题意 Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather…

P2865 [USACO06NOV]路障Roadblocks 最短路(次短路) 直接在dijkstra中维护2个数组:d1(最短路),d2(次短路),然后跑一遍就行了. attention:数据有不同权值的重边(40ptsQAQ) #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<cctype> using namespace std; t…

P2865 [USACO06NOV]路障Roadblocks 题目描述 Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided t…

P2865 [USACO06NOV]路障Roadblocks 题目描述 Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided t…

传送门 算法Dijkstra要求次短路 那么在不考虑重复走一条边的情况下 肯定是把最短路中的一段改成另一段 至少要换另一条边到路径里所以可以枚举所有不属于最短路的每条边(a,b) 那么dis(1,a)+(a,b)+ dis(b,n)就是一种可能的答案(记为S) 显然如果另一条不属于S的边更新S后会使S更长,就不可能为次短路了 那么只要对起点1和终点n分别跑Dijkstra就可以求出每个dis(1,a)和dis(b,n) 至于判断一条边是否在最短路上也很容易: 显然,如果dis(1,a)+(a,b…

题目链接 次短路模板题. 对每个点记录最短路和严格次短路,然后就是维护次值的方法了. 和这题一样. #include <cstdio> #include <queue> #include <cstring> using namespace std; inline int read(){ int s = 0, w = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') w = -…

链接:https://www.luogu.org/problemnew/show/P2865 题目描述 Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. S…

注意:如果是这么个写法,堆数组要开成n+m的. 为什么呢?设想一下从1到2有m条长度递减的路,这岂不是要入队m次-- #include <algorithm> #include <iostream> #include <cstring> #include <cstdio> using namespace std; struct Edge{ int too, nxt, val; }edge[200005]; struct Node{ int idd, val;…

·求1到n的严格次短路. [题解] dijktra魔改?允许多次入队,改了次短路的值也要入队. #include<cstdio> #include<algorithm> #define M 200010 #define N 5010 #define rg register using namespace std; int n,m,tot,last[N],dis[N],d2[N],pos[N]; struct edge{ int to,pre,dis; }e[M]; struct h…

http://poj.org/problem?id=3255 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15680   Accepted: 5510 Description Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get…