码代码的猿猿

kruscalAgri-NetTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 35233 Accepted: 14160DescriptionFarmer John has been elected mayor of his t

prim最小生成树里的最长边HighwaysTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 19843 Accepted: 9201DescriptionThe island nation of Flatopia is

tarjan。。。。。迷宫城堡Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5854    Accepted Submission(s): 2572Problem Description为了训练

抽象成：在一个有向图中，找所有的点都可以到达的点的个数根据题意结果必在一个强连通分量里（所有的答案之间肯定互相能到达）。用tarjan找出所有的连通分量，如果某个连通分量里有连到外面的边，则这个分量肯定不会是答案（分量图是有向的无环图，从分量1里连线到分量2中的边，分量2肯定没有变连回分量1，也就是1到达了2但是2不能到达1）。如果没有向外连边的连通分量不是1个就没有答案。

Tarjan求割点数NetworkTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8440 Accepted: 3950DescriptionA Telephone Line Company (TLC) is establish

找割点和去掉割点产生的连通块数。。。。。SPFTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5367 Accepted: 2476DescriptionConsider the two networks shown below

/* 强连通分量个数tarjan算法*/#include #include #include #include using namespace std;const int maxV=1000,maxE=1000000;struct Edge{ int to,next;}edge[maxE];int Adj[maxV],Size;void ini

/*求割点，去掉割点连通块的个数求桥*/const int maxV=10010,maxE=100010;struct Edge{ int to,next; bool cut;}edge[maxE];int Adj[maxV],Size;void init(){ Size=0; memset(Adj,-1,sizeof(Adj));}

tarjan求桥Critical LinksTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu[Submit]   [Go Back]   [Status]  Description  C

bellman_ford判断正环A - Currency ExchangeTime Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64uSubmit StatusAppoint description: System Crawler  (2013-10-

spfa找负环。。。。B - WormholesTime Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit StatusAppoint description: System Crawler  (2013-11-26)Descri

Floyd。。。。FroggerTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 22920 Accepted: 7448DescriptionFreddy Frog is sitting on a stone in the middle o

非常老的题目了，枚举层次dijkstra，G++AC昂贵的聘礼Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33747 Accepted: 9652Description年轻的探险家来到了一个印第安部落里。在那里他和酋长的女儿相爱了，于是

枚举每个起始点的最短路。。。Stockbroker GrapevineTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 24657 Accepted: 13579DescriptionStockbrokers are known

bellmanford最长路判正环ArbitrageTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13947 Accepted: 5876DescriptionArbitrage is the use of discrepancies i

判断次小生成树是否和最小生成树相等The Unique MSTTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 18632 Accepted: 6515DescriptionGiven a connected undirected g

次小生成树。。。Qin Shi Huang's National Road SystemTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2735    Accepted Submission(s): 983

Conference CallTime Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu[Submit]   [Go Back]   [Status]  DescriptionGood News! Utopia Polycom begi

将k换成2进制，构造分层的图B. Fox and Minimal pathtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputFox Ciel

tarjan+缩点+spfa最长路Instantaneous TransferenceTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 5443 Accepted: 1190DescriptionIt was long ago w

次短路。。。RoadblocksTime Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %I64d & %I64u[Submit]   [Go Back]   [Status]  DescriptionBessie has move

最小生成树。。。。ConscriptionTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6815 Accepted: 2347DescriptionWindy has a country, and he wants to bu

二分图染色。。。。CatchTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1214    Accepted Submission(s): 579Problem DescriptionA thie

判断一个集合内的点能否以花费为0互相抵达。。。。求集合间的最短路D. Dima and Bacteriatime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outp

经典2sat裸题,dfs的2sat可以方便输出字典序最小的解...Peaceful CommissionTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1578    Accepted Submission(s): 4

树的直径。。。。Park VisitTime Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2225    Accepted Submission(s): 982Problem DescriptionCl

2set的tarjan写法.....PartyTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3841    Accepted Submission(s): 1244Problem Description

二分时间+2sat边加多了....RE了好久......Nowor laterTime Limit: 9000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu[Submit]   [Go Back]   [Status]  D

裸的tarjianLCA。。。。Closest Common AncestorsTime Limit: 2000MS Memory Limit: 10000KTotal Submissions: 14328 Accepted: 4609DescriptionWrite a program

tarjanLCA求树上任意两点间的距离。。。。任意选一个点当根节点，dist[a,b]=   dist[a] +dist[b] -2* lca[a,b]How far away ？Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submissi

2sat....  队长离开时两个队员必须同时留下,两个队员中只要有人离开队长就得留下...Let's go homeTime Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1304    Accepted Submissio

2sat....可看出每个宇航员可以有两种任务选择 (A/B) 或 C对于每一对有矛盾的宇航员,如果属于同一年龄段 那么这两个人选择的任务不应该相同如果属于不同的年龄段,则俩个人不能同时选择任务CAstronautsTime Limit: 3000MS Memory Limit: Unknown 64bit IO Format: %l

二分距离2satBomb GameTime Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3382    Accepted Submission(s): 1161Problem DescriptionR

每条边有两种连法,在圆的外侧或圆的内侧. 对于有可能相交的两条边,不能在同一侧. 建图判断2sat Ikki's Story IV - Panda's TrickTime Limit: 1000MS Memory Limit: 131072KTotal Submissions: 7692 Accepted: 284

有两个权值,距离相等的时候判断一下花费最短路径问题Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11075    Accepted Submission(s): 3348Problem Descript

缩块找奇环.....Knights of the Round TableTime Limit: 7000MS Memory Limit: 65536KTotal Submissions: 9128 Accepted: 2943DescriptionBeing a knight is a very

畅通工程续Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 24979    Accepted Submission(s): 8900Problem Description某省自从实行了很多年的畅通工程计划后，

转自:http://blog.youkuaiyun.com/jarjingx/article/details/8521690

题意：给一个无向图，敌人还可以再随机建一条边，而你可以拆掉一条边并产生一定的花费。问至少要多少花费才能保证拆掉一条边之后图不联通。tarjan缩点。。。。如果最大度否则找最小的某个节点的到子节点的第3小边（第1小，和第2小的边是可能被连到一起的），具体做法：重新构图的时候找到最小的边，然后从最小的边的两个端点分别dfs找第二小的边。。。注意可能出现重边，建图的时候要给每

六度分离Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3859    Accepted Submission(s): 1567Problem Description1967年，美国著名的社会学家斯坦利·米尔