3592 e05

A map with a number indicates that there are digits and minerals on it. * indicates that this point can be transferred to another point. # indicates that it cannot be taken. Only the lower and right sides of a vertex can be reached. Now, starting from (0, 0), ask how many minerals can be collected at most. Train of Thought: This question must be illustrated, and then the longest path, the key is that with the transfer, it is possible to form a right ring, and then it will die in the spfa proces

+memset (Dist,0x3f,sizeof(Dist));//Distance Adequeint> Que (1, ST); theinq[st]=1; +dist[st]=0; - while(!que.empty ()) $ { $ intt=Que.front (); Que.pop_front (); -inq[t]=0; node E; - for(intI=HEAD[T]; i!=-1; I=e.next) the { -E=Edge[i];Wuyi if(dist[e.to]>dist[t]+E.dis) the { -dist[e.to]=dist[t]+E.dis; Wu if(!inq[e.to])//not in the queue - { About if(++cnt[e.to]>n)//too many times in the q

Click to open linkTest instructions: There are n individuals, and then x relationship and y relationship, the x relationship represents that the distance of these two people can not exceed c,y represents the distance of these two people is greater than or equal to C, if not meet all the output-1, if the position of 1 and n can be infinitely large output-2, otherwise the maximum distance to output two peopleIdea: is to check the model of the constraint, x relationship according to the location of

],num[n];intDis[n];structedge{intTo,next,val;} edge[4*N];voidAdd_edge (intSintEintval) {edge[cnt].to=e; Edge[cnt].next=Head[s]; Edge[cnt].val=Val; Head[s]=cnt++;} Queueint>Qu;voidSPFA () { while(!qu.empty ()) Qu.pop (); MST (Vis,0); MST (Dis,inf); MST (NUM,0); Qu.push (1); dis[1]=0; vis[1]=1; while(!Qu.empty ()) { intFr=Qu.front (); Qu.pop (); NUM[FR]++; if(num[fr]>N) {printf ("-1\n"); return ; } for(inti=head[fr];i!=-1; i=Edge[i].next) { intx=edge[i].to; if(

, output-1. If Person 1 and N can is arbitrarily far apart, output-2. Otherwise output The greatest possible distance between person 1 and N.Sample Input14 2 11 3 82 4 152 3 4Sample Output19Authoralpc20Source2010 acm-icpc multi-university Training Contest (--host by Nudt)/* ***********************************************author:ckbosscreated time:2015 July 30 Thursday 08:16 29 seconds file Name : hdoj3592.cpp************************************************ * * #include Copyright NOTICE: This art

coordinates of the arrival of the minerals owned by the edge of the weight value. The question then becomes: the longest path the minecart can reach from node 0. But in addition to the right and down the edge, considering the transmission point and the target coordinates are composed of edges, the original image will be much more back to the edge, forming a lot of the forward ring. The emergence of a forward ring, so that the mine can be picked up a part of the mineral, as long as it can go int

HDOJ 3592 World Exhibition difference constraint, hdoj3592 World Exhibition Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 1373 Accepted Submission (s): 673 Problem DescriptionNowadays, please people want to go to Shanghai to visit the World Exhibition. so there are always a lot of people who are standing along a straight line waiting for entering. assume that there are N (2 There is something

World Exhibition Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 568 accepted submission (s): 264Problem description Nowadays, Please people want to go to Shanghai to visit the world exhibition.

Question: There are several points on the plane, and the minimum spanning tree is obtained. There are a maximum of 8 packages. Each package has a price and a number of points. Once purchased, the points in the package will be connected to each other.

World ExhibitionTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 931 Accepted Submission (s): 468 Problem DescriptionNowadays, please people want to go to Shanghai to visit the World Exhibition.

It is very difficult for others to answer questions... Alas   First, this question is abstracted into a directed graph. Due to the existence of magic power, this graph has a ring, so I think of the method of reducing points with strongly connected

/* Question: Given a matrix, the southwest corner is the starting point. Each unit has a valuable gold mine (# indicates rock, inaccessible, * indicates space-time gate, and can reach a specified Unit) what benefits can be obtained at most? Strong

Click to open link Test instructions: There are n individuals, and then x relationship and y relationship, the x relationship represents that the distance of these two people can not exceed c,y represents the distance of these two people is greater

Oracle tree query. If you need it, refer to it. The code is messy. Sorry. Oracle tree query. If you need it, refer to it. The code is messy. Sorry. Format: SELECT column FROM table_name Start with column = value Connect by prior parent primary key = Child foreign key Select lpad ('', 4 * (level-1) | name, job, id, super from emp Start with super is null Connect by prior id = super Example: Raw data: select no, q from a_example2 NO NAME ---------------------------------------- 001 a01 001 a02 00

Format: Select column From table_name Start with column = Value Connect by prior parent primary key = Child foreign key Select lpad ('', 4 * (level-1) | Name, job, ID, super from EMP Start with super is null Connect by prior id = super Example: Raw data: Select No, Q from a_example2 No name ---------------------------------------- 001 A01 001 A02 001 A03 001 a04 001 a05 002 B01 003 c01 003 C02 004 d01 005 e01 005 e02 005 e03 005 e04 005 e05 The result

Format: SELECT column From table_name START with Column=value CONNECT by PRIOR Parent PRIMARY KEY = child foreign key Select Lpad (', 4* (level-1)) | | Name Name,job,id,super from EMP Start with super is null Connect by Prior Id=super Example: Raw data: Select No,q from A_example2 NO NAME ---------- ------------------------------ 001 A01 001 A02 001 A03 001 A04 001 A05 002 B01 003 C01 003 C02 004 D01 005 E01 005 E02 005 E03 005 E04 005 E05 The results

Format:Select columnFrom table_nameStart with column = ValueConnect by prior parent primary key = Child foreign key Select lpad ('', 4 * (level-1) | Name, job, ID, super from EMPStart with super is nullConnect by prior id = super Example:Raw data: Select No, Q from a_example2 No name----------------------------------------001 A01001 A02001 A03001 a04001 a05002 B01003 c01003 C02004 d01005 e01005 e02005 e03005 e04005 e05 The result is as follows:001 A

, row_number () over (order by no, name desc) rn from a_example2))Start with rn1 is null connect by rn1 = prior rnGroup by noStatement Analysis:1. select no, name, row_number () over (order by no, name desc) rn from a_example2Sort by NO in ascending order, and sort by NAME in descending order to generate pseudo columns to form a Tree StructureNO NAME RN001 a03 1001 a02 2001 a01 3002 b01 4003 c02 5003 c01 6004 d01 7005 e05 8005 e04 9005 e03 10005 e02 1

Note: 1. "A" in from (select distinct classname from #) A must be available. The record set obtained from (select distinct classname from #) is treated as a new table. 2. When a group by statement is used in an SQL statement, functions such as Count (), sum (), AVG (), max (), and min () are generally used; However, if you directly use these functions without using the group by statement, all datasets are treated as a group for processing. Question 2: The schema of a relational database is kn