到下一个一个比较)。给出一个beautifulbeautiful的n∗nn∗n的矩阵...