Summary


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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。。）

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
  • 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
  • 其他

题号以作业次数为准

STL

  • stack
  • queue
  • priority_queue
  • sort
  • map
  • set

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

时间： 2024-10-14 08:33:51

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