模板
bool isPrime(int n) {
if (n == 0 || n == 1) return false;
for (int i = 2; i * i <= n; i++)
if (n % i == 0) return false;
return true;
}
中序加后序建树  
Djstl最短路:
const int N = 1010, INF = 0x3f3f3f3f;
int g[N][N];
int a[N];
int n, m;
int check()
{
int used[N] = {0},dist[N] = {0};
memset(dist,0x3f,sizeof dist);
dist[a[1]] = 0;
for(int i = 1 ; i <= n; i ++)
{
int u = -1;
for(int v = 1 ; v <= n ; v ++)
if(!used[v] && (u == -1 || dist[u] > dist[v] )) u = v;
if(dist[u] != dist[a[i]])
return 0;
used[u] = 1;
for(int v = 1 ; v <= n; v ++)
if(v == u) continue;
else dist[v] = min(dist[v],dist[u]+g[u][v]);
}
return 1;
}
int main()
{
scanf("%d %d", &n, &m);
memset(g, 0x3f, sizeof(g));
for(int i = 0 ; i < m ; i++)
{
int a, b, c;
scanf("%d %d %d", &a, &b, &c);
g[a][b] = g[b][a] = c;
}
int k;
scanf("%d",&k);
for(int i = 0 ; i < k ; i++)
{
for(int j = 1 ;j <= n ; j++)
scanf("%d", &a[j]);
if(check()) printf("Yes\n");
else printf("No\n");
}
return 0;
}
 
DFS
 
二叉树
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vector<int> t[100];
int d[100]={0},manh=0;
void dfs(int index,int depth){
manh=max(manh,depth);
if(t[index].size()==0){
d[depth]++;
return;
}
for(int i=0;i<t[index].size();i++)
dfs(t[index][i],depth+1);
}
dfs(1,1);
BFS
 
最短路
Dijkstra迪杰斯特啦
  
并查集

string类
 c_str() 以 char* 形式传回 string 内含字符串 如果一个函数要求char*参数,可以使用c_str()方法: string s = “Hello World!”; printf(“%s”, s.c_str()); //输出 “Hello World!” string反转: string s = "hello";reverse(s.begin(),s.end());
vector
 v1.resize(len,value) :长度为len,len个元素值都为value 
 
set

map
 如果偶尔刷题时候?map或者set超时了,可以考虑? unordered_map(或者unordered_set)缩短代码运?时间、提?代码效率~?于?法和map、set map.find()!=map.end() 是?样的~
else if (num == 3) {
unordered_map<string, int> m;
for (int j = 0; j < n; j++)
if (v[j].t.substr(4, 6) == s) m[v[j].t.substr(1, 3)]++;
for (auto it : m) ans.push_back({it.first, it.second});
}
stack

queue

sort
  对vector排序: std::sort(s.begin(),s.end(),cmp); vector<string> s;
直接set.find()!=set.end() 不就好了吗??vector遍历又麻烦又超时!
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tostring
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stoi、stod可以将字符串string转化为对应的int型、double型变量
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