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…

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…

给一手链接 https://www.luogu.com.cn/problem/P2865 这道题其实就是在维护最短路的时候维护一下次短路就okay了 #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<queue> using namespace std; ,inf=1e9; int read(){ ,f=,c=getchar(); ; c…

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…

·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…

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…

一个次短路的问题,可以套用dijkstra求最短路的方法,用dis[0][i]表示最短路:dis[1][i]表示次短路,优先队列中存有最短路和次短路,然后每次找到一条道路对他进行判断,更新最短或次短路, 注意求次短路时不要打标记,因为有可能再次访问到该节点. 最后的答案就是dis[1][n]. 1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 200010 4 int tot,head[N],to[N],w[N],nxt[…

315. [POJ3255] 地砖RoadBlocks ★★★   输入文件:block.in   输出文件:block.out   简单对比时间限制:1 s   内存限制:128 MB 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 to her old home too q…

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 = -…