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
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
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;
}