Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4425    Accepted Submission(s): 1263 Problem Description After World War X, a lot of cities have been seriously damag…

Connections between cities Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 896 Accepted Submission(s): 236   Problem Description After World War X, a lot of cities have been seriously damaged, an…

Connections between cities [题目链接]Connections between cities [题目类型]LCA Tarjan &题意: 输入一个森林,总节点不超过N(N<10000),由C次询问(C<1000000),每次询问两个点,如果来联通输出,两点之间的距离,如果不来联通,输出"Not connected" &题解: md,就没人吐槽这题询问时有相等的情况吗?我被这个坑了一天= = 最后把ans数组置为-1才过的,以前一直初…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some mat…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11964    Accepted Submission(s): 2786 Problem Description After World War X, a lot of cities have been seriously damag…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4057    Accepted Submission(s): 1178 Problem Description After World War X, a lot of cities have been seriously damage…

题目链接: Connections between cities Time Limit: 10000/5000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However,…

题目链接 Connections between cities LCA的模板题啦. #include <bits/stdc++.h> using namespace std; #define REP(i,n) for(int i(0); i < (n); ++i) #define rep(i,a,b) for(int i(a); i <= (b); ++i) #define dec(i,a,b) for(int i(a); i >= (b); --i) #define for…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 11927    Accepted Submission(s): 2775 Problem Description After World War X, a lot of cities have been seriously dama…

Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So we need to transport these materials from city to city.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题目: Problem Description After World War X, a lot of cities have been seriously damaged, and we need to rebuild those cities. However, some materials needed can only be produced in certain places. So…

dfs找出所有节点所在树及到树根的距离及深度及父亲. i和j在一棵树上,则最短路为dis[i]+dis[j]-dis[LCA(i,j)]*2. #include <cstring> #include <cstdio> #define N 10005 #define add(u,v,w) e[++cnt]=(edge){v,head[u],w};head[u]=cnt using namespace std; struct edge{ int to,next,w; }e[N<&…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 这题有不连通的情况,特别注意. 觉得是存query的姿势不对,用前向星存了一遍,还是T…… /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┃┃┃┃┃┃ ┻┻┻┻┻┻ */ #include <algorithm>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意是给你n个点,m条边(无向),q个询问.接下来m行,每行两个点一个边权,而且这个图不能有环路.然后接下来q行,每行给你两个点,问你这两个点的最短距离是多少,要是不相连,则输出一串英文. 首先想到的是用(二分)倍增LCA,但是这题的坑点是两个点可能不在同一个图中,所以我dfs的时候用block[i]标记这个点属于哪一个图中,要是这个点在同一个图中,答案就是cost[u] + cost[v]…

题意: 城市 道路  没有环 不一定连通的树 求两城市的最短距离 设想一下就是很多小树  那好办 思路: lca离线算法 然后有个技巧就是 每次我们tarjan一棵树不是最后树的节点都访问过并且孩子全能找到老爸嘛 那么我们只要做做做做 做到全部的城市都访问过了  就行了 反正你做这颗小树的时候又不会影响到其他树的 #include <cstdio> #include <cstring> #include <algorithm> using namespace std;…

思路:LCA裸题.本来是帮pechpo调错,结果自己写了半天… 设$dis_x$是点$x$到根结点距离,不难想到两点$u$.$v$之间最短距离等于$dis_u+dis_v-dis_{LCA(u,v)}\times 2$. 然后我们可以用Tarjan做,然后发现MLE了. 以为是这题卡vector的内存,于是改成了链式前向星,还是MLE. 后来发现题目的内存限制只有32M,算了算,如果将数据离线保存下来,大约有20000K左右,再加上函数里面的栈,似乎确实有点危险. 最后改成用ST做,只用了585…

http://acm.hdu.edu.cn/showproblem.php?pid=2874 题意: 求两个城市之间的距离. 思路: LCA题,注意原图可能不连通. 如果不了解离线算法的话,可以看我之前博客写的解释http://www.cnblogs.com/zyb993963526/p/7295894.html #include<iostream> #include<algorithm> #include<cstring> #include<cstdio>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2874 题目大意:给出n个点,m条边,q个询问,每次询问(u,v)的最短距离,若(u,v)不连通即不在同一颗树上则输出“Not connected”. 解题思路:这题也是模板题,有所不同的是这次给出的是森林而不是一棵树,所以vis数组得稍作修改,标记vis数组的是当前树的编号.下面给出Tarjan和倍增法两种解法. Tarjan(离线)写法,被MLE坑了,离线写法必须要用静态邻接表,因为虽然n不大,但…

题意: 给出n个点m条边的图,c次询问 求询问中两个点间的最短距离. 解析: Floyd会T,所以用到了最短路树..具体思想为: 设k为u和v的最近公共祖先 d[i] 为祖结点到i的最短距离  则dis[u][v] = d[u] + d[v] - 2*d[k] 用tarjan的lca求即可 把这题代码当作模板就好啦 #include <iostream> #include <cstdio> #include <sstream> #include <cstring&…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=2874 [题目大意] 有n个村庄,m条路,不存在环,有q个询问,问两个村庄是否可达, 如果可达则输出最短路. [题解] 因为不存在环,所以是森林,我们计算每个连通块的dfs序,计算块内每个点到根距离 当两个点在同一个连通块时,我们输出其dis值之和减去其LCA的dis值, 否则输出不想连. [代码] #include <cstdio> #include <algorithm> #in…

题意:n棵树,求任意两点的最短距离. 解题关键:并查集判断两点是否位于一棵树上,然后求最短距离即可.此题可以直接对全部区间直接进行st表,因为first数组会将连接的两点的区间表示出来. //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib&…

第一次按常规的方法求,将所有的查询的u,v,和最近公共祖先都保存起来,然后用tarjan+并查集求最近公共祖先.因为询问的次数过多,所以在保存查询的时候总是MLE,后来参考了一下别人的代码,才突然觉悟,可以先将u,v,和其最近公共祖先保存到数组,然后再求结果,为什么不能直接保存其结果了.如果只保存结果的话,保存查询操作就可以节约1/3的内存,所以基本可以过了. 代码如下: 方法一: #include <iostream> #include <cstdio> #include <…

图不一定联通,所以用并查集找各个联通块的祖先分别建图,之后就和LCA的步骤差不多了 #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #include<algorithm> #include<vector> using namespace std; inline int read(){ int sum=0,x=1; char ch=getc…

Connections between cities Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8857    Accepted Submission(s): 2151 Problem Description After World War X, a lot of cities have been seriously damag…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…

强连通 迷宫城堡 Proving Equivalences Equivalent Sets Summer Holiday Intelligence System The King's Problem Hawk-and-Chicken Cactus 仙人掌图 [双连通]: 2242 考研路茫茫--空调教室 双联通缩点+树形DP 2460 Network 边双连通 3849 By Recognizing These Guys, We Find Social Networks Useful 双连通求桥…

什么是最近公共祖先? 在一棵没有环的树上,每个节点肯定有其父亲节点和祖先节点,而最近公共祖先,就是两个节点在这棵树上深度最大的公共的祖先节点. 换句话说,就是两个点在这棵树上距离最近的公共祖先节点. 所以LCA主要是用来处理当两个点仅有唯一一条确定的最短路径时的路径. 常用来求LCA的算法有:Tarjan/DFS(离线),ST/倍增(在线). 1,Tarjan tarjan的算法复杂度为$O(n+q)$. 思路:每进入一个节点u的深搜,就把整个树的一部分看作以节点u为根节点的小树,再搜索其他的节…

=============================以下是最小生成树+并查集====================================== [HDU] How Many Tables 基础并查集★ 小希的迷宫 基础并查集★ &&poj1308 Is It A Tree? 基础并查集★ More is better 基础并查集★ Constructing Roads 基础最小生成树★ 畅通工程 基础并查集★ 还是畅通工程 基础最小生成树★ 畅通工程 基础最小生成树★ 畅通…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30147   Accepted: 15413 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, eac…