Problem Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he…

睡觉啦 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ,M=2e6+,INF=1e7; inline int read(){ ,f=; ;c=getchar();} +c-';c=getchar();} return x…

Building roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6229   Accepted: 2093 Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some ro…

HDU 1815, POJ 2749 Building roads pid=1815" target="_blank" style="">题目链接HDU 题目链接POJ 题意: 有n个牛棚, 还有两个中转站S1和S2, S1和S2用一条路连接起来. 为了使得随意牛棚两个都能够有道路联通,如今要让每一个牛棚都连接一条路到S1或者S2. 有a对牛棚互相有仇恨,所以不能让他们的路连接到同一个中转站. 还有b对牛棚互相喜欢,所以他们的路必须连到同一个中专站.…

Building roads Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 34 Accepted Submission(s): 13   Problem Description Farmer John's farm has N barns, and there are some cows that live in each barn.…

Building roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8153   Accepted: 2772 Description Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some ro…

题目连接 http://poj.org/problem?id=3625 Building Roads Description Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already conn…

计算距离时平方爆了int结果就WA了一次...... ----------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #include<vector> #include<cmath…

P2872 [USACO07DEC]道路建设Building Roads kruskal求最小生成树. #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<queue> #include<algorithm> #define re register using namespace std; typedef double ld; t…

1626: [Usaco2007 Dec]Building Roads 修建道路 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John最近得到了一些新的农场,他想新修一些道路使得他的所有农场可以经过原有的或是新修的道路互达(也就是说,从任一个农场都可以经过一些首尾相连道路到达剩下的所有农场).有些农场之间原本就有道路相连. 所有N(1 <= N <= 1,000)个农场(用1..N顺次编号)在地图上都表示为坐标为(X_i,…

P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms…

洛谷 P2872 [USACO07DEC]道路建设Building Roads 洛谷传送门 JDOJ 2546: USACO 2007 Dec Silver 2.Building Roads JDOJ传送门 Description Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any o…

P2872 [USACO07DEC]道路建设Building Roads 题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms…

把爱恨和最大距离视为限制条件,可以知道,最大距离和限制条件多少具有单调性 所以可以二分最大距离,加边+check #include<cstdio> #include<algorithm> #include<cstring> #include<vector> #include<stack> #define N 5010 #define INF 4000000 using namespace std; ],love[N][],indx,inst[N]…

嘟嘟嘟 最近把21天漏的给不上. 今天重温了一下2-SAT,感觉很简单.就是把所有条件都转化成如果--必然能导出--.然后就这样连边建图,这样一个强连通分量中的所有点必然都是真或者假.从而根据这个点拆点后的两个点是否在一个强连通分量里判断是否有解. 这题人很容易想到拆点:\(i\)表示\(i\)连向\(s_1\),\(i + n\)表示\(i\)连向\(s_2\). 这道题关键在于距离这个限制.我们肯定能想到二分,但是接下来我就没想出来,因为2-SAT的图的边权是没有意义的.但实际上这个也是可以…

题目 Farmer John's farm has N barns, and there are some cows that live in each barn. The cows like to drop around, so John wants to build some roads to connect these barns. If he builds roads for every pair of different barns, then he must build N * (N…

传送门 Time Limit: 3000MS Description There is a magic planet in the space. There is a magical country on the planet. There are N cities in the country. The country is magical because there are exactly N −1 magic roads between the N cities, and from eac…

Description Farmer John最近得到了一些新的农场,他想新修一些道路使得他的所有农场可以经过原有的或是新修的道路互达(也就是说,从任一个农场都可以经过一些首尾相连道路到达剩下的所有农场).有些农场之间原本就有道路相连. 所有N(1 <= N <= 1,000)个农场(用1..N顺次编号)在地图上都表示为坐标为(X_i, Y_i)的点(0 <= X_i <= 1,000,000:0 <= Y_i <= 1,000,000),两个农场间道路的长度自然就是代…

二分答案 + 2-SAT验证 POJ 稳过,HDU C++ 超时,G++ 550ms左右AC #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<stack> #include<algorithm> using namespace std; +; int N,A,B; int left,right,mid; int s1x,s1y,s…

题意 有 N 个牛栏,现在通过一条通道(s1,s2)要么连到s1,要么连到s2,把他们连起来,他们之间有一些约束关系,一些牛栏不能连在同一个点,一些牛栏必须连在同一个点,现在问有没有可能把他们都连好,而且满足所有的约束关系,如果可以,输出距离最大的两个牛栏之间距离最小值(两点距离是指哈密顿距离) Sol 二分答案+\(2-SAT\)判定 每次二分答案,把枚举两个点距离\(>mid\)的就连边限制 傻逼到没输出-1,WA无数遍 # include <iostream> # include…

http://poj.org/problem?id=2749 (这个约翰的奶牛真多事…………………………) i表示u与s1连,i+n表示u与s2连. 老规矩,u到v表示取u必须取v. 那么对于互相打架的奶牛u,v,有: add(u,v+n);add(v,u+n); add(u+n,v);add(v+n,u); 对于互为朋友的奶牛u,v,有: add(u,v);add(v,u); add(u+n,v+n);add(v+n,u+n); 但这远远不够,我们需要求最大值最小…… 二分?但是我们怎么边处理…

题解: 考虑用线段树维护楼的最大值,然后这个问题就很简单了. 每次可以向左二分出比x高的第一个楼a,同理也可以向右二分出另一个楼b,如果a,b都存在,答案就是b-a-1. 注意到二分是可以直接在线段树上进行的,所以复杂度是O(nlogn). 当然这里是用分块做的,更暴力一些. #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace st…

原题 2-SAT+二分答案! 最小的最大值,这肯定是二分答案.而我们要2-SATcheck是否在该情况下有可行解. 对于目前的答案limit,首先把爱和恨连边,然后我们n^2枚举每两个点通过判断距离来实现连边,然后跑2-SAT判断是否有可行解 O(n^2logn) 想起来和听起来都很难写,事实上还好吧- #include<cstdio> #include<algorithm> #include<stack> #include<cstring> #define…

  Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10803   Accepted: 3062 Description Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a se…

题目描述 Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms. Each of the N (1 ≤ N ≤ 1,000) farms…

Farmer John had just acquired several new farms! He wants to connect the farms with roads so that he can travel from any farm to any other farm via a sequence of roads; roads already connect some of the farms. Each of the N (1 ≤ N ≤ 1,000) farms (con…

题目 //最小生成树,只是变成二维的了 #define _CRT_SECURE_NO_WARNINGS #include<stdlib.h> #include<stdio.h> #include<iostream> #include<string.h> #include<algorithm> #include<math.h> using namespace std; #define inf 999999999 #define M 10…

感慨一下,区域赛的题目果然很费脑啊!!不过确实是一道不可多得的好题目!! 题目大意:给你一棵有n个节点的树,让你移动树中一条边的位置,即将这条边连接到任意两个顶点(边的大小不变),要求使得到的新树的直径最小. 解题思路:此题先求出原始树的直径maxr1,并记录直径上的各个节点.很容易想到要移动的边一定是直径上的边,只有这样才有可能使树的直径减小!! 接着就是枚举直径上的每条边,并用这条边作为分隔将原始树分割成两棵子树(即子树一和子树二),然后分别求子树一的直径maxr2 和子树二的直径maxr3…

Description Farmer John最近得到了一些新的农场,他想新修一些道路使得他的所有农场可以经过原有的或是新修的道路互达(也就是说,从任一个农场都可以经过一些首尾相连道路到达剩下的所有农场).有些农场之间原本就有道路相连. 所有N(1 <= N <= 1,000)个农场(用1..N顺次编号)在地图上都表示为坐标为(X_i, Y_i)的点(0 <= X_i <= 1,000,000:0 <= Y_i <= 1,000,000),两个农场间道路的长度自然就是代…

http://www.lydsy.com/JudgeOnline/problem.php?id=1626 依旧是水题..太水了.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> using namespace std; #define rep(i, n) f…