BFS. /* 1484 */ #include <iostream> #include <queue> #include <string> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef struct { int x, y; string s; } node_t; ][][]; ][]; ] = "…

Basic wall maze Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 168    Accepted Submission(s): 52 Special Judge Problem Description In this problem you have to solve a very simple maze consisti…

这道题目巨坑啊,注意__int64,int wa了一个下午. #include <cstdio> #include <cstring> #include <cstdlib> #define MAXN 105 char vals[] = "0123456789abcdef"; char s[MAXN]; int l; bool flag; bool check(char ch, int base) { ) ); else ) ); } int getv…

Basic Wall Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3384   Accepted: 1525   Special Judge Description In this problem you have to solve a very simple maze consisting of: a 6 by 6 grid of unit squares 3 walls of length between…

[LeetCode]227. Basic Calculator II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/basic-calculator-ii/description/ 题目描述: Implement a basic calculator to evaluate a simple exp…

[LeetCode]554. Brick Wall 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/brick-wall/description/ 题目描述: There is a brick wall in front of you. The wall is rectangular and has…

计算几何-凸包模板题目,Graham算法解. /* 1348 */ #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> using namespace std; #define MAXN 1005 typedef struct Point_t { double…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

DP/四边形不等式 裸题环形石子合并…… 拆环为链即可 //HDOJ 3506 #include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define…

DP/四边形不等式 这题跟石子合并有点像…… dp[i][j]为将第 i 个点开始的 j 个点合并的最小代价. 易知有 dp[i][j]=min{dp[i][j] , dp[i][k-i+1]+dp[k+1][j-(k-i+1)]+w(i,k,j)} (这个地方一开始写错了……) 即,将一棵树从k处断开成(i,k)和(k+1,i+j-1)两棵树,再加上将两棵树连起来的两条树枝的长度w(i,k,j) 其中,$ w(i,k,j)=x[k+1]-x[i]+y[k]-y[i+j-1] $ 那么根据四边形…