有用的板子

常用技巧

inline ll read(){
	ll x = 0, w = 1;
	char ch = getchar();
	while(ch < '0' || ch > '9'){
		if(ch == '-')	w = -1;
		ch = getchar();
	}
	while(ch >= '0' && ch <= '9'){
		x = (x << 3) + (x << 1) + (ch - '0');
		ch = getchar();
	} 
	return x * w;
} // 快读

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
}// 关同步，可以用 cin 输入，用 printf 输出

//对拍
#include <iostream>
#include <cstdio>
#include <stdlib.h>
using namespace std;

ll cnt=0;

int main(){
	while(true){
		cnt ++ ;
		system("data.exe > T.in");
		system("baoli.exe < T.in > baoli.out");
		system("std.exe < T.in > std.out");
		if(system(fc std.out baoli.out)){
			cerr << "WA on test" << cnt << endl;
			system("pause");
		}
		cerr << "AC on test" << cnt << endl;
	}
}

// 离散化
std::vector<ll> base, pos;
std::sort(pos.begin(), pos.end());
pos.erase(std::unique(pos.begin(), pos.end()), pos.end());
for(int i = 0; i < n; ++i){
    base[i] = std::lower_bound(pos.begin(), pos.end(), base[i]) - pos.begin();
}

图论相关

最短路

struct edge{
	ll v, w;
};
struct node{
	ll dis,u;
	bool operator < (const node& a) const{
		return dis > a.dis;
	}
}

vector<edge> e[maxn];
ll dis[maxn], vis[maxn];
priority_queue<node> q;

void dijkstra(ll n,ll s){
	for(int i = 1; i <= n; i++){
		dis[i] = INF, vis[i] = false;
	}
	dis[s] = 0; q.push((node){0,s});
	while(!q.empty()){
		ll u = q.top(); q.pop();
		if(vis[u])	continue;
		vis[u] = 1;
		for(auto to : e[u]){
			ll v = to.v, w = to.w;
			if(dis[v] > dis[u] + w){
				dis[v] = dis[u] + w;
				p.push(node{dis[v],v});
			}
		}
	}
}// Dijkstra

// Floyd
for(int k = 1; k <= n; k++){
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            f[i][j] = std::min(f[i][j], f[i][k] + f[k][j]);
        }
    }
}

树的直径

void dfs(ll u,ll fa){
	for(auto v : e[u]){
		if(v == fa)	continue;
		d[v] = d[u] + 1;
		if(d[v] > d[c]) c = v;
		dfs(v, u);
	}
}
int main(){
	dfs(1,0);
	d[c] = 0;
	dfc(c,0);
} // 两次dfs

void dfs(ll u, ll fa){
	d1[u] = d2[u] = 0;
	for(auto v : e[u]){
		if(v == fa)	continue;
		dfs(v, u);
		ll tmp = d1[v] + 1;
		if(t > d1[u]){
			d2[u] = d1[u];
			d1[u] = tmp;
		}
		else if(tmp > d2[u]){
			d2[u] = tmp;
		}
	}
	d = d > d1[u] + d2[u] ? d : d1[u] + d2[u];
}
int main(){
	dfs(1,0);
	cout << d;
}// 树形DP

数学相关

inline ll binpow(ll a,ll b){
    a %= mod;
	ll res = 1;
    while(b > 0){
        if(b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return r
} // 快速幂