hdu 3435 A new Graph Game】的更多相关文章

http://acm.hdu.edu.cn/showproblem.php?pid=3435 #include <cstdio> #include <iostream> #include <cstring> #include <queue> #include <algorithm> #define inf 0x3f3f3f3f #define maxn 54444 using namespace std; queue<int>q; s…
http://acm.hdu.edu.cn/showproblem.php?pid=3435 题意:有n个点和m条边,你可以删去任意条边,使得所有点在一个哈密顿路径上,路径的权值得最小. 思路: 费用流,注意判断重边,否则会超时. #include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<queue> using namespace std; ty…
Problem Description An undirected graph is a graph in which the nodes are connected by undirected arcs. An undirected arc is an edge that has no arrow. Both ends of an undirected arc are equivalent--there is no head or tail. Therefore, we represent a…
A new Graph Game Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1849    Accepted Submission(s): 802 Problem Description An undirected graph is a graph in which the nodes are connected by undir…
A new Graph Game Problem Description An undirected graph is a graph in which the nodes are connected by undirected arcs. An undirected arc is an edge that has no arrow. Both ends of an undirected arc are equivalent--there is no head or tail. Therefor…
6343.Problem L. Graph Theory Homework 官方题解: 一篇写的很好的博客: HDU 6343 - Problem L. Graph Theory Homework - [(伪装成图论题的)简单数学题] 代码: //1012-6343-数学 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset&g…
和HDU 3488一样的,只不过要判断一下是否有解. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; + ; const int INF = 0x3f3f3f3f; int n, m; int W[maxn][maxn], lft[maxn]; int slack…
http://acm.hdu.edu.cn/showproblem.php?pid=5876 Sparse Graph Problem Description   In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6343 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Problem DescriptionThere is a complete graph containing n vertices, the weight of the i-th vertex is wi.The length…
Rikka with Graph 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5631 Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct gra…
Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 182    Accepted Submission(s): 95 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation,…
http://acm.hdu.edu.cn/showproblem.php?pid=5424 哈密顿通路:联通的图,访问每个顶点的路径且只访问一次 n个点n条边 n个顶点有n - 1条边,最后一条边的连接情况: (1)自环(这里不需要考虑): (2)最后一条边将首和尾连接,这样每个点的度都为2: (3)最后一条边将首和除尾之外的点连接或将尾和出尾之外的点连接,这样相应的首或尾的度最小,度为1: (4)最后一条边将首和尾除外的两个点连接,这样就有两个点的度最小,度都为1 如果所给的图是联通的话,那…
Problem Description   As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and n edges. Now he wants you to tell him…
Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and m edges. The length of each edge . Now…
Rikka with Graph 思路: 官方题解: 代码: #include<bits/stdc++.h> using namespace std; #define ll long long int main() { int t; scanf("%d",&t); while(t--) { ll n,m,ans; scanf("%lld%lld",&n,&m); ) { ans=m*+(m-)*m*+(n*(n-)-m*-(m-)…
题目链接 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: For an undirected graph G with n nodes and m edges, we can define the distance bet…
In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G. Now you are given an undirected graph G of N nodes and M bidirectional edges o…
A new Graph Game Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2360    Accepted Submission(s): 951 Problem Description An undirected graph is a graph in which the nodes are connected by undire…
Rikka with Graph II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1051    Accepted Submission(s): 266 Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situati…
Rikka with Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is…
此题是道bfs搜索的题目.bfs的精髓就是找到下一步的所有可能然后存储起来,有点暴力的感觉,这题就是每步中 所有的可能都入队,然后一一 判断.这道题的题意是 : 给你一幅完全图,再给你三个盘,目的是把这三个盘移动到一个点上,输出最少步数!盘移动的时候有要求,比如移第一个盘,把1盘移动到2这个位置,(1,2)这个点有颜色标记,另外两个盘比如说在3,4两点,那么1盘能移动到2这个点的条件是(1,2)这个点的颜色要与(3,4)这点的颜色相同!! #include"iostream" #inc…
题目大意: 在 N 个点 N 条边组成的图中判断是否存在汉密尔顿路径. 思路:忽略重边与自回路,先判断是否连通,否则输出"NO",DFS搜索是否存在汉密尔顿路径. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<queue> #include<algorithm> #include<cmath&g…
n个点最少要n-1条边才能连通,可以删除一条边,最多删除2条边,然后枚举删除的1条边或2条边,用并查集判断是否连通,时间复杂度为O(n^3) 这边犯了个错误, for(int i=0;i<N;i++){ fa[i]=i; } 这个将i<=N,导致错误,值得注意 AC代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio>…
如果原图不连通,直接输出0. 如果原图连通,删除X条边之后要保证新图连通,再看数据是n+1条边-->因此,最多只能删去两条边. 因为n=100,可以枚举进行验证,枚举删去每一条边是否连通,枚举删去每两条边是否连通,验证是否连通可以用并查集,可以BFS. #include<cstdio> #include<cstring> #include<cmath> #include<string> #include<vector> #include&l…
<题目链接> 题目大意: 给你一个连通的无向图,问你删除每一条边后,是否能够出现一对(u,v),使得u,v不连通,且u<v,如果有多对u,v,则输出尽量大的u,和尽量小的v. 解题分析: 首先要明确,因为该图是连通的无向图,所以删除的边是桥才能够使至少两点不连通.但是对于删除桥的情况,如何输出尽可能大的u和尽可能小的v呢? 我们要知道,删除一个桥,是将整张图分成两部分,这两部分的点仍然是连通的,并且,由于题目要求u<v,且u尽可能的大,v尽可能的小,所以,我们可以推断出,u与n一定…
将一个无向图分成许多回路,回路点交集为空,点幷集为V.幷最小化回路边权和. #include <cstdio> #include <cstring> #include <queue> #include <vector> #define maxn 2010 #define oo 0x3f3f3f3f using namespace std; struct Edge { int u, v, c, f; Edge( int u, int v, int c, int…
题意: 给一个图一些边,保证图连通 问对于每条边,如果去除该边后使得图中一些点不连通.设这些点(u,v),要求使u尽量小,v尽量大,输出这样的(u,v).否则输出0 0. #include <bits/stdc++.h> using namespace std; ; typedef pair <int, int>pii; vector<pii>G[MAXN]; bool isBridge[MAXN]; int clk, pre[MAXN], low[MAXN]; int…
[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] DP 设f[i][j]表示前i个操作,已经匹配了的点的状态集合为j的方案数 对于+操作 有两种情况. 1.这条边作为匹配的边 2.这条边没有作为匹配边 f[i][j] = f[i-1][j-(u,v)] + f[i-1][j] 作为匹配边,转化一下就是这条边的两个点连上了.也即被匹配了. 对于-操作 考虑前i-1个操作. 会发现+操作的先后顺序不影响前i-1个操作之后的结果. 因此我们干脆就认为第i-1个操作就是和- u,v对应…
题目地址:HDU 3435 这题刚上来一看,感觉毫无头绪. .再细致想想.. 发现跟我做的前两道费用流的题是差点儿相同的. 能够往那上面转换. 建图基本差点儿相同.仅仅只是这里是无向图.建图依旧是拆点,推断入度出度.最后推断是否满流,满流的话这时的费用流是符合要求的.输出.不能满流的话,输出NO. 代码例如以下: #include <iostream> #include <stdio.h> #include <string.h> #include <stdlib.…
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…