K短路模板,A*+SPFA求K短路.A*中h的求法为在反图中做SPFA,求出到T点的最短路,极为估价函数h(这里不再是估价,而是准确值),然后跑A*,从S点开始(此时为最短路),然后把与S点能达到的点加入堆中,维护堆,再从堆顶取当前g值最小的点(此时为第2短路),再添加相邻的点放入堆中,依此类推······保证第k次从堆顶取到的点都是第k短路(至于为什么,自己想)其实就是A*算法,这里太啰嗦了 1 #include<queue> 2 #include<cstdio> #includ…

采用A*算法的k短路模板 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> #include <queue> using namespace std; const int MAXN=200005; int init(){ int rv=0,fh=1; char c=getchar(); whi…

求第k短路模板 先逆向求每个点到终点的距离,再用dij算法,不会超时(虽然还没搞明白为啥... #include<iostream> #include<cstdio> #include<cmath> #include<queue> #include<vector> #include<string.h> #include<cstring> #include<algorithm> #include<set&g…

解题关键:k短路模板题,A*算法解决. #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<queue> using namespace std; typedef long long ll; ; ; const int inf=1e9; str…

第k*短路模板(单项边) #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> #define Max 100005 #define inf 1<<28 using namespace std; int S,T,K,n,m; int head[Max],rehead[Max]; int num,ren…

  Time Limit: 4000MS   Memory Limit: 65536K Total Submissions:32863   Accepted: 8953 Description "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a st…

题目链接:http://poj.org/problem?id=2449 Time Limit: 4000MS Memory Limit: 65536K Description "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly touching his little ducks' head, he told them a story. &quo…

题目链接 K短路居然用A*……奇妙. 先建反图从终点(1)跑一遍最短路,再A*,用堆存当前点到终点距离+从起点到当前点距离. 每次取出终点都可以视为发现了一个新的最短路. #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<cctype> #include<queue> #define maxn 1020 #define m…

Remmarguts' Date http://poj.org/problem?id=2449 Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 30772   Accepted: 8397 Description "Good man never makes girls wait or breaks an appointment!" said the mandarin duck father. Softly tou…

1.到底如何求k短路的? 我们考虑,要求k短路,要先求出最短路/次短路/第三短路……/第(k-1)短路,然后访问到第k短路. 接下来的方法就是如此操作的. 2.f(x)的意义? 我们得到的f(x)更小,优先访问这个f(x)的点. 我们可以定义一组数{p,g,h},p是某一个点,g是估价,h是实际,那么g+h更小的点p会优先访问. 为什么呢?因为假设我们求出了w短路,接下来要求(w+1)短路,就要求最小的另一条路径. 应该易理解. 3.为什么选择最短路来估价? 很简单的选择,我们既然要求最短了,当…