remind slack

HDU_2853 We can think that if we start matching from scratch and if the final part of matching changes but does not affect the final result, we do not need to change it into a new matching method. Therefore, we consider adding a "preference" to the original edge to ensure that it matches the original edge first without affecting the result. When processing data, we can expand each edge to a certain value (multiply by 10 in my program), and then add the original edge weight to 1, in this way, w

HDU_3772 We can split a string into two vertices to represent the inbound and outbound degrees, and then use the KM algorithm to perform Optimal Matching for the bipartite graph. #include#include#define MAXD 210#define INF 1000000000char b[MAXD][1010];int G[MAXD][MAXD] , yM[MAXD], N;int A[MAXD], B[MAXD], slack;int visx[MAXD], visy[MAXD];int check(char *str1, char *str2){int i, j, num = 0; i =strlen(str1) - 1;for(j = 0; str1[i] == str2[j] i >= 0

Uva_000046 This topic can be done with the minimum cost and the maximum flow. In order to review the optimal matching of the bipartite graph, we wrote a KM program. At first, because I had not processed the double KM before, I was going to multiply the decimal number by a certain power of 10 into an integer, but I found that WA was always there, and then I simply wrote it using double, and then the AC. #include#include#include#define MAXD 30#define INF 1000000000int N, M, xM[MAXD], yM[MAXD] ,vis

of their income. (Villagers who have the money to buy a house but not necessarily can buy it depends on what the village leader assigns).Input data contains multiple sets of test cases, the first row of each set of data input n, indicating the number of houses (also the number of people home), followed by n rows, the number of n per row represents the price of the room of the second village name (nOutput make the maximum revenue value for each set of data, one row for each set of outputs.Sample

intmaxn= -+Ten;Ten One intn,m; A intLX[MAXN],LY[MAXN],VISX[MAXN],VISY[MAXN]; - intLINK[MAXN],SLACK[MAXN],W[MAXN][MAXN]; - the intDfsintx) - { -visx[x]=1; - for(inty=1; yif(w[x][y]!=-1) + { - if(Visy[y])Continue; + intt=lx[x]+ly[y]-W[x][y]; A if(t==0) at { -visy[y]=1; - if(link[y]==-1||DFS (Link[y])) - { -link[y]=x; - return 1; in } - } to El

A tough question First, the input is very painful. According to the online experts, the question matrix is reversed. Then I gave it back. If it is a match between n and n, there is no pressure to directly obtain the negative value of KM. However, if the points on both sides are different, it is said that there will be problems. [Cpp]# Include # Include # Include # Include # Include # Include # Define maxn505# Define MAXM 555555# Define INF 1000000000Using namespace std;Int n, m, ny, nx;Int w [

20005Using namespace std;Struct Point {Double x, y;} Ant [105], tree [105];Double path [1, 101] [2, 101];Int cnt;Double lx [101], ly [101];Int match [101];Double slack;Bool v_x [101], v_y [101];Double dist (Point p1, Point p2 ){Return sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y ));}Bool dfs (int k ){V_x [k] = true;Double temp;For (int I = 0; I If (! V_y [I]) {Temp = lx [k] + ly [I]-path [k] [I];If (zero (temp )){V_y [I] = true;If (match

why calculate Setup time of the Slack need to consider adding cycles, Hold Time when you don't need it? Summary one:Because the data delay,launch edge and capture edge do not correspond to the same clock edge of the clock signal source clocks, due to the existence of the time of Setup, the addition cycle needs to be considered. When a single clock cycle is checked, the tool defaults to capture Edge-launch edge=1 cycles.Hold time because the two corre

easily solved. (It seems that a template like this, basic all similar problems can be resolved)1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defineMAXN 1008 #defineINF 1000000009 Ten intCases,n; One intA[MAXN][MAXN],B[MAXN][MAXN],VAL[MAXN][MAXN],SLACK[MAXN],VALX[MAXN],VALY[MAXN],LINKY[MAXN]; A BOOLVISX[MAXN],VISY[MAXN]; - - structpoint{ the intx, y; - }ans; - -Pointoperator-(point A,point b) {return(point) {a.x-b.

1 intG[n][n];2 intLx[n], ly[n];3 intSlack[n];4 intMatch[n];5 BOOLVisitx[n], visity[n];6 intN;7 8 BOOLHungary (intu)9 {TenVisitx[u] =true; One for(inti =0; I i) A { - if(Visity[i]) - Continue; the Else - { - if(Lx[u] + ly[i] = =G[u][i]) - { +Visity[i] =true; - if(Match[i] = =-1||Hungary (Match[i])) + { AMatch[i] =u; at return true; - } - } -

HDU_3488 It is worth mentioning that if we use the KM algorithm to find the perfect matching of minimum weights, we need to initialize the edge weight to the MAX-G [I] [j] and then find the perfect matching of the maximum right, and then convert the result back. If you initialize G [I] [j] to-G [I] [j] to find the perfect matching of the maximum permission, the program I write Will time out, at the moment, I have no idea whether this idea is a problem or my writing is a problem. #include#inclu

HDU_3718 First, we need to convert the characters in the string into the number of [0, k-1] with the same meaning, and then scan the array sequentially, we can get the maximum value of various matching between the two sets of numbers, this completes the graph creation. Then use the KM algorithm to find the optimal match. #include#include#define MAXD 30#define MAXN 10010#define INF 1000000000char a[MAXN][5], b[MAXN][5];int visa[MAXN], visb[MAXN], x[MAXN], y[MAXN];int G[MAXD][MAXD], yM[MAXD], N,

HDU_3395 We can split a fish into two points, which respectively represent attack and attack, and then set the Edge Weight of G [I] [j] = 1 to value [I].^ Value [j], and then use the KM algorithm to find the optimal matching of the Bipartite Graph. #include#include#define MAXD 110#define INF 1000000000int yM[MAXD], G[MAXD][MAXD], N, value[MAXD];int A[MAXD], B[MAXD];int visx[MAXD], visy[MAXD], slack;char b[MAXD];int init(){int i, j; scanf("%d", N);

tree, The Edge (I,J) may be added to the equal sub-graphIn order for the A[i]+b[j]>=w (I,J) to always be true, and at least one edge is added to the equal sub-graph, d=min{a[i]+b[j]-w (i,j)},i in the interlaced tree, J is not in the interlaced treetime complexity: need to find an O (n) time augmentation path. Each augmentation requires an O (n) sub-index change. Each time you change the top, enumerate the edges to find the D value, the complexity is O (N2), and the total complexity is O (N4). S

distance s. When 1 times slack is made on each side, the branches that start from S and have a maximum level of 1 are generated. That is, the shortest path to the vertices with a maximum of 1 edges associated with S is found, and the 2nd-pass relaxation of each edge creates a 2nd-level branch, which means that the shortest path of those vertices connected by 2 edges is found .... Because the shortest path contains up to |v|-1 edges only, you only nee

';"; Exequery ($connection, $query 1); } return $query; } function Affair_sms () { Include_once ("inc/utility_sms1.php"); Global $connection; Global $LOGIN _user_id; $CUR _date = DATE ("y-m-d", Time ()); $CUR _time = Date ("Y-m-d h:i:s", Time ()); $query = "SELECT * from Affair where User_id= '". $LOGIN _user_id. "' and Begin_time $cursor = Exequery ($connection, $query); while ($ROW = Mysql_fetch_array ($cursor)) { $AFF _id = $ROW [' aff_id ']; $USER _id = $ROW [' user_id '];

for relaxation until the queue is empty.The expected time complexity O (ke), where k is the average number of incoming teams for all vertices, can prove that K is generally less than or equal to 2.Implementation method:Set up a queue that initially has only a starting point in the queue, and then establish a table that records the shortest path from the starting point to all points (the initial value of the table is assigned a maximum value, and the point to his own path is assigned to 0). Then

* wait_address, poll_table * P) { -StructPoll_table_entry * entry = poll_get_entry (P ); + StructPoll_wqueues * pwq = container_of (p, struct poll_wqueues, pt ); + StructPoll_table_entry * entry = poll_get_entry (pwq ); If(! Entry) Return; Get_file (filp ); Entry-> filp= Filp; Entry-> wait_address= Wait_address; -Init_waitqueue_entry ( Entry-> wait,Current ); + Init_waitqueue_func_entry ( Entry-> wait,Pollwake ); + Entry-> wait. Private= Pwq; Add_wait_queue (wait_address, Entry-> wait ); } + In

decide whether to re-call nanosleep Based on the returned value to continue the execution of the remaining latency. The following is the hrtimer_nanosleep code: long hrtimer_nanosleep(struct timespec *rqtp, struct timespec __user *rmtp, const enum hrtimer_mode mode, const clockid_t clockid){struct restart_block *restart;struct hrtimer_sleeper t;int ret = 0;unsigned long slack;slack = current->timer_s

))return -EINVAL;return hrtimer_nanosleep(tu, rmtp, HRTIMER_MODE_REL, CLOCK_MONOTONIC);} Hrtimer_nanosleep is implemented as follows: long hrtimer_nanosleep(struct timespec *rqtp, struct timespec __user *rmtp, const enum hrtimer_mode mode, const clockid_t clockid){struct restart_block *restart;struct hrtimer_sleeper t;int ret = 0;unsigned long slack;slack = current->timer_slack_ns;if (rt_task(current)