题目传送门 Roadblocks 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 quickly, because she likes the scenery along the way. She has decided to take…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7760   Accepted: 2848 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…

题目大意:求图的严格次短路. 方法1: SPFA,同时求单源最短路径和单源次短路径.站在节点u上放松与其向量的v的次短路径时时,先尝试由u的最短路径放松,再尝试由u的次短路径放松(该两步并非非此即彼). 由u的最短路径放松: if(u->Dist + e->Weight < v->Dist) v->Dist2=v->Dist; //此处隐藏最短路放松.次短路不在此固定,Dist2可能在由u次短路放松时被放松得更短 if(u->Dist + e->Weight…

Dijkstra最短路算法 --转自啊哈磊[坐在马桶上看算法]算法7:Dijkstra最短路算法 上节我们介绍了神奇的只有五行的Floyd最短路算法,它可以方便的求得任意两点的最短路径,这称为“多源最短路”.本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做“单源最短路径”.例如求下图中的1号顶点到2.3.4.5.6号顶点的最短路径. 与Floyd-Warshall算法一样这里仍然使用二维数组e来存储顶点之间边的关系,初始值如下. 我们还需要用一个一维数组dis来存储1号顶点到其余各…

dijkstra(最短路)和Prim(最小生成树)下的堆优化 最小堆: down(i)[向下调整]:从第k层的点i开始向下操作,第k层的点与第k+1层的点(如果有)进行值大小的判断,如果父节点的值大于子节点的值,则修改,并继续对第k+1层与第k+2层的点进行判断和修改,否则不修改,且退出.当点向下移动到树的最后一层,没有子节点供判断与修改,停止操作. 树最多有log(n) 层[log(n)=log2n,一般省略数字2],时间复杂度log(n)次. up(i)[向上调整]:同理,时间复杂度log(…

       上周我们介绍了神奇的只有五行的Floyd最短路算法,它可以方便的求得任意两点的最短路径,这称为“多源最短路”.本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做“单源最短路径”.例如求下图中的1号顶点到2.3.4.5.6号顶点的最短路径.           与Floyd-Warshall算法一样这里仍然使用二维数组e来存储顶点之间边的关系,初始值如下.           我们还需要用一个一维数组dis来存储1号顶点到其余各个顶点的初始路程,如下.          …

题目: POJ3255 洛谷2865 分析: 这道题第一眼看上去有点懵-- 不过既然要求次短路,那估计跟最短路有点关系,所以就拿着优先队列优化的Dijkstra乱搞,搞着搞着就通了. 开两个数组:\(dis\)存最短路,\(dis2\)存次短路 在松弛的时候同时更新两个数组,要判断三个条件 (\(u\)是当前考虑的点,\(v\)是与\(u\)有边相连的点,\(d(u,v)\)表示从\(u\)到\(v\)的边长) 1.如果\(dis[v]>dis[u]+d(u,v)\),则更新\(dis[v]\)…

Roadblocks 直接翻译了 Descriptions Bessie搬到了一个新的农场,有时候他会回去看他的老朋友.但是他不想很快的回去,他喜欢欣赏沿途的风景,所以他会选择次短路,因为她知道一定有一条次短路.这个乡村有R(1<=R<=100000)条双向道路,每一条连接N(1<=N<=5000)个点中的两个.Bessie在1号节点,他的朋友家是n号节点Input第一行:两个整数N和R接下来R行:每行包含三个整数,A,B,D,表示一条连接A与B的长度为D的路径Output输出1到…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10098   Accepted: 3620 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…

Roadblocks Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13594   Accepted: 4783 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…

题目描述 现在是晚餐时间,而母牛们在外面分散的牧场中. 农民约翰按响了电铃,所以她们开始向谷仓走去. 你的工作是要指出哪只母牛会最先到达谷仓(在给出的测试数据中,总会有且只有一只最快的母牛). 在挤奶的时候(晚餐前),每只母牛都在她自己的牧场上,一些牧场上可能没有母牛. 每个牧场由一条条道路和一个或多个牧场连接(可能包括自己). 有时,两个牧场(可能是字母相同的)之间会有超过一条道路相连. 至少有一个牧场和谷仓之间有道路连接. 因此,所有的母牛最后都能到达谷仓,并且母牛总是走最短的路径. 当然,…

最朴素的做法o(V*V/2+2E)~O(V^2)#include<iostream>using namespace std;#include<vector>#include<algorithm>#include<string>#include<string.h>const int MAX = 2002;int n;int graph[MAX][MAX];int dis[MAX];bool vis[MAX];const int INF = 0X7F…

上周我们介绍了神奇的只有五行的 Floyd 最短路算法,它可以方便的求得任意两点的最短路径,这称为"多源最短路".本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做"单源最短路径".例如求下图中的 1 号顶点到 2.3.4.5.6 号顶点的最短路径. 与 Floyd-Warshall 算法一样这里仍然使用二维数组 e 来存储顶点之间边的关系,初始值如下. 我们还需要用一个一维数组 dis 来存储 1 号顶点到其余各个顶点的初始路程,如下. 我们将此时 d…

背景 USACO OCT09 9TH 描述 德克萨斯纯朴的民眾们这个夏天正在遭受巨大的热浪!!!他们的德克萨斯长角牛吃起来不错,可是他们并不是很擅长生產富含奶油的乳製品.Farmer John此时以先天下之忧而忧,后天下之乐而乐的精神,身先士卒地承担起向德克萨斯运送大量的营养冰凉的牛奶的重任,以减轻德克萨斯人忍受酷暑的痛苦. FJ已经研究过可以把牛奶从威斯康星运送到德克萨斯州的路线.这些路线包括起始点和终点先一共经过T (1 <= T <= 2,500)个城镇,方便地标号為1到T.除了起点和终…

Description During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s class a large bag of candies and had flymouse distribute them. All the kids loved candies very much and ofte…

上周我们介绍了神奇的只有五行的Floyd最短路算法,它可以方便的求得任意两点的最短路径,这称为“多源最短路”.本周来来介绍指定一个点(源点)到其余各个顶点的最短路径,也叫做“单源最短路径”.例如求下图中的1号顶点到2.3.4.5.6号顶点的最短路径. <ignore_js_op>          与Floyd-Warshall算法一样这里仍然使用二维数组e来存储顶点之间边的关系,初始值如下. <ignore_js_op>          我们还需要用一个一维数组dis来存储1号…

Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road i re…

Telephone Lines Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7214   Accepted: 2638 Description Farmer John wants to set up a telephone line at his farm. Unfortunately, the phone company is uncooperative, so he needs to pay for some of…

感觉自己太懒了,以后每天更博客激励自己吧. //时间复杂度O(n*n)的最短路算法 //首先需要设置一个访问数组v[maxn],一个数组d[maxn], memset(v,,sizeof(v)); ;i<n;i++) d[i]=(i==?:inf); ;i<n;i++) { ;y<n;y++) { int x,m=inf; if(!v[y]&&d[y]<=m) m=d[x=y]; v[x]=; ;y<n;y++) if(d[y]>d[x]+w[x][y]…

Dijkstra(最短路求解) 模板: #include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; typedef __int64 LL; const int maxn = 2005; const int INF = 0x3f3f3f3f; struct Edge{ int u,v,ne…

Domino Effect Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9335   Accepted: 2325 Description Did you know that you can use domino bones for other things besides playing Dominoes? Take a number of dominoes and build a row by standing t…

Roadblocks http://poj.org/problem?id=3255 Time Limit: 2000MS   Memory Limit: 65536K       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 quick…

题目链接:http://poj.org/problem?id=2253 Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 25773   Accepted: 8374 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on an…

1018. Public Bike Management (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue There is a public bike service in Hangzhou City which provides great convenience to the tourists from all over the world. One may rent a bike at any…

1003. Emergency (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount…

UVA - 11374 Airport Express Time Limit:1000MS   Memory Limit:Unknown   64bit IO Format:%lld & %llu [Submit]  [Go Back]  [id=22966" style="color:rgb(106,57,6); text-decoration:none">Status] Description ProblemD: Airport Express In a s…

·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…

题目描述: Jzzhu and Cities time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Jzzhu is the president of country A. There are n cities numbered from 1 to n in his country. City 1 is the capital of…

点这里看题目 3228K 485MS G++ 2453B 根据题意和测试用例知道这是一个求次短路径的题目.次短路径,就是比最短路径长那么一丢丢的路径,而题中又是要求从一点到指定点的次短路径,果断Dijkstra. R (1 ≤ R ≤ 100,000,N (1 ≤ N ≤ 5000) ,length D (1 ≤ D ≤ 5000),所以我用链式向前星方法存储,这个不知道的点这里(我转载别人的,讲的挺详细). 用一个二维数组dist[MAXN][2],去记录i->j的最短路径和次短路径,第二维是…

好久没有看图论了,就从最短路算法开始了. dijkstra算法的本质是贪心.只适用于不含负权的图中.因为出现负权的话,贪心会出错. 一般来说,我们用堆(优先队列)来优化,将它O(n2)的复杂度优化为O((m+n)logn) 模板链接 套用dijkstra模板即可.给出范例: #include<cstdio> #include<iostream> #include<queue> using namespace std; int n,m,s,tot,head[500000]…