Zenith_Habitant的博客

题目链接：https://www.luogu.com.cn/problem/P2657解题思路：数位DP模板题，按位进行记忆优化搜索，详情见代码const int N = 15;int len;string L , R, v;ll dp[N][N][2][2];//cur表示当前位，x代表上一位填的什么//f表示从[0,cur-1]前缀是否和给定上界相同//g表示是否还在填充前导0ll dfs (int cur ,int x, bool f, bool g){ if (cu.

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&.

状态压缩DP观察数据范围提取题目中状态关系设计状态转移方程-代码套路先计算出所有合法状态再根据情况处理出所有合法状态可以转移到的合法状态经典例题 (棋盘式)SGU223//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define sz(a) (int)a.size()

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

//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define sz(a) (int)a.size()#define INF 0x3f3f3f3f#define DNF 0x7f7f7f7f#define DBG printf("this is a input\n")#define fi f.

//#define LOCAL#include <bits/stdc++.h>using namespace std;#define ll long long#define mem(a, b) memset(a,b,sizeof(a))#define sz(a) (int)a.size()#define INF 0x3f3f3f3f#define DNF 0x7f7f7f7f#define DBG printf("this is a input\n")#define fi f.

#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 DBG printf("this is a input\n")#define fi first#define se second#define mk(a, b) make_pair(a,b)#define p_queue priority_queue...

//#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...

dp[i][j]:整体经过状态为i时，终点在j的最短路#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 inp...

#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;struct node{ int x , y, index;}cu[100005];bool cmp(node a ,node b){ return a.x < b.x;}int dp[200005];int ca[200005];int ans[20000...

#include <algorithm>#include <iostream>#include <string.h>#include <stdio.h>#include <vector>#include <queue>#include <set>#include <map>using...

#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")string s;int match[800]...

#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")int num[100];ll dp[30...

#include&lt;stdio.h&gt;#include&lt;iostream&gt;#include&lt;algorithm&gt;#include&lt;string.h&gt;#include&lt;bits/stdc++.h&gt;using namespace std;int dp[20][20000],num[22][22], True[20000];int n;bool s...

#include&lt;stdio.h&gt;#include&lt;iostream&gt;#include&lt;algorithm&gt;#include&lt;string.h&gt;using namespace std;#define mod 100000000int n , m;int False[20000],True[20000],dp[14][20000];bool searc...

借鉴了其他博客的思路整理了一下代码#include&lt;stdio.h&gt;#include&lt;iostream&gt;#include&lt;algorithm&gt;#include&lt;string.h&gt;using namespace std;#define INF 9999999int dp[105][105];int num[105];int main(){    int...

不知道是听谁说的，这是一道区间DP的入门题，对于区间思想确实是一道入门题，但是DP方程个人觉得还是很难的，反正我这种蒟蒻肯定是想不出来的。嘤首先感谢资料：https://blog.youkuaiyun.com/libin56842/article/details/9722077来看一下题目：有一群人，按照给定输出从1到n排好了顺序，准备一个接一个的出去.此外旁边还一个小黑屋(一个类似于栈的东西),可以用...

#include&lt;stdio.h&gt;#include&lt;iostream&gt;#include&lt;string.h&gt;#include&lt;algorithm&gt;#include&lt;math.h&gt;using namespace std;int dp[105][105];int ans[200];int main(void){    str...