Summary

Posted 2020-10-17 Eventide

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暑假开始准备转移博客，试了几个都不怎么满意（我还去试了下LineBlog 不知道那时候在想什么。。），屯着一堆文章，，到时候一起发了
现在暂时转移至WordPress，不过还在完善中，预计。。算了不瞎预计的好。。

课上说最好做个代码集，嗯嗯我也觉得挺有必要的
毕竟现在我连Floyd怎么写都忘了无脑SPFA_(:з」∠)_

反正有用没用都稍微写一下，暂定是目录这些，有些还在找例题、整理代码什么的，所以还是空的。
GItHub上还欠了几题，之后会补上来。

~~我做的二级目录到博客园就被无视了，，将就看看吧~~
感觉实在简陋了些啊。。

STL
- stack
- queue
- priority_queue
- sort
- map
- set
功能函数
- MAX
- MIN
- 最大公约数
基础算法与数据结构
其他

题号以作业次数为准

stack

头文件

#include<stcak>
using namespace std;

声明

stack<数据类型> 变量名;

a.empty() 判断栈是否为空
a.pop() 移除栈顶元素
a.push(b) 将元素b压入栈中
a.size() 返回栈中元素个数
a.top() 返回栈顶元素

queue

头文件

#include<queue>
using namespace std;

声明

queue<数据类型> 变量名;

a.empty() 判断队列是否为空
a.pop() 将队头元素出队
a.push(b) 将元素b入队
a.size() 返回队列中元素个数
a.front() 返回队头元素
a.back() 返回队尾元素

priority_queue

头文件

#include<queue>
using namespace std;

声明

priority_queue<数据类型> 变量名;

a.empty() 判断队列是否为空
a.pop() 移除队头元素
a.push(b) 将元素b入队
a.size() 返回队列中元素个数
a.top() 返回队头元素

//默认从大到小
//从小到大&&多关键字
struct t
{
	int p, q;
};
priority_queue<t> a[n];
bool operator < (t x, t y)
{
	return x.p < y.p;
}

sort

头文件

#include<algorithm>
using namespace std;

//从小到大
int a[n];
sort(a,a+n);

//从大到小
int compare(int x, int y)
{
	return x > y;
}
sort(a, a + 3, compare);

//多关键字
struct t
{
	int p, q;
};
t a[n];
int compare(t x, t y)
{
	if (x.p == y.p) return x.q > y.q;
	else return x.p > y.p;
}
sort(a, a+n, compare);

功能函数

MAX
MIN
最大公约数

MAX

int max(int x, int y)
{
    return x > y ? x : y;
}

MIN

int min(int x, int y)
{
    return x < y ? x : y;
}

最大公约数

int gcd(int x, int y)
{
    if (y == 0) return x;
    else return gcd(y, x%y);
}

基础算法与数据结构

快速排序
归并排序
表达式求值（调度场算法）
线段树求区间和
AVL树（不包含删除操作）
k叉哈夫曼树（求合并n个数的最小代价）
并查集（求图的连通性）
SPFA求负权环
SPFA求多源点最短路径（可直接作单源点用）
Dijkstra
Floyd
字典树
哈希表
堆
优先队列
深搜
广搜
双向广搜
红黑树
Prim
Kruskal
KMP
ac自动机
快速幂

快速排序

#include<iostream>
using namespace std;

int i, j, k, n, m, s, t, a[1000];

void q(int l, int r)
{
	int i, j, x, t;
	i = l;
	j = r;
	x = a[(i + j) / 2];
	do
	{
		while (a[i] < x) i++;
		while (a[j] > x) j--;
		if (i <= j)
		{
			t = a[i];
			a[i] = a[j];
			a[j] = t;
			i++;
			j--;
		}
	} while (i <= j);
	if (j > l) q(l, j);
	if (i < r) q(i, r);
}

int main()
{
	cin >> n;
	for (i = 1; i <= n; i++)
		cin >> a[i];
	q(1, n);
	for (i = 1; i <= n; i++)
		cout << a[i] << \' \';
	return 0;
}

归并排序

2.1 nxd

给定 n 个数 a1,a2,...,an，求满足条件的(i,j)数量： i < j 且 a[i] < a[j]

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int a[200000], b[200000];
__int64 s;

void p(int l, int m, int r)
{
	int i = l;
	int j = m + 1;
	int k = l;
	while (i <= m && j <= r)
	{
		if (a[i] < a[j])
		{
			b[k++] = a[j++];
			s += m - i + 1;
		}
		else
		{
			b[k++] = a[i++];
		}
	}
	while (i <= m) b[k++] = a[i++];
	while (j <= r) b[k++] = a[j++];
	for (i = l; i <= r; i++)
		a[i] = b[i];
}

void q(int l, int r)
{
	if (l < r)
	{
		int m = (l + r) >> 1;
		q(l, m);
		q(m + 1, r);
		p(l, m, r);
	}
}

int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 0; i<n; i++)
		scanf("%d", &a[i]);
	s = 0;
	q(0, n - 1);
	printf("%I64d", s);
	return 0;
}

表达式求值（调度场算法）

3.2 calculator

#include<stdio.h>
#include<string.h>

int i, j, k, n, m, s, t, a[1000];
char b[2000], c[2000], d[2000];

int main()
{
	scanf("%s", &b);
	i = 0;
	j = 0;
	k = 0;
	n = strlen(b);
	//中缀转后缀
	while (i < n)
	{
		if ((b[i] >= \'0\') && (b[i] <= \'9\'))
		{
			while ((b[i] >= \'0\') && (b[i] <= \'9\'))
			{
				c[j++] = b[i++];
			}
			c[j++] = \'!\';
		}
		if ((b[i] == \'+\') || (b[i] == \'-\'))
		{
			while ((k > 0) && (d[k - 1] != \'(\'))
			{
				c[j++] = d[k - 1];
				k--;
			}
			d[k++] = b[i];
		}
		if ((b[i] == \'*\') || (b[i] == \'/\'))
		{
			while ((k > 0) && (d[k - 1] != \'(\') && ((d[k - 1] == \'*\') || (d[k - 1] == \'/\')))
			{
				c[j++] = d[k - 1];
				k--;
			}
			d[k++] = b[i];
		}
		if (b[i] == \'(\')
		{
			d[k++] = b[i];
		}
		if (b[i] == \')\')
		{
			while ((k > 0) && (d[k - 1] != \'(\'))
			{
				c[j++] = d[k - 1];
				k--;
			}
			if (k > 0) k--;
		}
		i++;
	}
	while (k > 0)
	{
		c[j++] = d[k - 1];
		k--;
	}
	//计算后缀
	c[j] = \'\\0\';
	i = 0;
	j = -1;
	while (c[i] != \'\\0\')
	{
		if ((c[i] >= \'0\') && (c[i] <= \'9\'))
		{
			double x = 0;
			while ((c[i] >= \'0\') && (c[i] <= \'9\'))
			{
				x = 10 * x + c[i] - \'0\';
				i++;
			}
			j++;
			a[j] = x;
		}
		else
		{
			j--;
			switch (c[i])
			{
			case \'+\':
			{
				a[j] += a[j + 1];
				break;
			}
			case \'-\':
			{
				a[j] -= a[j + 1];
				break;
			}
			case \'*\':
			{
				a[j] *= a[j + 1];
				break;
			}
			case \'/\':
			{
				a[j] /= a[j + 1];
				break;
			}
			}
		}
		i++;
	}
	printf("%d", a[j]);
	return 0;
}

线段树求区间和

5.2 bubble_sort

#include<stdio.h>

int i, j, k, n, m, s, t, a[300001], b[100001], c[100001];

int min(int x, int y)
{
	return x < y ? x : y;
}
int max(int x, int y)
{
	return x > y ? x : y;
}
int p(int l, int r)
{
	int s;
	s = 0;
	l += m - 1;
	r += m + 1;
	while ((l^r != 1) && (l != r))
	{
		if (l & 1 == 0) s += a[l ^ 1];
		if (r & 1 == 1) s += a[r ^ 1];
		l >>= 1;
		r >>= 1;
	}
	return s;
}

void q(int k)
{
	k >>= 1;
	while (k > 1)
	{
		a[k] = a[k << 1] + a[(k << 1) + 1];
		k >>= 1;
	}
}

int main()
{
	scanf("%d", &n);
	for (i = 1; i <= n; i++)
		scanf("%d", &b[i]);
	m = 1;
	while (m <= n) m <<= 1;
	for (i = m + 1; i <= m + n; i++)
		a[i] = 1;
	for (i = m - 1; i >= 1; i--)
		a[i] = a[i << 1] + a[(i << 1) + 1];
	for (i = 1; i <= n; i++)
	{
		t = p(1, b[i] - 1) + i;
		c[b[i]] = max(b[i], max(t, i)) - min(b[i], min(t, i));
		a[m + b[i]] = 0;
		q(m + b[i]);
	}
	printf("%d", c[1]);
	for (i = 2; i <= n; i++)
		printf(" %d", c[i]);
	return 0;
}

AVL树（不包含删除操作）

8.1 wbhavl

#include<stdio.h>
#include<stdlib.h>

int i, j, k, n, m, s, t, a[100001];

struct node
{
	int dep;
	int val;
	node *p;
	node *l;
	node *r;
};

node* insert(node *tree, int value);
void updata(node *tree);
int depth(node *tree);
node* aaaavl(node *tree, node *newp);
int papa(node *tree);
node* leftSingle(node *tree);
node* rightSingle(node *tree);
node* leftDouble(node *tree);
node* rightDouble(node *tree);
int haha(node *tree, int pp);

node* insert(node *tree, int value)
{
	node *newp, *nowp;
	newp = new node;
	newp->val = value;
	newp->p = NULL;
	newp->l = NULL;
	newp->r = NULL;
	if (tree == NULL)
	{
		newp->dep = 1;
		tree = newp;
	}
	else
	{
		nowp = tree;
		while (1 > 0)
		{
			if (newp->val <= nowp->val)
			{
				if (nowp->l == NULL)
				{
					nowp->l = newp;
					newp->p = nowp;
					break;
				}
				else
				{
					nowp = nowp->l;
					continue;
				}
			}
			else
			{
				if (nowp->r == NULL)
				{
					nowp->r = newp;
					newp->p = nowp;
					break;
				}
				else
				{
					nowp = nowp->r;
					continue;
				}
			}
		}
		updata(newp);
		tree = aaaavl(tree, newp);
	}
	return tree;
}

void updata(node *tree)
{
	if (tree == NULL) return;
	else
	{
		int l, r;
		l = depth(tree->l);
		r = depth(tree->r);
		tree->dep = 1 + (l > r ? l : r);
	}
}

int depth(node *tree)
{
	if (tree == NULL) return 0;
	else return tree->dep;
}

node* aaaavl(node *tree, node *newp)
{
	int pa;
	while (newp != NULL)
	{
		updata(newp);
		pa = papa(newp);
		if ((pa < -1) || (pa > 1))
		{
			if (pa > 1)
			{
				if (papa(newp->r) > 0)
				{
					newp = leftSingle(newp);
				}
				else
				{
					newp = leftDouble(newp);
				}
			}
			if (pa < -1)
			{
				if (papa(newp->l) < 0)
				{
					newp = rightSingle(newp);
				}
				else
				{
					newp = rightDouble(newp);
				}
			}
			if (newp->p == NULL) tree = newp;
			break;
		}
		newp = newp->p;
	}
	return tree;
}

int papa(node *tree)
{
	if (tree == NULL) return 0;
	else return depth(tree->r) - depth(tree->l);
}

node* leftSingle(node *tree)
{
	node *newroot, *mature;
	mature = tree->p;
	newroot = tree->r;
	if (newroot->l != NULL)
	{
		newroot->l->p = tree;
	}
	tree->r = newroot->l;
	updata(tree);
	newroot->l = tree;
	newroot->p = mature;
	if (mature != NULL)
	{
		if (mature->l == tree)
		{
			mature->l = newroot;
		}
		else
		{
			mature->r = newroot;
		}
	}
	tree->p = newroot;
	updata(newroot);
	return newroot;
}

node* rightSingle(node *tree)
{
	node *newroot, *mature, *naive;
	mature = tree->p;
	newroot = tree->l;
	if (newroot->r != NULL)
	{
		newroot->r->p = tree;
	}
	tree->l = newroot->r;
	updata(tree);
	newroot->r = tree;
	newroot->p = mature;
	if (mature != NULL)
	{
		if (mature->l == tree)
		{
			mature->l = newroot;
		}
		else
		{
			mature->r = newroot;
		}
	}
	tree->p = newroot;
	updata(newroot);
	return newroot;
}

node* leftDouble(node *tree)
{
	rightSingle(tree->r);
	return leftSingle(tree);
}

node* rightDouble(node *tree)
{
	leftSingle(tree->l);
	return rightSingle(tree);
}

int haha(node *tree, int pp)
{
	node *nowp;
	int qq;
	qq = 1;
	nowp = tree;
	while (nowp)
	{
		if (nowp->val > pp)
		{
			nowp = nowp->l;
			qq++;
		}
		else
		{
			if (nowp->val < pp)
			{
				nowp = nowp->r;
				qq++;
			}
			else break;
		}
	}
	return qq;
}

int main()
{
	node *tree, *now;
	int val;
	tree = NULL;
	scanf("%d", &n);
	for (i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
		tree = insert(tree, a[i]);
	}
	printf("%d", haha(tree, a[0]));
	for (i = 1; i < n; i++)
		printf(" %d", haha(tree, a[i]));
	return 0;
}

k叉哈夫曼树（求合并n个数的最小代价）

也可用堆或优先队列

9.1 hbsz

#include<stdio.h>
#include<algorithm>
using namespace std;

int i, j, k, n, m, s, t, b[100002];
short int a[100002];

int main()
{
	scanf("%d", &n);
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	sort(a, a + n);
	t = 0;
	i = 0;
	j = 0;
	s = 0;
	while (n - i + t - j > 1)
	{
		m = 0;
		for (k = 0; k < 2; k++)
		{
			if (i == n)
			{
				m += b[j];
				j++;
			}
			else
				if (j == t)
				{
					m += a[i];
					i++;
				}
				else
					if (a[i] < b[j])
					{
						m += a[i];
						i++;
					}
					else
					{
						m += b[j];
						j++;
					}
		}
		s += m;
		b[t] = m;
		t++;
	}
	printf("%d", s);
	return 0;
}

并查集（求图的连通性）

10.2 friends

#include<stdio.h>

struct node
{
	int x, y;
};

node e[50010];

int i, j, k, n, m, s, t, x, y, d, l, a[50010], b[50010], f[50010], c[50010], p[50010], q[50010];

int aaaa(int x)
{
	return f[x] == x ? x : f[x] = aaaa(f[x]);
}

void qqq(int x)
{
	int i, pp, qq;
	pp = aaaa(x);
	i = a[x];
	while (i != 0)
	{
		if (p[e[i].y])
		{
			qq = aaaa(e[i].y);
			if (pp != qq)
			{
				t--;
				f[qq] = pp;
			}
		}
		i = e[i].x;
	}
}

int main()
{
	scanf("%d%d", &n, &m);
	for (i = 0; i < n; i++)
	{
		f[i] = i;
	}
	l = 0;
	for (i = 0; i < m; i++)
	{
		scanf("%d%d", &x, &y);
		l++;
		e[l].x = a[x];
		a[x] = l;
		e[l].y = y;
		l++;
		e[l].x = a[y];
		a[y] = l;
		e[l].y = x;
	}
	scanf("%d", &d);
	for (i = 1; i <= d; i++)
	{
		scanf("%d", &b[i]);
		c[b[i]] = 1;
	}
	t = 0;
	for (i = 0; i < n; i++)
	{
		if (!c[i])
		{
			t++;
			qqq(i);
			p[i] = 1;
		}
	}
	q[d + 1] = t;
	for (i = d; i >= 1; i--)
	{
		t++;
		qqq(b[i]);
		p[b[i]] = 1;
		q[i] = t;
	}
	for (i = 1; i <= d + 1; i++)
	{
		printf("%d\\n", q[i]);
	}
	return 0;
}

SPFA求负权环

11.1 CrazyScientist

#include<stdio.h>

int i, j, k, n, m, s, t, p, a[2010], b[80010][3], c[2010];
bool d[2010];

void q(int k)
{
	int i, j;
	d[k] = true;
	i = a[k];
	while (i != 0)
	{
		j = b[i][0];
		if (c[k] + b[i][1] < c[j])
		{
			c[j] = c[k] + b[i][1];
			if ((d[j] == true) || (p == 1))
			{
				p = 1;
				if (d[s] == true)
				{
					t = 1;
				}
				break;
			}
			q(j);
		}
		i = b[i][2];
	}
	d[k] = false;
}

int main()
{
	scanf("%d%d", &n, &m);
	for (i = 1; i <= n; i++)
	{
		a[i] = 0;
		c[i] = 0;
		d[i] = false;
	}
	s = 0;
	for (i = 1; i <= m; i++)
	{
		scanf("%d%d%d", &j, &k, &t);
		s++;
		b[s][0] = k;
		b[s][1] = t;
		b[s][2] = a[j];
		a[j] = s;
	}
	scanf("%d", &s);
	t = 0;
	for (i = 1; i <= n; i++)
	{
		p = 0;
		q(i);
		if (t == 1) break;
	}
	if (t == 1)
		printf("EL PSY CONGROO");
	else
		printf("ttt");
	return 0;
}

SPFA求多源点最短路径（可直接作单源点用）

11.2 FuYihao

#include<stdio.h>
#include<string.h>

int i, j, k, n, m, s, t, q, a[410][410] = { 0 }, b[410][410] = { 0 }, c[410], d[200010], e[410][410];
bool f[410];

void sasasa(int k)
{
	int i, j, h, t;
	if (k > 1)
	{
		j = 1;
		for (i = 2; i < k; i++)
			if (e[i][k] < e[j][k]) j = i;
		for (i = 1; i <= n; i++)
			e[k][i] = e[j][k] + e[j][i];
	}
	e[k][k] = 0;
	f[k] = true;
	d[1] = k;
	h = 0;
	t = 1;
	while (h < t)
	{
		h++;
		j = d[h];
		f[j] = false;
		for (i = 1; i <= n; i++)
		{
			if (e[k][i] > e[k][j] + a[j][i])
			{
				e[k][i] = e[k][j] + a[j][i];
				if (f[i] == false)
				{
					t++;
					d[t] = i;
					f[i] = true;
				}
			}
		}
	}
}

int main()
{
	memset(a, 0x3f, sizeof(a));
	memset(e, 0x3f, sizeof(e));
	scanf("%d%d", &n, &m);
	for (i = 0; i < m; i++)
	{
		scanf("%d%d%d", &j, &k, &t);
		if ((a[j][k] != 0) && (t > a[j][k])) continue;
		a[j][k] = t;
		a[k][j] = t;
	}
	scanf("%d", &q);
	for (i = 1; i <= n; i++)
	{
		memset(f, 0, sizeof(f));
		sasasa(i);
	}
	while (q--)
	{
		scanf("%d%d", &j, &k);
		if (e[j][k] != 0x3f3f3f3f)
		{
			if (q > 0) printf("%d\\n", e[j][k]);
			else printf("%d", e[j][k]);
		}
		else
		{
			if (q > 0) printf("-1\\n");
			else printf("-1");
		}
	}
	return 0;
}

Dijkstra

直接手打的

#include<iostream>
#include<cstring>
using namespace std;

int i, j, k, n, m, s, t, x, y, a[100][100], b[100] = { 0 }, d[100];

int main()
{
	memset(a, 0x3f, sizeof(a));
	cin >> n >> m;
	for (i = 0; i < m; i++)
	{
		cin >> x >> y >> t;
		a[x - 1][y - 1] = t;
		a[y - 1][x - 1] = t;
	}
	cin >> x >> y;
	x--;
	y--;
	for (i = 0; i < n; i++)
		d[i] = a[x][i];
	b[x] = 1;
	d[x] = 0;
	for (i = 0; i < n - 1; i++)
	{
		t = 0x3f3f3f3f;
		k = -1;
		for (j = 0; j < n; j++)
			if ((b[j] == 0) && (d[j] < t))
			{
				k = j;
				t = d[j];
			}
		if (k == -1) break;
		b[k] = 1;
		for (j = 0; j < n; j++)
			if (d[k] + a[k][j] < d[j])
				d[j] = d[k] + a[k][j];
	}
	cout << d[y];
	return 0;
}

Floyd

11.2 FuYihao

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

int i, j, k, n, m, s, t, a[410][410];

int main()
{
	memset(a, 0x3f, sizeof(a));
	cin >> n >> m;
	for (i = 1; i <= n; i++)
		a[i][i] = 0;
	for (i = 0; i < m; i++)
	{
		scanf("%d%d%d", &j, &k, &t);
		if (t < a[j][k])
		{
			a[j][k] = t;
			a[k][j] = t;
		}
	}
	for (k = 1; k <= n; k++)
		for (i = 1; i <= n; i++)
			for (j = 1; j <= n; j++)
				if (a[i][j] > a[i][k] + a[k][j])
				{
					a[i][j] = a[i][k] + a[k][j];
				}
	cin >> t;
	for (i = 0; i < t; i++)
	{
		cin >> j >> k;
		if (a[j][k] == 0x3f3f3f3f) printf("%d\\n", -1);
		else printf("%d\\n", a[j][k]);
	}
	return 0;
}

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