勿忘初衷

确定比赛名次Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4   Accepted Submission(s) : 1Font: Times New Roman | Verdana | GeorgiaFont

畅通工程再续Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 3   Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont

过山车Time Limit : 1000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 2   Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Siz

棋盘游戏Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 4   Accepted Submission(s) : 2Font: Times New Roman | Verdana | GeorgiaFont Si

Uncle Tom's Inherited Land*Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 967    Accepted Submission(s): 422Special JudgeProblem De

Children of the Candy CornTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)Total Submission(s) : 6   Accepted Submission(s) : 6Problem DescriptionThe co

Catch That CowTime Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)Total Submission(s) : 25   Accepted Submission(s) : 9Problem DescriptionFarmer John has

Choose the best routeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4040    Accepted Submission(s): 1260Problem DescriptionOne

Bus SystemTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4239    Accepted Submission(s): 1066Problem DescriptionBecause of the

Minimal Ratio TreeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1285    Accepted Submission(s): 383Problem DescriptionFor a tr

Problem A: The MonocycleA monocycle is a cycle that runs on one wheel and the one we will be considering is a bit more special. It has a solid wheel colored with five different colors as sho

Given a sequence of integers, a1, a2,...,an , we define its sign matrix S such that, for1ijn , Sij = `` + " if Sij = `` - " if ai +...+aj ; and Sij = ``0" otherwise.For example, if (a1,

Problem B: Fire! Joe works in a maze. Unfortunately, portions of the maze have caught on fire, and the owner of the maze neglected to create a fire escape plan. Help Joe escape the maze.Given Joe'

Problem C: A Walk Through the ForestJimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To m

Problem D: Airport ExpressIn a small city called Iokh, a train service, Airport-Express, takes residents to the airport more quickly than other transports. There are two types of tra

The army of United Nations launched a new wave of air strikes on terrorist forces. The objective of the mission is to reduce enemy's logistical mobility. Each air strike will destroy a path and theref

TLE 啊 TLE，我的想法和讨论版里的是一样的，为什么就是一直TLE呢。是每次查询都重做一遍dijkstra导致的吗？怎么还有人也那样却ac了呢。求大神解释啊。代码啊，恶心的代码啊#include #include #include #include #include //#include using namespace std;const int max

差分约束这块内容，花了我好几个小时才算真正看懂了。我觉得不是我理解能力差，是网上和书上的资料都没说的很清楚开始时一直不能理解那个d[v]因为我们在求最短路判断时写的是>=后来才明白那个不等式指的是做完最短路后，每个点满足的性质。。。然后下面是两个我觉得比较好的链接，作为查分约束的入门和题集http://zakir.is-programmer.com/posts/21699.

Problem GDoves and BombsInput: Standard InputOutput: Standard Output Time Limit: 2 SecondsIt is the year 95 ACM (After the Crash of Microsoft). After many years of peace, a war has broken

Problem CSAM I AMInput: Standard InputOutput: Standard Output The world is in great danger!! Mental's forces have returned to Earth to eradicate humankind. Our last hope to stop this great evi

Hua and Shen have invented a simple solitaire board game that they call ``The Great Wall Game." The game is played withn stones on ann×n grid. The stones are placed at random in the squares of the g

Problem C - Cat vs. DogTime limit: 2 secondsThe latest reality show has hit the TV: ``Cat vs. Dog''. In this show, a bunch of cats and dogs compete for the very prestigiousBest Pet Ever title. I

这道题是uva 1349 的简化版，那题没过，不知道为什么。我觉得那题就是多了一个先判断他最大匹配数是不是n，是的话，再找最优匹配。回到这题，匹配问题，又是有向图，直接想到了拆点法。然后发现若每个点都恰好属于一个环，即是每个点的初度入度都为1，所以只要这个二分图有完美匹配，就是满足题目意思的。这样的话就是找最小匹配了，处理方法，每条边的距离用一个INF值减，最后累加时别忘了再用INF减回来

Problem IRevolC FaeLoNHopefully, you can remember Koorosh from one of the previous problems. Fortunately, Koorosh has solved Basm’s problem without your help! Now, he is a high-ranked advisor of B

Problem BClaw DecompositionInput: Standard InputOutput: Standard Output A claw is defined as a pointed curved nail on the end of each toe in birds, some reptiles, and some mammals. However, if

Problem B: The Largest CliqueGiven a directed graph G, consider the following transformation. First, create a new graphT(G) to have the same vertex set as G. Create a directed edge between two v

#include #include #include using namespace std;const int maxn = 500 + 10;const int INF = 1000000000;int W[maxn][maxn], n;int Lx[maxn], Ly[maxn]; // 顶标int left[maxn]; // lef

感觉这个问题还是蛮有趣的，照着书上敲了一遍，时间复杂度O（n2）。留存作模版#include #include using namespace std;const int maxn = 1000 + 10;int pref[maxn][maxn],order[maxn][maxn],next[maxn];int future_husband[maxn],future_w

#include#include#include#include#includeusing namespace std;const int maxn = 100 + 10;const int INF = 1000000000;struct Edge { int from, to, cap, flow;};struct Dinic {

#include#include#include#include#includeusing namespace std;const int maxn = 202 + 10;const int INF = 1000000000;struct Edge { int from, to, cap, flow, cost;};struct MCMF

#include#include#include#include#includeusing namespace std;const int maxn = 1000 + 10;struct Edge { int x, y; double d; bool operator < (const Edge& rhs) const { retur

#include#include#include#include#includeusing namespace std;const int maxn = 1000 + 10;struct Edge { int u, v; };int pre[maxn], iscut[maxn], bccno[maxn], dfs_clock, bcc_cnt; // 割

#include#include#include#include#includeusing namespace std;const int maxn = 1000 + 10;//图中节点编号从0开始，scc从1开始vector G[maxn];int pre[maxn], lowlink[maxn], sccno[maxn], dfs_clock, scc_

const int maxn = 500 + 5; // 单侧顶点的最大数目// 二分图最大基数匹配，邻接矩阵写法struct BPM { int n, m; // 左右顶点个数 int G[maxn][maxn]; // 邻接表 int left[maxn]; // left[i]为右边第i个点的匹配点编号，-1

Problem E: WeddingUp to thirty couples will attend a wedding feast, at which they will be seated on either side of a long table. The bride and groom sit at one end, opposite each other, and the brid

csdn真是神坑，刚写的解题竟然被吞了，尼玛，害我重写。本题的突破口在每个投票人都要有超过一半的建议被采纳。所以对于k=3的，只有一个建议不能采纳。都得采纳的，我用了一个must数组记录，然后dfs找出所有开始就定下来的建议的值；只有一个不能采纳，即每两个建议间都至少有一个被采纳。最后输出要判断“？”，我的方法是对每个建议检查是否无论它采不采纳都有解。开始忘了dfs找所有应该must的

Problem IAcme CorporationInput: Standard InputOutput: Standard Output Wile E. Coyote is back. He is back in the business. The business of capturing the road runner. Being the most loyal custom

Problem B - March of the PenguinsTime limit: 4 secondsSomewhere near the south pole, a number of penguins are standing on a number of ice floes. Being social animals, the penguins would like to ge

#include #include #include #include using namespace std;const int maxn = 1000 + 5; // 单侧顶点的最大数目// 二分图最大基数匹配struct BPM { int n, m; // 左右顶点个数 vector G[maxn]; //

You are appointed director of a famous concert hall, to save it from bankruptcy. The hall is very popular, and receives many requests to use its two fine rooms, but unfortunately the previous director