题意是给你点与点之间的距离,求来回到点1的最短路中的边权和。 因为边很大,不能用原来的dijkstra什么的,所以用spfa来做。并且注意要用long long int 来存储。 稍微改了一下学长的模板。 stack stl 实现代码: #include<stdio.h>#include<stack>using namespace std;const int M
传送门:【HDU】4876 ZCC loves cards 题目分析: 这题无力吐嘈。。。。能想到的优化都用上吧。。。 代码如下: #include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define REP( i , n ) for ( int i =
题意 有 n n n 个元素,第 i i i 个元素有两个权值 a i a_i ai 和 b i b_i bi;有 m m m 次操作,每次操作会交换两个元素的位置,且都需要回答:是否存在一种方案,使得每个元素各选择一个权值后,组成的序列从左到右单调不降。 解法 完全可以把交换操作看作两次单点修改,每次只需要考虑一个元素的变化对答案的影响即可。对于一个区间中的元素,显然开
Patchouli's Spell Cards Time Limit: 7 Seconds Memory Limit: 65536 KB Patchouli Knowledge, the unmoving great library, is a magician who has settled down in the Scarlet Devil Mansion (紅魔館)
原题: Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck: Thro
//用vector写邻接表无情的TLE,只好用静态邻接表,两次SPFA分别求去和回的最短路 #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<vector> using namespace std; const int INF=0x3f3f3f3f; const int maxn=1000
Nauuo and Cards 传送门 Nauuo is a girl who loves playing cards. One day she was playing cards but found that the cards were mixed with some empty ones. There are n n n cards numbered from 1 1 1 to
A. Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output There are n cards (n is even) in the deck. Each card has a po
题目描述 Eddy likes to play cards game since there are always lots of randomness in the game. For most of the cards game, the very first step in the game is shuffling the cards. And, mostly the randomnes
传送门 Patchouli Knowledge, the unmoving great library, is a magician who has settled down in the Scarlet Devil Mansion (紅魔館). Her specialty is elemental magic employing the seven elements fire, wat
文章目录 一、题目二、题解 一、题目 There are several cards arranged in a row, and each card has an associated number of points. The points are given in the integer array cardPoints. In one step, you can ta