Deleting Edges                                                                                  Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)                                                                    …

https://vjudge.net/problem/HDU-6026 我一直想不明白的是,它的乘法是如何保证n-1条边的.后来画了一张图大概能明白了. 结合最后的乘法二层循环的代码来看,当i=4的时候,j有1和3满足条件,3满足不用说是迪杰斯特拉本身跑出来的,1满足条件,而中间的点2.3也可以有自己本身的路径达到,最后满足共n-1条边. #include<iostream> #include<cstdio> #include<queue> #include<cs…

Deleting Edges Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 567    Accepted Submission(s): 210 Problem Description Little Q is crazy about graph theory, and now he creates a game about grap…

题意 有一个n个节点的无向图,结点编号从0-n-1,每条边的长度时1to9的一个正整数.现在要删除一些边(或者不删),使得到的新图满足下面两个要求. 1.新图是一颗树有n-1条边2.对于每个结点v(0ton-1)从0到v的距离要等于原图上从0到v的最短路 现在计算有多少种不同的图满足上面两个条件 分析 统计每个点可以到达最短路的入度值,然后相乘.这个题因为spfa写错了调了老久··· #include <cstdio> #include <cstring> #include <…

最短路. 先建一个只包含最短路的有向无环图,每一个点选择任意一条入边即可生成一个树形图,那么树的种类就等于每个点的入度乘积. #include <bits/stdc++.h> using namespace std; ; int n; ][]; ],f[]; ]; void spfa() { queue<int>Q; ;i<n;i++) f[i]=,dis[i] = 0x7FFFFFFF; dis[]=; Q.push(); f[]=; while(!Q.empty()) {…

Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.…

主题链接:http://acm.hdu.edu.cn/showproblem.php?pid=4496 D-City Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Lux…

D-City Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 5911    Accepted Submission(s): 2064 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-…

Problem Description Luxer is a really bad guy. He destroys everything he met.  One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants…

D-City Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3838    Accepted Submission(s): 1379 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-…

点击打开链接 D．D-City Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line  connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-ci…

D-City Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 1315    Accepted Submission(s): 496 Problem Description Luxer is a really bad guy. He destroys everything he met.  One day Luxer went to D…

Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants…

D-City Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city. D-cit…

D-City Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 18    Accepted Submission(s): 15 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city…

D-City Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2933    Accepted Submission(s): 1038 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-…

D-City Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 2448 Accepted Submission(s): 862 Problem Description Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city.…

D-City Time Limit: 1000ms Memory Limit: 65535KB This problem will be judged on HDU. Original ID: 449664-bit integer IO format: %I64d      Java class name: Main   Luxer is a really bad guy. He destroys everything he met. One day Luxer went to D-city.…

此题卡Dijkstra... Code: #include <bits/stdc++.h> #define N 30005 #define M 4000000 #define ll long long #define inf 100000000 #define E 14300000 #define setIO(s) freopen(s".in","r",stdin) using namespace std; ll d[M]; int n,m,block,…

https://vjudge.net/problem/Aizu-2249 感觉这题和2017女生赛的Deleting Edge思路很像,都是先找最短路,然后替换边的. 但是这题用最朴素的dijkstra的话memory limit exceed了,看了下超了约7倍无望... 不得已这时候不得不学dijkstra的堆优化了. #include<iostream> #include<cstdio> #include<queue> #include<cstring>…

该文章可能存在硬伤与不妥,不能作为教程阅读.(因为我真的鶸 Dij作为单源最短路算法,需要先确定一个起点.Dij的函数主体为维护每个节点的dis和vis两个变量.dis表示该点距离起点的最短路权值和,vis存储该点是否被访问过. 设点数为n,无向边数为m,则要进行n次循环,每次循环中找出一个dis最小且vis不为真的点进行松弛操作. 这里抄一段百科: 松弛操作是指对于每个顶点v∈V,都设置一个属性g[v],用来描述从源点s到v的最短路径上权值的上界,称为最短路径估计(shortest-path…

题目链接 \(Description\) 给定\(n,m,C\)及大小为\((n+1)(m+1)\)的矩阵\(c[i][j]\).平面上有\((n+1)(m+1)\)个点,从\((0,0)\)编号到\((n,m)\). 在任意时刻,你可以选择在当前点\((x,y)\)获取任意实数个单位的能量,获取每一单位需花费时间\(c[x][y]\): 也可以选择从一个点移动到另一个点,花费的能量是移动路线的欧几里得距离(只可以沿与坐标轴平行或与坐标轴夹角\(45^{\circ}\)的方向走),不花费时间.…

题目链接 题意可以看做,用\(a_1,a_2,...,a_n\),能组成多少个\([L,R]\)中的数. (40分就是个完全背包) 首先如果\(k*a_i+x\)可以组成,那么\((k+1)*a_i+x\)自然也可以组成(\(k\geq 0,0\leq x<a_i\)). 即我们选取一个\(a_i\)做模数,若\(d\equiv x(mod\ a_i)\),则\(d+a_i\equiv x(mod\ a_i)\),那么对于当前余数\(x\)我们可以算出\([0,R]\)中有多少个数模\(a_i\…

题目链接 \(Description\) 给出一个N个点M条边的无向图,经过一个点的代价是进入和离开这个点的两条边的边权的较大值,求从起点1到点N的最小代价.起点的代价是离开起点的边的边权,终点的代价是进入终点的边的边权. \(Solution\) 最直接的方法是把每条边作为一个点,对于连接同一个点的两条边连一条新边,最后把连接1和n的点分别连S.T,跑最短路 但这样边数是O(m^2)的 对于路径上相邻两条边\((i,j,v1)\)和\((j,k,v2)\),v1<v2,考虑如何构图把v1比v2…

题目链接:http://icpc.njust.edu.cn/Problem/Hdu/6026/ 题意大致是:给定一个图,要求删边使他变成树,使得每个点到0的距离就是原图中0到这个点的最短路径.其实就是最短路树. 证明1: 对于每个结点我们只要知道有多少条路径到它的距离是最短路长度,记为cnt,那么就有cnt条边连着前驱结点,首先,能保证的是每个点都是0能够到达的,所以前驱结点一定是全部都能到达的,就是他们的d一定存在,所以我们只要删除cnt-1条边,留下一条就能建成最短路树.对于每个结点,我们扫…

本来就是水题一道. 题意:一个人要从点1去到点2,中间还有很多点和很多条边.问你如果他每次走的边(a,b)都满足:a点到目标点的最短距离<b点到目标点的最短距离,那么他从点1出发到点2总共有多少条路径. 思路:先用Dijkstra求最短路,然后创一个图,对于满足条件的边(a,b)就加一条有向边,由于是严格按照小于加的边,图中绝对没有环,是个DAG.接下来dp就行了. dp[i]表示i点出发有多少条路径,dp[i]=sum(dp[j]). #include <cstdio> #includ…

Dijkstra算法: 解决的问题: 带权重的有向图上单源最短路径问题.且权重都为非负值.如果采用的实现方法合适,Dijkstra运行时间要低于Bellman-Ford算法. 思路: 如果存在一条从i到j的最短路径(Vi.....Vk,Vj),Vk是Vj前面的一顶点.那么(Vi...Vk)也必定是从i到k的最短路径.为了求出最短路径,Dijkstra就提出了以最短路径长度递增,逐次生成最短路径的算法.譬如对于源顶点V0,首先选择其直接相邻的顶点中长度最短的顶点Vi,那么当前已知可得从V0到达Vj…

下一节<Cocos2d-x 地图行走的实现2:SPFA算法>: http://blog.csdn.net/stevenkylelee/article/details/38440663 本文乃Siliphen原创,转载请注明出处:http://blog.csdn.net/stevenkylelee 本文的实现使用的环境是:Cocos2d-x 3.2.VS2013 本文,我们终于实现的地图行走效果例如以下2图: 以下是2张屏幕录制的gif动绘图,有点大.看不到的话.耐心等待一下.或者刷新页面试试.…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1535 Problem Description In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all,…

概述 与前面说的Floyd算法相比,Dijkstra算法只能求得图中特定顶点到其余所有顶点的最短路径长度,即单源最短路径问题. 算法思路 1.初始化,集合K中加入顶点v,顶点v到其自身的最短距离为0,到其它所有顶点为无穷. 2.遍历与集合K中结点直接相邻的边(U,V,C),其中U属于集合K,V不属于集合K,计算由结点v出发,按照已经得到的最短路径到达U,再由U经过该边达到V时的路径长度.比较所有与集合K中结点直接相邻的非集合K结点该路径长度,其中路径长度最小的顶点被确定为下一个最短路径确定的结点…