2019牛客多校第八场
A. All-one Matices
solved at 01:58(+2)
求一个(n*m)的(01)矩阵的极大全(1)子矩阵数目
悬线法处理出(d)数组（从这个位置最多向上延伸多少个(1)），然后单调栈处理出每个位置的(d)能延伸的左右最远位置，(vis)打标记的时候如果发现标记不是(i-1)也不是(0)说明这里之前有一个极大矩阵
#include <bits/stdc++.h>
using namespace std;
const int N = 3010;
char a[N][N];
int d[N][N], s[N][N], s2[N][N], n, m, vis[N][N], ans;
int main() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i) {
scanf("%s", a[i] + 1);
}
for(int j = 1; j <= m; ++j) {
for(int i = 1; i <= n; ++i) {
if(a[i][j] == '0')
d[i][j] = 0;
else
d[i][j] = d[i - 1][j] + 1;
}
}
for(int i = 1; i <= n; ++i) {
stack<int> st;
while(!st.empty()) st.pop();
d[i][m + 1] = -1;
st.push(m + 1);
for(int j = m; j; --j) {
while(d[i][st.top()] >= d[i][j]) st.pop();
s[i][j] = st.top() - 1;
st.push(j);
}
}
for(int i = 1; i <= n; ++i) {
stack<int> st;
while(!st.empty()) st.pop();
d[i][0] = -1;
st.push(0);
for(int j = 1; j <= m; ++j) {
while(d[i][st.top()] >= d[i][j]) st.pop();
s2[i][j] = st.top() + 1;
st.push(j);
}
}
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= m; ++j) {
if(a[i][j] == '0') continue;
int r = s[i][j], l = s2[i][j];
if(vis[l][r] == i) continue;
if(vis[l][r] != i - 1 && vis[l][r]) ans++;
vis[l][r] = i;
}
}
for(int i = 1; i <= m; ++i)
for(int j = i; j <= m; ++j)
ans += vis[i][j] != 0;
printf("%d
", ans);
return 0;
}
B. Beauty Values
solved at 00:13
定义beauty values为数组中不同的数的数量，求所有区间的beauty values 之和
期望的线性性
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, a[N];
long long ans;
vector<int> pos[N];
int main() {
scanf("%d", &n);
for(int i = 1; i < N; ++i) pos[i].push_back(0);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
pos[a[i]].push_back(i);
}
for(int i = 1; i < N; ++i) {
for(int j = 1; j < pos[i].size(); ++j) {
ans += 1LL * (pos[i][j] - pos[i][j - 1]) * (n - pos[i][j] + 1);
}
}
printf("%lld
", ans);
return 0;
}
C. CDMA
solved at 01:30
构造一个(m*m)的矩阵，元素均为正负一，要求满足任意两行的相应位置乘积之和为(0)，(m)为2的幂次
首先手推出(m=2)的解（其实样例已经给了），然后把它复制三边得到一个(4*4)的矩阵，把右下角的(2*2)个元素取反，就得到了(m=4)的解，以此类推即可(我是用的(m=4)作为初始矩阵往下推得)
#include <bits/stdc++.h>
using namespace std;
int d[4][4] = {
{1, 1, 1, 1},
{1, 1, -1, -1},
{1, -1, 1, -1},
{1, -1, -1, 1}
};
int m;
int f(int i, int j, int mod) {
if(mod == 4) return 1;
mod /= 2;
return (i >= mod && j >= mod ? -1 : 1) * f(i % mod, j % mod, mod);
}
int main() {
scanf("%d", &m);
if(m == 2) {
puts("1 1
1 -1");
return 0;
}
for(int i = 0; i < m; ++i) {
for(int j = 0; j < m; ++j) {
printf("%d ", f(i, j, m) * d[i % 4][j % 4]);
}
puts("");
}
return 0;
}
D. Distance
upsolved
有一个三维空间，有两种操作，给一个点打上标记，或者查询给定点到所有已标记点中的最小曼哈顿距离，保证三维空间长宽高乘积不超过(1e5)
可以定期重构(bfs)，但是显然三维树状数组更短
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, h, q, op, x, y, z;
struct BIT {
static int lowbit(int x) {return x & -x;}
static void mmax(int &x, int y) {x = (x > y ? x : y);}
static int index(int i, int j, int k) {return (i - 1) * m * h + (j - 1) * h + k;}
int c[N];
void init() {memset(c, 0xc0, sizeof(c));}
void update(int x, int y, int z, int val) {
for(int i = x; i <= n; i += lowbit(i))
for(int j = y; j <= m; j += lowbit(j))
for(int k = z; k <= h; k += lowbit(k))
mmax(c[index(i, j, k)], val);
}
int query(int x, int y, int z) {
int res = -1e9;
for(int i = x; i; i -= lowbit(i))
for(int j = y; j; j -= lowbit(j))
for(int k = z; k; k -= lowbit(k))
res = max(res, c[index(i, j, k)]);
return res;
}
}T[8];
int main() {
scanf("%d%d%d%d", &n, &m, &h, &q);
for(int i = 0; i < 8; ++i) T[i].init();
while(q--) {
scanf("%d%d%d%d", &op, &x, &y, &z);
if(op == 1) {
for(int mask = 0; mask < 8; ++mask) {
int tx = x, ty = y, tz = z, vx = x, vy = y, vz = z;
if(mask & 1) tx = n - tx + 1, vx = -vx;
if(mask & 2) ty = m - ty + 1, vy = -vy;
if(mask & 4) tz = h - tz + 1, vz = -vz;
T[mask].update(tx, ty, tz, vx + vy + vz);
}
}
else {
int res = 1e9;
for(int mask = 0; mask < 8; ++mask) {
int tx = x, ty = y, tz = z, vx = x, vy = y, vz = z;
if(mask & 1) tx = n - tx + 1, vx = -vx;
if(mask & 2) ty = m - ty + 1, vy = -vy;
if(mask & 4) tz = h - tz + 1, vz = -vz;
res = min(res, vx + vy + vz - T[mask].query(tx, ty, tz));
}
printf("%d
", res);
}
}
return 0;
}
E. Explorer
upsolved
无向图，每条边有上下界限制，只有在上下界之中的数字可以通过这条边，询问有多少种数字可以从(1)走到(n)
时间分治线段树+可撤销并查集
把size限制看成时间限制就好了
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int u[N], v[N], l[N], r[N], b[N << 1], n, m, tot, fa[N], sz[N];
vector<int> g[N << 3];
stack<pair<int*, int>> stk[22];
long long ans;
int find(int x) {
return x== fa[x] ? x : find(fa[x]);
}
void update(int rt, int l, int r, int L, int R, int id) {
if(L <= l && r <= R) {g[rt].push_back(id); return;}
int mid = l + r >> 1;
if(L <= mid) update(rt << 1, l, mid, L, R, id);
if(R > mid) update(rt << 1 | 1, mid + 1, r, L, R, id);
}
void dfs(int rt, int l, int r, int dep) {
while(!stk[dep].empty()) stk[dep].pop();
for(auto i: g[rt]) {
int x = find(u[i]), y = find(v[i]);
if(x == y) continue;
if(sz[x] < sz[y]) swap(x, y);
stk[dep].push(make_pair(&sz[x], sz[x]));
stk[dep].push(make_pair(&fa[y], fa[y]));
sz[x] += sz[y];
fa[y] = x;
}
if(l == r) {
if(find(1) == find(n))
ans += b[r + 1] - b[l];
}
else {
int mid = l + r >> 1;
dfs(rt << 1, l, mid, dep + 1);
dfs(rt << 1 | 1, mid + 1, r, dep + 1);
}
while(!stk[dep].empty()) {
*(stk[dep].top().first) = stk[dep].top().second;
stk[dep].pop();
}
}
int main() {
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) {
scanf("%d%d%d%d", &u[i], &v[i], &l[i], &r[i]);
b[2 * i - 1] = l[i];
b[2 * i] = ++r[i];
}
sort(b + 1, b + 2 * m + 1);
tot = unique(b + 1, b + 2 * m + 1) - b - 1;
for(int i = 1; i <= m; ++i) {
l[i] = lower_bound(b + 1, b + tot + 1, l[i]) - b;
r[i] = lower_bound(b + 1, b + tot + 1, r[i]) - b - 1;
update(1, 1, tot, l[i], r[i], i);
}
for(int i = 1; i <= n; ++i) fa[i] = i, sz[i] = 1;
dfs(1, 1, tot, 0);
printf("%lld
", ans);
return 0;
}
G. Gemstones
solved at 00:17
签到题，用栈贪心即可