题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1-F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to ta…

传送门 解题思路 刚开始是找的桥,后来发现这样不对,因为一条链就可以被卡.后来想到应该缩点后找到度数为1 的点然后两两配对. #include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<set> using namespace std; ; ; inline int rd(){ ,f=;char ch=getchar(); :;ch=getcha…

题目链接:https://www.luogu.org/problemnew/show/P2860 考虑在无向图上缩点. 运用到边双.桥的知识. 缩点后统计度为1的点. 度为1是有一条路径,度为2是有两条路径. #include <stack> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; cons…

题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to t…

P2860 [USACO06JAN]冗余路径Redundant Paths 题目描述 为了从F(1≤F≤5000)个草场中的一个走到另一个,贝茜和她的同伴们有时不得不路过一些她们讨厌的可怕的树．奶牛们已经厌倦了被迫走某一条路,所以她们想建一些新路,使每一对草场之间都会至少有两条相互分离的路径,这样她们就有多一些选择． 每对草场之间已经有至少一条路径．给出所有R(F-1≤R≤10000)条双向路的描述,每条路连接了两个不同的草场,请计算最少的新建道路的数量, 路径由若干道路首尾相连而成．两条路径相…

题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to t…

题目链接 https://www.luogu.org/problemnew/show/P2860 思路 缩点,之后就成了个树一般的东西了 然后(叶子节点+1)/2就是答案,好像贪心的样子,lmc好像讲过诶 cpp #include <iostream> #include <cstdio> #include <vector> #include <bits/stdc++.h> #define iter vector<int>::iterator us…

题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to t…

题解: 首先要边双缩点这很显然 然后变成树上问题 发现dp,dfs好像不太对 考虑一下度数 发现只要在度数为1的点之间连边 但我好像不太会证明这个东西.. 网上也没有看到比较正确的证明方法和连边策略.. 另外从桥上考虑这题是错的 举出链的反例就很容易能发现了…

P2860 [USACO06JAN]冗余路径Redundant Paths 为了从F(1≤F≤5000)个草场中的一个走到另一个,贝茜和她的同伴们有时不得不路过一些她们讨厌的可怕的树．奶牛们已经厌倦了被迫走某一条路,所以她们想建一些新路,使每一对草场之间都会至少有两条相互分离的路径,这样她们就有多一些选择． 每对草场之间已经有至少一条路径．给出所有R(F-1≤R≤10000)条双向路的描述,每条路连接了两个不同的草场,请计算最少的新建道路的数量, 路径由若干道路首尾相连而成．两条路径相互分离,是…

Luogu2860 [USACO06JAN]冗余路径Redundant Paths 给定一个连通无向图,求至少加多少条边才能使得原图变为边双连通分量 \(1\leq n\leq5000,\ n-1\leq m\leq10^4\) tarjan 边双无疑不用考虑,于是就可以边双缩点成一棵树 令现在要连的边为 \((u,\ v)\) ,当前树上 \(bl_u\) 到 \(bl_v\) 的链将会变为一个新的点双,可以将他们看为一个新的点 可以贪心地连边使得每次连边后,不复存在的点尽量多,当只剩一个点时…

题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to t…

(写题解不容易,来我的博客玩玩咯qwq~) 该题考察的知识点是边双连通分量 边双连通分量即一个无向图中,去掉一条边后仍互相连通的极大子图.(单独的一个点也可能是一个边双连通分量) 换言之,一个边双连通分量中不包含桥. 例如下图(样例)中的边双连通分量有(1),(2,3,5,6),(4),(7) 不难发现,在一个边双连通分量中,任意两点都存在至少两条互相分离的路径:(如1->2与1->3->2) 如若不在一个边双连通分量中,则可能经过桥(甚至不联通)如:2->4. 由于桥是必须通过的…

原题链接 题意实际上就是让你添加尽量少的边,使得每个点都在至少一个环上. 显然对于在一个边双连通分量里的点已经满足要求,所以可以用\(tarjan\)找边双并缩点. 对于缩点后的树,先讲下我自己的弱鸡做法,每次找直径,因为将直径改为环显然使得新添的边贡献最大,这样贪心的连下去,直到所有点满足要求为止. 而实际上有一个简单结论,该题答案就是缩点后树的叶子节点个数除\(2\)(向上取整). 代码是用的我自己的弱鸡做法. #include<cstdio> using namespace std; c…

题目链接 https://www.luogu.org/problemnew/show/P2860 https://www.lydsy.com/JudgeOnline/problem.php?id=1718 分析 首先这题目的意思就是让任意两点之间至少有两条没有重复道路的路径,很显然,如果这个图不存在桥,就一定满足上述条件. 于是我们就是要求使这个图不存在桥需要连接的最小边数 如果把桥从图中去掉,很显然剩余的联通块中任意两点之间至少有两条没有重复道路的路径(当然也可能不是联通块而是孤立的点),对答…

http://poj.org/problem?id=3177 这个妹妹我大概也曾见过的~~~我似乎还没写过双联通分量的blog,真是智障. 最少需要添多少条边才能使这个图没有割边. 边双缩点后图变成一棵树,( 树上度数为1的点的数目+1 ) / 2就是答案. 注意: 1.直接缩成一个点的时候特判一下(不需要加边). 2.找割边同时用栈缩点的话要注意需要缩成一个点的是割边后面所有的点,能缩的时候直接判断末尾有没有到当前点就完事了,如果判断low来找割边后面所有的点是不准确的. #include<i…

Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forc…

题意:给你一个无向连通图,每次加一条边后,问图中桥的数目. 思路:先将图进行双联通缩点,则缩点后图的边就是桥,然后dfs记录节点深度,给出(u,v)使其节点深度先降到同一等级,然后同时降等级直到汇合到同一个点为止.途中直接进行删边处理且桥的数目减少. 代码: #include<iostream> #include<cstring> #include<cstdio> #include<vector> using namespace std; #define M…

#pragma comment(linker,"/STACK:102400000,102400000")//总是爆栈加上这个就么么哒了 #include<stdio.h> #include<queue> #include<string.h> using namespace std; #define N 210000 #define inf 99999999 struct node { int u,v,w,next; }bian[N*20],biant…

Redundant Paths Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of…

题目大意:给定一个 N 个点,M 条边组成的无向图,求至少在图中加入几条边才能使得整个图没有割边. 题解:求出该无向图的所有边双联通分量,每个边双联通分量可以理解成无向图的一个极大环,对该无向图进行缩点,形成一棵树.至少加入的边数和树中入度为 1 的节点个数有关,找找规律即可求得结果. 代码如下 #include <bits/stdc++.h> #define fi first #define se second #define pb push_back #define mp make_pai…

A /*Huyyt*/ #include<bits/stdc++.h> #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; typedef unsigned long long ull; ][] = {{, }, {, }, {, -}, { -, }, {, }, {, -}, { -, -}, { -, }}; , gakki = + + + + 1e9; , MAXM = 2e…

题目链接:http://lightoj.com/volume_showproblem.php?problem=1291 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<queue> #include<vector> using namespace std; ; const i…

Traffic Real Time Query System Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1929    Accepted Submission(s): 380 Problem Description City C is really a nightmare of all drivers for its traffi…

#include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long ll; ; ; int dfn[maxn], low[maxn], head[maxn]; ll ans[maxn], siz[maxn]; int n, m, tot, num, root; bool cut[maxn]; struct edge{ int to, next; }…

在看了春晚小彩旗的E技能(旋转)后就一直在lol……额抽点时间撸一题吧…… Redundant Paths Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8159   Accepted: 3541 Description In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to anot…

加一条边后最少还有多少个桥,先Tarjan双联通缩点, 然后建树,求出树的直径,在直径起点终点加一条边去的桥最多, #pragma comment(linker, "/STACK:1024000000,1024000000") #include<stdio.h> #include<string.h> #include<stack> #define N 200001 using namespace std; int belong[N],head[N],…

在给出的两个点上加一条边,求剩下桥的数量,,不会LCA在线,就用了最普通的,先Tarjan双联通缩点,然后将缩完的图建成一棵树,树的所有边就是桥了,如果在任意两点间加一条边的话,那么从两点到最近公共祖先的所有边都不是桥了...... #pragma comment(linker, "/STACK:10240000000000,10240000000000") #include<stdio.h> #include<stack> #include<string…

/*先求出双联通缩点,然后进行树形dp*/ #include<stdio.h> #include<string.h> #include<math.h> #define inf 0x3fffffff #define N 11000 struct node { int u,v,next; } bian[N*4],edge[N*4]; int head[N],yong,dfn[N],low[N],index,f[N*4],cnt,n,num[N]; int yon; int…

Redundant Paths 分离的路径 题目描述 In order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of o…