HDU 4292 food 37th ACM/ICPC Chengdu division Network Competition 1005 question (maximum Stream)

Food

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 76 accepted submission (s): 47

Problem description you, a part-time dining service worker in your college's dining hall, are now confused with a new problem: serve as your people as possible.
The issue comes up as people in your college are more and more difficult to serve with meal: they eat only some certain kinds of food and drink, and with requirement unsatisfied, go away directly.
You have prepared F (1 <= F <= 200) kinds of food and D (1 <= d <= 200) kinds of drink. each kind of food or drink has certain amount, that is, how many people cocould this food or drink serve. besides, you know there're n (1 <= n <= 200) people and you too can tell people's personal preference for food and drink.
Back to your goal: to serve as your people as possible. so you must decide a plan where some people are served while requirements of the rest of them are unmet. you shoshould notice that, when one's requirement is unmet, he/she woshould just go away, refusing any service.

 

Input there are several test cases.
For each test case, the first line contains three numbers: N, F, D, denoting the number of people, food, and drink.
The second line contains F integers, the ith number of which denotes amount of representative food.
The third line contains d integers, the ith number of which denotes amount of representative drink.
Following is N line, each consisting of a string of length F. E jth character in the ith one of these lines denotes whether people I wowould accept food J. "Y" for yes and "N" for no.
Following is N line, each consisting of a string of length D. E jth character in the ith one of these lines denotes whether people I wowould accept drink J. "Y" for yes and "N" for no.
Please process until EOF (end of file ).

 

Output for each test case, please print a single line with one integer, the maximum number of people to be satisfied.

 

Sample input4 3 3 1 1 1 1 1 yyn nyy yny yyn NNY

 

Sample output3

 

Source2012 ACM/ICPC Asia Regional Chengdu online

 

Recommendliuyiding It is very bare and the biggest stream question. Similar to poj 3182. Use SAPAlgorithmIt does not time out and is more efficient.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Include <Iostream> Using   Namespace  STD; Const   Int Maxn = 11000  ;  Const   Int Maxm = 405000  ;  Const   Int INF = 0x3f3f3f  ;  Struct  Node {  Int   From , To, next;  Int  CAP;} edge [maxm];  Int  Tol;  Int  Head [maxn];  Int  Dep [maxn];  Int  Gap [maxn];  Int  N;  Void  Init () {Tol = 0 ; Memset (Head, - 1 , Sizeof  (Head ));}  Void Addedge ( Int U, Int V, Int  W) {edge [tol].  From = U; edge [tol]. = V; edge [tol]. Cap = W; edge [tol]. Next =Head [u]; head [u] = Tol ++ ; Edge [tol].  From = V; edge [tol]. = U; edge [tol]. Cap = 0  ; Edge [tol]. Next = Head [v]; head [v] = Tol ++ ;}  Void BFS ( Int Start, Int  End) {memset (DEP, -1 , Sizeof  (DEP); memset (gap,  0 , Sizeof  (GAP); Gap [  0 ] = 1  ;  Int  Que [maxn];  Int  Front, rear; front = Rear = 0  ; Dep [end] =0  ; Que [rear ++] = End;  While (Front! = Rear ){  Int U = que [Front ++ ];  If (Front = maxn) Front = 0  ;  For ( Int I = head [u]; I! =- 1 ; I = Edge [I]. Next ){ Int V = Edge [I].;  If (Edge [I]. Cap! = 0 | Dep [v]! =- 1 ) Continue  ; Que [rear ++] = V;  If (Rear> = maxn) rear = 0  ; Dep [v] = Dep [u] + 1  ; ++Gap [Dep [v] ;}}  Int SAP ( Int Start, Int  End ){  Int Res = 0  ; BFS (START, end );  Int  Cur [maxn];  Int  S [maxn];  Int Top = 0  ; Memcpy (cur, Head, Sizeof  (Head ));  Int U = Start;  Int  I;  While (DEP [start] < N ){  If (U = End ){  Int Temp = INF;  Int  Inser;  For (I = 0 ; I <top; I ++ )  If (Temp> Edge [s [I]. Cap) {temp = Edge [s [I]. CAP; inser = I ;}  For (I = 0 ; I <top; I ++ ) {Edge [s [I]. Cap -= Temp; edge [s [I] ^ 1 ]. Cap + =Temp;} res + = Temp; top = Inser; u = Edge [s [Top]. From  ;}  If (U! = End & gap [Dep [u]- 1 ] = 0 ) //  Fault occurred, no zengguang Road             Break  ;  For (I = cur [u]; I! =-1 ; I = Edge [I]. Next)  If (Edge [I]. Cap! = 0 & Amp; Dep [u] = Dep [edge [I]. To] + 1  )  Break  ;  If (I! =- 1  ) {Cur [u] = I; s [Top ++] = I; u =Edge [I]. ;}  Else  {  Int Min = N;  For (I = head [u]; I! =- 1 ; I = Edge [I]. Next ){  If (Edge [I]. Cap = 0 ) Continue  ;  If (Min>Dep [edge [I]. To]) {min = Dep [edge [I]. To]; cur [u] = I ;}} -- Gap [Dep [u]; Dep [u] = Min + 1  ; ++ Gap [Dep [u];  If (U! = Start) u = edge [s [-- top]. From  ;}}  Return  Res ;} Int G [ 2000 ] [ 2000  ];  Char STR [ 1200  ];  Int  Main (){  Int  Start, end;  Int  N, F, D;  Int  U;  Int I;  While (Scanf ( "  % D  " , & N, & F, & D )! = EOF) {memset (G,  0 , Sizeof  (G); Init (); n = F + D + 2 * N; Start = 0  ; End = N + 1 ;  For (I = 1 ; I <= f; I ++ ) {Scanf (  "  % D  " , & G [ 0  ] [I]); addedge (  0 , I, G [ 0  ] [I]);}  For (I = F + 2 * N + 1 ; I <= F + 2 * N + D; I ++ ) {Scanf (  "  % D  " ,& G [I] [end]); addedge (I, end, G [I] [end]);}  For (I = 1 ; I <= N; I ++ ) Addedge (F + 2 * I- 1 , F + 2 * I, 1 );  For (I = 1 ; I <= N; I ++ ) {Scanf (  "  % S  " ,& Str );  For ( Int J = 0 ; J <F; j ++ ){  If (STR [J] = ' Y  '  ) {Addedge (J + 1 , F + 2 * I- 1 , 1  );}}}  For (I = 1 ; I <= N; I ++ ) {Scanf (  "  % S  " ,&Str );  For ( Int J = 0 ; J <D; j ++ ){  If (STR [J] = '  Y  '  ) {Addedge (F + 2 * I, F + 2 * N + J + 1 , 1 ) ;}} Start = 0  ; End = N + 1  ; N + = 2  ; Printf (  "  % D \ n  "  , SAP (START, end ));}  Return   0  ;}