Dij:贪心思想的单源最短路,时间复杂度O(n^2). Dij算法流程: d数组记录源点s到每个点的距离,若无边则设为inf,标记源点: 选出d数组中未标记的最小值,该节点记为k,并标记k为已求出最短路: 枚举每个节点(记为j),若经过k到达j的路径<d[j]且未标记,则更新d[j]; 重复2.3步n次: d[v]为s-v的最短路: 堆优Dij:即用堆优化的dij算法,时间复杂度O(nlogn);(但是据说跑起来比spfa快?求神犇解释) 堆优Dij算法流程: q为priority_queue,…

#include<cstdio> #include<vector> #include<queue> using namespace std; struct edge { int to,val; }; priority_queue<pair<int,int>,vector<pair<int,int> >,greater<pair<int,int> > >q; vector<edge>e[2…

//归并排序求逆序对 #include<bits/stdc++.h> #define ll long long using namespace std; ]; ll ans; ]; void merge_sort(int l,int r) { if(l==r)return; ; ,cnt=l; merge_sort(l,mid); merge_sort(mid+,r); while(i<=mid&&j<=r) { if(a[i]<=a[j])b[cnt++]=…

HDU Today Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6147    Accepted Submission(s): 1485 Problem Description 经过锦囊相助,海东集团最终度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强.这时候.…

我仅仅是贴一下手写堆优化的dij模板.尽管.它.TLE了--**** #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define N 2001000 #define inf 0x3f3f3f3f #define longlong int using namespace std; struct Katarina { int v,next; longl…

BZOJ4152 The Captain 题面很简洁: 给定平面上的n个点,定义(x1,y1)到(x2,y2)的费用为min(|x1-x2|,|y1-y2|),求从1号点走到n号点的最小费用. 很明显这是道最短路 我们知道在求最短路时本身就要不断求min 所以相对于拐弯抹角的横纵差的较小值,我们完全可以无视这个求min 转为建|x1-x2|,|y1-y2|两条边 最后跑一遍最短路即可 (ps.这题是卡SPFA的) IO优化.头文件啥的就自己yy一下吧 #define int long long…

Background Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place…

题目描述 You are given a weighed undirected connected graph, consisting of n vertices and mm edges. You should answer q queries, the i-th query is to find the shortest distance between vertices ui and vi. Input The first line contains two integers n and…

#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <cstring> #include <cmath> #include <vector> #include <queue> #include <stack> #include <set> #include <m…

题意:有n个电梯,目的地是第K层(起点是第0层),给出每个电梯的速度,以及每个电梯能到达的层数,如果中途需要换电梯的话,时间需要+60,求到达目的地的最短时间: 思路:Dij求最短路.如果是另一条路比较短的话,相当于乘别的电梯,也就是再+60,自己不可能和自己比,即d[e.v] 和 d[e.u] + d + 60比较即可: #include <iostream> #include <cstdio> #include <cstring> #include <stri…