acwing算法基础课I
作者：互联网
acwing 算法基础课I 基础算法.
排序: 快排, 归并排序, 主要思想.
模板 能够默写出来
重复写3-5次
排序
快速排序: 分治
- 确定分界点 取 左边界
q[l] q[(l+r)/2]q[r] - 根据
x的值 重新调整区间 . 左边小于等于分界点, 右边大于等于分界点 - 递归处理左右两遍
void qsort(int q[], int l, int r) {
if (l >= r) return;
int x = q[l + r >> 1], i = l - 1, j = r + 1;// 注意取值.
while (i < j) {
while(q[++i] < x); // 这样就舒服多了吧.
while(q[--j] > x);
if (i < j) swap(q[i], q[j]);
}
qsort(q, l, j);
qsort(q, j + 1, r); //注意最后递归的边界问题.
}归并排序:
分治...
void msort(int l, int r) {
if (l >= r) return;
int mid = l + (r - l) / 2;
msort(l, mid), msort(mid + 1, r);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
if (a[i] < a[j]) tmp[k++] = a[i++];
else tmp[k++] = a[j++];
while (i <= mid) tmp[k++] = a[i++]; //这种放后面写清晰点.
while (j <= r) tmp[k++] = a[j++];
for (i = l, j = 0;i <= r;i++, j++) a[i] = tmp[j];
}逆序对的数量
void msort(int l, int r) {
if (l >= r) return;
int k = 0;
int mi = l + (r - l) / 2;
msort(l, mi), msort(mi + 1, r);
int i = l, j = mi + 1;
while (i <= mi && j <= r) {
if (a[i] <= a[j])tmp[k++] = a[i++];
else tmp[k++] = a[j++], res += mi - i + 1; //只需要这里加上就行
}
while (i <= mi) tmp[k++] = a[i++];
while (j <= r) tmp[k++] = a[j++];
rep(i, 0, k) a[i + l] = tmp[i];
}
int main() {
int n = read();
rep(i, 0, n) a[i] = read();
msort(0, n - 1);
da(a, n);
printf("%llu", res);
return 0;
}二分
整数二分
acwing解法
- 右半边满足 : 比如
x <= a[mi]r = mid, l = mid - 1mi = l + r >> 1
- 左半边满足: 比如
x <= a[mi]l = mid, r = mid +1
// 二分就是最左和最右的问题, 只有两个地方不一样.
//左边界
int lo = 0, hi = n - 1;
while (lo <= hi) { //注意一定是<=
int mid = lo + (hi - lo) / 2;
if (a[mid] < target) lo = mid + 1;
else if (target < a[mid]) hi = mid - 1;
else if (a[mid] == target) hi = mid - 1; // 1.
}
return lo; // 2.
//右边界
while (lo <= hi) {
int mid = lo + (hi - lo) / 2;
if (a[mid] < target) lo = mid + 1;
else if (target < a[mid]) hi = mid - 1;
else if (a[mid] == target) lo = mid + 1; // 1.
}
return hi; // 2.35. Search Insert Position
这个题是寻找符合c++ 规范的那个位置
int searchInsert(vector<int>& nums, int a) {
// 这个题应该是找右边的最左边j.
int l = 0, r = nums.size(); // 注意这里是可以取到 nums.size()的, 所以直接这样.
//不过这样居然不会出现问题...
while(l < r){
int mi = l + r >> 1;
if(nums[mi] >= a) r = mi;
else l = mi + 1;
}
return l;
}275. H-Index II
这种右边边界插入的 和 定义相同的题, 直接让 r = n 才能过
int hIndex(vector<int>& a) {
int n = a.size();
int l = 0, r = n;
while(l < r){
int mi = l + r >> 1;
if(a[mi] >= n - mi) r = mi;
else l = mi + 1;
}
return n - l;
}162. Find Peak Element
一个倒过来 V 的数组 寻找最高点. 利用两边的性质就好.
int findPeakElement(vector<int>& a) {
int n = a.size();
int l = 0, r = n - 1;
while(l < r){
int mi = (l + r) >> 1;
if(mi + 1< n && a[mi] > a[mi + 1]) r = mi;
else l = mi + 1;
}
return l;
}高精度
大整数的存储. 逆序的.
加法
vector<int> add(vector<int>& a, vector<int>& b) {
vector<int> res;
int t = 0;
for (int i = 0;i < a.size() || i < b.size();i++) {
if (i < a.size()) t += a[i];
if (i < b.size()) t += b[i];
res.push_back(t % 10);
t /= 10;
}
if (t) res.push_back(t);
return res;
}减法 ★★★
减法的判断要麻烦一些.
string s, ss;
vector<int> a, b;
vector<int> sub(vector<int>& a, vector<int>& b) {
vector<int> c;
int t = 0;
rep(i, 0, a.size()) {
t = a[i] - t;
if (i < b.size()) t -= b[i];
c.push_back((t + 10) % 10);
t = t < 0 ? 1 : 0;
}
while (c.size() > 1 && c.back() == 0) c.pop_back(); //去除前导0
return c;
}
int main() {
cin >> s >> ss;
per(i, s.size(), 0) a.push_back(s[i] - '0');
per(i, ss.size(), 0) b.push_back(ss[i] - '0');
bool gre = false;
if (a.size() < b.size()) gre = true;
else if (a.size() == b.size()) {
per(i, a.size(), 0) if (a[i] != b[i]) { gre = a[i] < b[i];break; }
}
if (gre) swap(a, b), printf("-");
auto c = sub(a, b);
per(i, c.size(), 0) printf("%d", c[i]);
return 0;
}乘法 ★
vector<int> mul(vector<int>& A, int b) {
if (b == 0) return { 0 };
vector<int> C;
int t = 0; //进位
for (int i = 0; i < A.size() || t; i++)
{
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
return C;
}除法 ★
vector<int> div(vector<int>& A, int b, int& r) {
vector<int> C;
r = 0;
per(i, A.size(), 0) {
r = r * 10 + A[i];
C.push_back(r / b);
r = r % b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}前缀和 & 差分
前缀和
Q: 下标从 0 还是 1 开始? 从 1 开始.
求[l,r]的话, 直接求 \(S_r-S_{l-1}\)
int main() {
int n = read(), q = read(), l, r;
rep(i, 1, n + 1) a[i] = read();
rep(i, 1, n + 1) s[i] += a[i] + s[i - 1];
while (q--) {
l = read(), r = read();
printf("%d\n", s[r] - s[l - 1]);
}
return 0;
}矩阵前缀和
Q: 从 1 还是 0 开始: 1 开始.
int a[N3][N3], s[N3][N3];
int main() {
int n = read(), m = read(), q = read();
rep(i, 1, n + 1) rep(j, 1, m + 1) a[i][j] = read();
rep(i, 1, n + 1) rep(j, 1, m + 1) {
s[i][j] = a[i][j] + s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1]; //前缀和
}
da2(s, n + 1, m + 1);
while (q--) {
int x1 = read(), y1 = read(), x2 = read(), y2 = read();
cout << s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1] << endl; // 求子数组的和.
}
return 0;
}差分 ★★
一个数组实现差分
// 如何通过单个数组做出来这个差分?
int s[N6];
void insert(int l, int r, int c) {
s[l] += c;
s[r + 1] -= c;
}
int main() {
int n = read(), q = read();
rep(i, 1, n + 1) { insert(i, i, read()); }
da(s, n + 1);
while (q--) {
int l = read(), r = read(), c = read();
insert(l, r, c);
da(s, n + 1);
}
rep(i, 1, n + 1) s[i] += s[i - 1], O(s[i]);
return 0;
}差分矩阵 ★★
差分矩阵如何构造和实现.
int s[N3][N3];
void insert(int x1, int y1, int x2, int y2, int c) {
s[x1][y1] += c;
s[x2 + 1][y1] -= c;
s[x1][y2 + 1] -= c;
s[x2 + 1][y2 + 1] += c;
}
int main() {
int n = read(), m = read(), q = read();
rep(i, 1, n + 1)rep(j, 1, m + 1) insert(i, j, i, j, read());
while (q--) {
int x1 = read(), y1 = read(), x2 = read(), y2 = read(), c = read();
insert(x1, y1, x2, y2, c);
}
rep(i, 1, n + 1) {
rep(j, 1, m + 1)
s[i][j] += s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1], O(s[i][j]);
puts("");
}
return 0;
}双指针
通用模板
for(int i = 0, j = 0; i < n; i++){
while(j < i && check(i, j)) j++; //进行判断 缩短区间的长度.
//题目的逻辑
}核心思想是将两重循环优化到O(n)去.
双指针输出每个字符
//双指针输出每个字符;
// 其实和前面有点不一样了, 因为循环中改变i了
int main() {
char str[1000];
cin.getline(str, 1000);
int n = strlen(str);
for (int i = 0; i < n;i++) {
int j = i;
while (j < n && str[j] != ' ') j++;
rep(k, i, j) cout << str[k];
cout << endl;
i = j; // i 等于当前的空格, 便于下一次++以后跳过这个空格
}
return 0;
}799. 最长连续不重复子序列
int a[N5], b[N5];
int main() {
int n = read();
rep(i, 0, n) {
a[i] = read();
}
int res = 0;
for (int i = 0, j = 0; i < n; i++) {
b[a[i]]++;
while (j < i && b[a[i]] > 1) b[a[j++]]--; //进行判断 缩短区间的长度.
res = max(res, i - j + 1);
}
O(res);
return 0;
}位运算 常用操作:
返回第n位数字: a >> n & 1;
返回最后一位 lowbit(n) = n & -n ;
补码的来源:
-x = 0 - x = 000000 - x =1000000 - x = ~x + 1
// 二进制中1的个数.
int main() {
int n = read();
rep(i, 0, n) {
int a = read();
int cnt = 0;
while (a) cnt++, a -= a & -a;
O(cnt);
}
return 0;
}整数离散化
区间和 ★★★
Q: 无限长数轴, 其中 n 个位置数是C, 询问 [l, r] 之间所有数的和
如果有\(10^5\)个数字, 但是值在\(10^9\)之间. 需要进行映射.
- 可能有重复元素, 进行去重
- 如何找下标, 二分.
- 前缀和可以用 0 开始的 ★★★
vector<PII> q, p;
vector<int> ord, s;
int find(int x) {
int l = 0, r = ord.size() - 1;
while (l < r) {
int mi = l + r >> 1;
if (ord[mi] >= x) r = mi;
else l = mi + 1;
}
return l;
}
int main() {
int n = read(), m = read();
rep(i, 0, n) {
int a = read(), b = read();
ord.push_back(a), q.push_back({ a,b });
}
rep(i, 0, m) {
int a = read(), b = read();
ord.push_back(a), ord.push_back(b);
p.push_back({ a,b });
}
sort(ord.begin(), ord.end());
ord.erase(unique(ord.begin(), ord.end()), ord.end());
s.resize(ord.size());
for (auto [k, v] : q) {
k = find(k);
s[k] += v;
}
for (int i = 1;i < s.size();i++) s[i] += s[i - 1];
for (auto [k, v] : p) {
int l = find(k), r = find(v);
printf("%d\n", l == 0 ? s[r] : s[r] - s[l - 1]);
}
return 0;
}区间合并
Q: 给定有 n 个区间, 合并有交集的区间.
vector<PII> a;
int main() {
int n = read();
rep(i, 0, n) a.push_back({ read(),read() });
sort(a.begin(), a.end());
int end = -2e9; // 写成 2e-9 WA了一次...
int res = 0;
for (auto [l, r] : a) {
if (l > end) res++;
end = max(r, end);
}
O(res);
return 0;
}每日一题的...
AcWing 1826. 农田缩减
排序后进行判断是否是边界就行了, 不过这样需要4个数组了.
int x[N5], y[N5], a[N5], b[N5];
int main() {
int n = read();
rep(i, 0, n) {
a[i] = x[i] = read();
b[i] = y[i] = read();
}
sort(a, a + n);sort(b, b + n);
int res = 2e9;
rep(i, 0, n) {
int x1, x2, y1, y2;
x1 = x[i] == a[0] ? a[1] : a[0];
x2 = x[i] == a[n - 1] ? a[n - 2] : a[n - 1];
y1 = y[i] == b[0] ? b[1] : b[0];
y2 = y[i] == b[n - 1] ? b[n - 2] : b[n - 1];
res = min(res, (x2 - x1) * (y2 - y1));
}
O(res);
return 0;
}标签：return,int,mi,read,算法,vector,基础课,acwing,size 来源： https://www.cnblogs.com/benenzhu/p/15869911.html