hqd_acm的专栏

这道题目很无语，首先是输入，最后采取的输入方法是读入一个数，就把对应的那个数算出来。接着不管读入的是什么也不会影响到。接着这道题目用的竟然是FLOYED，但这个算法是经过改变，不需要讨论所有的，只有那种事0的外面才算的，好像是这样，但这道题目我觉得还是有歧义的。   StatusIn/OutTIME LimitMEMORY LimitSubmit T

 StatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s8192K322183StandardRisk is a board game in whic

 StatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout5s8192K228103StandardArthur想要将手下的战士按照能力进行排名, 选择出得力的助手进行冒险. 他收集了很多人之间比武的胜负情况，如A战

最小生成树。StatusIn/OutTIME LimitMEMORY LimitSubmit TimesSolved UsersJUDGE TYPEstdin/stdout3s8192K551233Standard1st Jilin University ACM

#include#includeint m,n;int map[20][20],degree[20];char s[20],list[20];bool visit[20];void dfs(int depth){ int i,j; if(depth>=m) {  list[depth]=0;  printf("%s/n",list);  return;  } for(i=1;i if(degree

const int maxn=605;const int inf=8000000;void dij(int n,int g[][maxn],int s,int d[]){ int i,r; int minc,used[n]; for(i=0;i {  d[i]=inf;used[i]=0; } d[s]=0; for(int k=0;k {  minc=inf;  for(i=0;i   if

　　求单源最短路的SPFA算法的全称是：Shortest Path Faster Algorithm。　　SPFA算法是西南交通大学段凡丁于1994年发表的.　　从名字我们就可以看出，这种算法在效率上一定有过人之处。　　很多时候，给定的图存在负权边，这时类似Dijkstra等算法便没有了用武之地，而Bellman-Ford算法的复杂度又过高，SPFA算法便派上用场了。　　简洁起见，

A Pair of GraphsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 357    Accepted Submission(s): 184Problem DescriptionWe say that two graph

<br />Frogger<br />Time Limit: 1000 ms     Case Time Limit: 1000 ms     Memory Limit: 65535 KB<br />Submit: 138     Accepted: 46Description<br /><br />一只叫Freddy的青蛙蹲坐在湖中的一块石头上。突然他发现一只叫Fiona的青蛙在湖中的另一块石头上。Freddy想要跟Fiona约会，但由于湖水太脏，他不想游泳过去而是跳过去找Fiona。<br />很不幸，

B: Cheering up the Cows<br />Time Limit: 1000 ms     Case Time Limit: 1000 ms     Memory Limit: 65536 KB<br />Submit: 13     Accepted: 8 <br />Description<br /><br />Farmer John has grown so lazy that he no longer wants to continue maintaining the cow path

<br />Genealogical treeTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 815 Accepted: 578 Special Judge<br />DescriptionThe system of Martians' blood relations is confusing enough. Actually, Martians bud when they want and where they want. They ga

<br /><br />The Bottom of a GraphTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3066 Accepted: 1227<br />DescriptionWe will use the following (standard) definitions from graph theory. Let V be a nonempty and finite set, its elements being ca

<br /><br />Popular CowsTime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9234 Accepted: 3690<br />DescriptionEvery cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M

<br />Proving EquivalencesTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 205    Accepted Submission(s): 89<br /><br /><br />Problem DescriptionConsider the following exercise, found in a generi

<br /><br />Network of SchoolsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3010 Accepted: 1160<br />DescriptionA number of schools are connected to a computer network. Agreements have been developed among those schools: each school maintai

World Islands<br />Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)<br />Total Submission(s): 206    Accepted Submission(s): 78<br /><br />Problem DescriptionDubai is a haven for the rich. The government of Dubai finds a

差分约束系统(System Of Difference Constraints)（本文假设读者已经有以下知识：最短路径的基本性质、Bellman-Ford算法。）    比如有这样一组不等式：    X1 - X2 X1 - X5 X2 - X5 X3 - X1 X4 - X1 X4 - X3 X5 - X3 X5 - X4     全都是两个未知数的差小于等于某个常数（大于等于也可以，因为左右乘以-1就可以化成小于等于）。这样的不等式组就称作差分约束系统。    这个不等式组要么无解，

鄙人第一个差分约束系统。看了N多的资料和别人的解题报告，发现下面用BELLMAN FORD 加稍微的剪枝时间可以飙到200MS以内，不错不错。就是那个FLAG，没了这个就TLE了。。。惨剧啊！！！！另外的话，这个题目的例子是给了5个区间，说5个区间包含了Z数组C个数，接着问你Z数组到底至少有几个数。数组不用连续主要是下面三点1。s[b]>=s[a-1]+c2.s[b-1]>=s[b]+(-1)3.s[b]>=s[b-1]维护这三个不等式即可，s[]表示从0到x含有的数字的个数如何维护呢？想像一下做图论的时候

SPFA算法为了简便，我们约定图中不存在负权回路，这可以通过一次拓扑排序知道。SPFA实际是Bellman-Ford算法的一种队列实现，用一个数组来保存最短路径的估计值，初始时将源加入队列，每次从队列中取队头元素，并对所有与其相邻的结点进行松弛操作，如果该点的估计值有所调整，且该点不在队列中，则入队，如此反复，直到队空。定理1：SPFA算法采用的动态优化方法，能够使得路径估计值变得越来越小，优化过程是正确的。证明：算法中每一步都是从该队取出点来优化其它各点。现假设算法结束时计算的结果不是最短路径，即从原点到

<br /><br />Posted on 2009-10-30 14:00 小强摩羯座 阅读(350) 评论(0) 编辑 收藏 所属分类: 算法编程 <br />最短路径 之 SPFA算法 (转载）(2009-05-06 20:41:51)<br />标签：spfa 杂谈   <br />求最短路径的算法有许多种，除了排序外，恐怕是OI界中解决同一类问题算法最多的了。最熟悉的无疑是Dijkstra，接着是Bellman-Ford，它们都可以求出由一个源点向其他各点的最短路径；如果我们想要求出每一对顶点

<br /><br />CandiesTime Limit: 1500MS Memory Limit: 131072KTotal Submissions: 12484 Accepted: 3226<br />Description<br />During the kindergarten days, flymouse was the monitor of his class. Occasionally the head-teacher brought the kids of flymouse’s cla

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2066 Accepted: 993DescriptionLike everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2

我自己的程序#include#includeconst int E=50005;const int V=50005;const int INF=1struct EDGE{    int link,val,next;}edge[E*3];int head[V],dist[V],q[E*3],e,rmin,rmax;bool vis[V];void addedge(int a,int b,int c){    edge[e].link=b;    edge[e].val=c;    edge[e].next=h

<br />Cashier EmploymentTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3409 Accepted: 1211<br />DescriptionA supermarket in Tehran is open 24 hours a day every day and needs a number of cashiers to fit its need. The supermarket manager has h

<br /><br />KingTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4080 Accepted: 1630<br />DescriptionOnce, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a so

KingTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 4087 Accepted: 1633<br />DescriptionOnce, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a sound king.'' Afte

Integer IntervalsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7404 Accepted: 2991<br />DescriptionAn integer interval [a,b], a < b, is a set of all consecutive integers beginning with a and ending with b. <br />Write a program that: finds the

<br /><br />Selecting CoursesTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4638 Accepted: 2009<br />DescriptionIt is well known that it is not easy to select courses in the college, for there is usually conflict among the time of the courses.

<br /><br />Machine ScheduleTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5816 Accepted: 2425<br />DescriptionAs we all know, machine scheduling is a very classical problem in computer science and has been studied for a very long history. S

一.最大独立点集的定义给定无向图G(V,E)，其中A为顶点集合V的子集，且E∩{(i,j) | i,j∈A}=空集.求A，使得|A|最大二.最小顶点覆盖集的定义给定无向图G(V,E)，一个顶点覆盖集(vertex covering)是指顶点集S(S为V的子集)，使得G中任一条边(x,y)至少有一端点在S中.求S，使得|S|最小三.最大完全子图的定义给定无向图G(V,E)，其中C为顶点集合V的子集，对于任意两个点i,j∈C，且i≠j，有(i,j)∈E.求C，使得|C|最大四.三者的等价性(1)最大独立点集与最

<br />什么是二分图，什么是二分图的最大匹配，这些定义我就不讲了，网上随便都找得到。二分图的最大匹配有两种求法，第 一种是最大流（我在此假设读者已有网络流的知识）；第二种就是我现在要讲的匈牙利算法。这个算法说白了就是最大流的算法，但是它跟据二分图匹配这个问题的 特点，把最大流算法做了简化，提高了效率。匈牙利算法其实很简单，但是网上搜不到什么说得清楚的文章。所以我决定要写一下。<br />最大流算法的核心问题就是找增广路径（augment path）。匈牙利算法也不例外，它的基本模式就是：<br /><b

<br />转自Matrix67神牛<br /> <br /> <br />如果你看不清楚第二个字母，下面有一个大号字体版本：<br /><br />二分图最大匹配的König定理及其证明<br /><br />    本文将是这一系列里最短的一篇，因为我只打算把König定理证了，其它的废话一概没有。<br />    以下五个问题我可能会在以后的文章里说，如果你现在很想知道的话，网上去找找答案：<br />    1. 什么是二分图；<br />    2. 什么是二分图的匹配；<br />    3.

<br /><br />AsteroidsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5980 Accepted: 3091<br />DescriptionBessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <= N <= 500). The grid contain

<br />KM算法是通过给每个顶点一个标号（叫做顶标）来把求最大权匹配的问题转化为求完备匹配的问题的。设顶点Xi的顶标为A[i]，顶点Yi的顶标为B[i]，顶点Xi与Yj之间的边权为w[i,j]。在算法执行过程中的任一时刻，对于任一条边(i,j)，A[i]+B[j]>=w[i,j]始终成立。KM算法的正确性基于以下定理：<br />若由二分图中所有满足A[i]+B[j]=w[i,j]的边(i,j)构成的子图（称做相等子图）有完备匹配，那么这个完备匹配就是二分图的最大权匹配。<br />　　这个定理是显然的

<br /><br />Gopher IITime Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3494 Accepted: 1461<br />DescriptionThe gopher family, having averted the canine threat, must face a new predator. <br /><br />The are n gophers and m gopher holes, each at dist

<br />最小路径覆盖<br />　　在一个ＰＸＰ的有向图中，路径覆盖就是在图中找一些路经，使之覆盖了图中的所有顶点，且任何一个顶点有且只有一条路径与之关联；（如果把这些路径中的每条路径从它的起始点走到它的终点，那么恰好可以经过图中的每个顶点一次且仅一次）；如果不考虑图中存在回路，那么每每条路径就是一个弱连通子集．　　由上面可以得出：　　１.一个单独的顶点是一条路径；　　２．如果存在一路径p1,p2,......pk，其中p1 为起点，pk为终点，那么在覆盖图中，顶点p1,p2,......pk不再与其它

<br /><br />    Air RaidTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 3329 Accepted: 1946<br />DescriptionConsider a town where all the streets are one-way and each street leads from one intersection to another. It is also known that starting f

Guardian of DecencyTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 3126 Accepted: 1308DescriptionFrank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is afraid that some of t

<br />Steady Cow AssignmentTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2134 Accepted: 730<br />DescriptionFarmer John's N (1 <= N <= 1000) cows each reside in one of B (1 <= B <= 20) barns which, of course, have limited capacity. Some cows re

<br /><br />【二分图】二分图是一种特殊的图结构，所有点分为两类，记做x和y，所有的边的两端分别在x和y，不存在两端同在x或y的边。<br /> <br />【最大匹配、完备匹配】<br />给定一个二分图(x,y)，找到一种匹配数最大的方案，记做最大匹配。|x|=|y|=匹配数时，我们称该匹配方案为完备匹配。<br />显然，解决了最大匹配也就解决了完备匹配。<br />解决二分图的最大匹配可以用网络流或者匈牙利算法，两者本质上是相同的，不过不论从编程复杂度还是运行效率来讲，匈牙利算法都更加优秀