Zenith_Habitant的博客

题目链接：https://codeforces.com/problemset/problem/1422/D稍微思考一下可以很直接的得到，最后的答案无非两种情况：从瞬移点普通的走到终点从起点普通的走到终点由此我们可以转换问题，我们只需要求出起点到所有瞬移点的最小距离就好了这里可以发现，从一个点到一个点和两个点之间纵坐标，横坐标的差值相关，这里我们把每个点拆成3个，一个点代表瞬移点的横坐标，一个点作为瞬移点的纵坐标，然后横纵坐标都链接瞬移点。瞬移点链接终点，起点链接瞬移点横坐标，瞬移点纵坐标，.

题目链接：https://codeforces.com/problemset/problem/1442/C解题思路：发现当反转18次的时候，代价是超过最大的m的，所以分开讨论，首先计算出从起点到达终点最少需要多少次反转，如果超过18次，按最少次数转移，同时计算出最少次数下的最短路，当小于18次时，这个时候，所需要的所有代价不会超过long long（所以取模后不会丢失大小关系），所以可以直接使用dijkstra求解//#define LOCAL#include <bits/stdc++.h&.

对于一道题目，如果涉及到了位运算，二进制，那么优先考虑二进制，观察其性质本题以对点权的二进制构造边，构造32个虚点，从而建立出图堆优化优先队列，一定不要出现爆LL的情况，这样极有可能MLE，TLE，因为入队次数会变得乱糟糟的//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#defi.

//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DNF 0x7f#define DBG printf("this is a input\n")#define fi first#define se second#define mk(.

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define p_queue priority_queue#define DBG printf("this is a input\n")#define fi first#define se second#define mk(a.

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define p_queue priority_queue#define DBG printf("this is a input\n")#define fi first#define se second#define mk(a.

对于差分约束的题目，重于挖掘题目暗含的约束条件, 当图存在负环时无解1.最大化xn-x1 正常建边，跑最短路2.最小化xn-x1 反向建边，跑最长路//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,...

//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DNF 0x7f#define DBG printf("this is a...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

struct node{ int v; ll value; node(int a, ll b):v(a),value(b){}; bool operator < (const node& no) const { return value > no.value; }};void dijkstra(){ m...

#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define DBG printf("this is a input\n")#define fi first#defi...

#include <queue>#include <iostream>#include <algorithm>#include <string.h>#include <stdio.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,...