1 基础算法

排序

快速排序

void quick_sort(int q[], int l, int r)
{
if (l >= r) return;
int i = l - 1, j = r + 1, x = q[l + r >> 1];
while (i < j)
{
do i ++ ; while (q[i] < x);
do j -- ; while (q[j] > x);
if (i < j) swap(q[i], q[j]);
}
quick_sort(q, l, j), quick_sort(q, j + 1, r);
}
分治的思想
  1. 确定分界点：q[l]、q[l + r >>1]、q[r]、随机
  2. 调整区间：≤x的在x左边，≥x的在x右边
  3. 递归：分别给左右边排序
void quick_sort(int q[],int l,int r)
{
int i = l, j = r, x = q[l];
while (i < j)
{
while (q[j] >= x && i < j) j -- ;
if (i >= j) break;
q[i] = q[j];
i ++ ;
while (q[i] <= x && i < j) i ++ ;
if (i >= j) break;
q[j] = q[i];
j -- ;
}
q[i] = x;
if (i > l + 1) quick_sort(q, l, i - 1);
if (j < r - 1) quick_sort(q, j + 1, r);
}

归并排序

void merge_sort(int q[], int l, int r)
{
if (l >= r) return;
int mid = l + r >> 1;
merge_sort(q, l, mid);
merge_sort(q, mid + 1, r);
int k = 0, i = l, j = mid + 1;
while (i <= mid && j <= r)
if (q[i] <= q[j]) tmp[k ++ ] = q[i ++ ];
else tmp[k ++ ] = q[j ++ ];
while (i <= mid) tmp[k ++ ] = q[i ++ ];
while (j <= r) tmp[k ++ ] = q[j ++ ];
for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}
  1. 确定分界点：mid = (l + r)/2
  2. 递归排序左右边
  3. 归并：合二为一
稳定性

二分

整数二分法

bool check(int x) {/* ... */} // 检查x是否满足某种性质
// 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用：
int bsearch_1(int l, int r)
{
while (l < r)
{
int mid = l + r >> 1;
if (check(mid)) r = mid; // check()判断mid是否满足性质
else l = mid + 1;
}
return l;
}
// 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
int bsearch_2(int l, int r)
{
while (l < r)
{
int mid = l + r + 1 >> 1;
if (check(mid)) l = mid;
else r = mid - 1;
}
return l;
}
有单调性一定能二分，能二分不一定有单调性。
本质：找到某种性质，将序列一分为二，一边满足，一边不满足。
  1. 满足左边性质
    1. mid = (l + r + 1)/2
    2. if(check(mid))
      1. true: [mid, r], l = mid
      2. false: [l, mid - 1], r = mid - 1
  2. 满足右边性质
    1. mid = (l + r)/2
    2. if(check(mid))
      1. true: [l, mid], r = mid
      2. false: [mid + 1, r], l = mid + 1

浮点数二分算法

bool check(double x) {/* ... */} // 检查x是否满足某种性质
double bsearch_3(double l, double r)
{
const double eps = 1e-6; // eps 表示精度，取决于题目对精度的要求
while (r - l > eps)
{
double mid = (l + r) / 2;
if (check(mid)) r = mid;
else l = mid;
}
return l;
}

大整数

大整数的存储
从头存个位，十位……，1个元素存1位

高精度加法

// C = A + B, A >= 0, B >= 0
vector add(vector &A, vector &B)
{
if (A.size() < B.size()) return add(B, A);
vector C;
int t = 0;
for (int i = 0; i < A.size(); i ++ )
{
t += A[i];
if (i < B.size()) t += B[i];
C.push_back(t % 10);
t /= 10;
}
if (t) C.push_back(t);
return C;
}

比大小

//是否A≥B
bool cmp(vector &A,vector &B)
{
if (A.size() != B.size()) return A.size()>B.size();
for (int i = A.size() - 1; i >= 0; i -- )
if (A[i] != B[i]) return A[i]>B[i];
return true;
}

高精度减法

// C = A - B, 满足A >= B, A >= 0, B >= 0
vector sub(vector &A, vector &B)
{
vector C;
for (int i = 0, t = 0; i < A.size(); i ++ )
{
t = A[i] - t;
if (i < B.size()) t -= B[i];
C.push_back((t + 10) % 10);
if (t < 0) t = 1;
else t = 0;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}
主函数
int main()
{
string a, b;
vector a, b;
int i;
cin >> a >> b;
for (i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
for (i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');
if (cmp(A, B))
{
auto C = sub(A, B);
for(i = C.size() - 1; i >= 0; i -- )printf("%d",C[i]);
}
else
{
auto C = sub(B, A);
printf("-");
for (i = C.size() - 1; i >= 0; i -- ) printf("%d",C[i]);
}
printf("\n");
return 0;
}

高精度乘以低精度

// C = A * b, A >= 0, b > 0
vector mul(vector &A, int b)
{
vector C;
int t = 0;
for (int i = 0; i < A.size() || t; i ++ )
{
if (i < A.size()) t += A[i] * b;
C.push_back(t % 10);
t /= 10;
}
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}

高精度除以低精度

//A/b，商是C，余数是r
vector div(vector &A, int b, int &r)//r是引用
{
vector C;
r = 0;
for (int i = A.size() - 1; i >= 0; i -- )
{
r = r * 10 + A[i];
C.push_back(r / b);
r %= b;
}
reverse(C.begin(), C.end());
while (C.size() > 1 && C.back() == 0) C.pop_back();
return C;
}

前缀和

一维前缀和

Si = a1 + a2 + … + ai
= Si-1 + ai
作用：快速求出一段的和
Sr - Sl-1 [l,r]

二维前缀和

S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]

差分

一维差分

bi = ai - ai-1
ai+c,i∈[l,r]→bl+c,br+1 - c
#define N 10010
int n, m;
int a[N], b[N];
void insert(int l, int r, int c)
{
b[l] += c;
b[r+1] -= c;
}
int main()
{
int i, l, r, c;
scanf("%d %d", &n, &m);
for (i = 1; i <= n; i ++ )
{
scanf("%d", &a[i]);
insert(i, i, a[i]);
}
while (m -- )
{
scanf("%d %d %d", &l, &r, &c);
insert(l, r, c);
}
for (i = 1; i <= n; i ++ )
{
b[i] += b[i - 1];
if (i != 1) printf(" ");
printf("%d", b[i]);
}
printf("\n");
return 0;
}

二维差分

b[x1,y1]+=c,b[x2 + 1,y1]-=c,b[x1,y2 + 1]-=c,b[x2 + 1,y2 +1]+=c
#define N 1010
int n, m, q;
int a[N][N], b[N][N];
void insert(int x1, int y1, int x2, int y2, int c)
{
b[x1][y1] += c;
b[x2 +1][y1] -= c;
b[x1][y2 + 1] -= c;
b[x2 + 1][y2 + 1] += c;
}
int main()
{
int i, j, x1, y1, x2, y2, c;
scanf("%d %d %d",&n, &m, &q);
for (i = 1; i <= n; i ++ )
{
for (j = 1; j <= m; j ++ )
{
scanf("%d", &a[i][j]);
insert(i, j, i, j, a[i][j]);
}
}
while (q -- )
{
scanf("%d %d %d %d %d", &x1, &y1, &x2, &y2, &c);
insert(x1, y1, x2, y2, c);
}
for (i = 1; i <= n; i ++ )
{
for (j = 1; j <= m; j ++ )
{
b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];
if (j > 1) printf(" ");
printf("%d", &a[i][j]);
}
printf("\n");
}
return 0;
}

双指针算法

for (int i = 0, j = 0; i < n; i ++ )
{
while (j < i && check(i, j)) j ++ ;
// 具体问题的逻辑
}
常见问题分类：
  1. 对于一个序列，用两个指针维护一段区间
  2. 对于两个序列，维护某种次序，比如归并排序中合并两个有序序列的操作

位运算

求n的第k位数字: n >> k & 1
返回n的最后一位1：lowbit(n) = n & -n 得到的是2^x

离散化

vector alls; // 存储所有待离散化的值
sort(alls.begin(), alls.end()); // 将所有值排序
alls.erase(unique(alls.begin(), alls.end()), alls.end()); // 去掉重复元素
// 二分求出x对应的离散化的值
int find(int x) // 找到第一个大于等于x的位置
{
int l = 0, r = alls.size() - 1;
while (l < r)
{
int mid = l + r >> 1;
if (alls[mid] >= x) r = mid;
else l = mid + 1;
}
return r + 1; // 映射到1, 2, ...n
}
序列中可能有重复元素，需要去重
如何计算出离散化后的数

区间合并

// 将所有存在交集的区间合并
void merge(vector &segs)
{
vector res;
sort(segs.begin(), segs.end());
int st = -2e9, ed = -2e9;
for (auto seg : segs)
if (ed < seg.first)
{
if (st != -2e9) res.push_back({st, ed});
st = seg.first, ed = seg.second;
}
else ed = max(ed, seg.second);
if (st != -2e9) res.push_back({st, ed});
segs = res;
}