HDU 3605 escape (max Stream)

Escape

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 2523 accepted submission (s): 691

Problem description2012 if this is the end of the world how to do? I do not know how. but now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. now scientists want your help, is to determine what all of people can live in these planets.

 

Inputmore set of test data, the beginning of each data is n (1 <= n <= 100000), m (1 <= m <= 10) N indicate there n people on the earth, M representatives M planet, planet and people labels are from 0. here are n lines, each line represents a suitable living conditions of people, each row has m digits, the ith digits is 1, said that a person is fit to live in the ith-planet, or is 0 for this person is not suitable for living in the ith planet.
The last line has m digits, the ith digit AI indicates the ith planet can contain AI people most ..
0 <= AI <= 100000

 

Outputdetermine whether all people can live up to these stars
If you can output Yes, otherwise output No.

 

Sample input1 1 1
1
1

2 2
1 0
1 0
1 1

 

Sample outputyes
No

 

ACM-ICPC multi-university training Contest (17) -- host by zstu

 

Recommendlcy is the most simple stream. However, the data is big and many templates have been tried, all of which are TLE. Later, we found that vertices can be merged because M <= 10. Therefore, binary records are used. In N points, if they are the same, they are merged. In this way, the maximum number is 1024 + m + 2 points. But this question is still very difficult .... In HDU, G ++ is used to deliver the data to TLE. To directly read data and output data, convert it to "C ++" and then "AC...

 //  ========================================================== ==========================================  //  Name: HDU. cpp  //  Author: //  Version:  //  Copyright: Your copyright notice  //  Description: Hello world in C ++, ANSI-style  //  ========================================================== ==========================================  # Include <Iostream> # Include < String . H> # Include <Algorithm> # Include <Stdio. h> Using   Namespace  STD;  Const   Int Maxn = 100110  ;  Const   Int Maxm = 4000110  ;  Const   Int INF = 0x3f3f3f  ;  Struct Node {  Int  To, next, cap;} edge [maxm];  Int  Tol;  Int  Head [maxn];  Int  Gap [maxn], DIS [maxn], pre [maxn], cur [maxn];  Void  Init () {Tol = 0  ; Memset (Head, - 1 , Sizeof (Head ));}  Void Addedge ( Int U, Int V, Int W, Int RW = 0  ) {Edge [tol]. = V; edge [tol]. Cap = W; edge [tol]. Next = head [u]; head [u] = tol ++ ; Edge [tol]. = U; edge [tol]. Cap = RW; edge [tol]. Next = head [v]; head [v] = tol ++ ;}  Int SAP ( Int Start,Int End, Int  Nodenum) {memset (DIS,  0 , Sizeof  (DIS); memset (gap,  0 , Sizeof  (GAP); memcpy (cur, Head,  Sizeof  (Head ));  Int U = pre [start] = start, maxflow = 0 , Aug =- 1 ; Gap [  0 ] = Nodenum;  While (DIS [start] < Nodenum) {loop:  For ( Int & I = cur [u]; I! =- 1 ; I = Edge [I]. Next ){  Int V = Edge [I].;  If (Edge [I]. Cap & dis [u] = dis [v] + 1 ){  If (Aug =- 1 | Aug> Edge [I]. Cap) Aug = Edge [I]. CAP; Pre [v] = U; u = V;  If (V = End) {maxflow + = Aug;  For (U = pre [u]; V! = Start; V = u, u =Pre [u]) {edge [cur [u]. Cap -= Aug; edge [cur [u] ^ 1 ]. Cap + = Aug;} Aug =- 1  ;}  Goto  Loop ;}}  Int Mindis = Nodenum;  For ( Int I = head [u]; I! =- 1 ; I = Edge [I]. Next ){  Int V = Edge [I].;  If (Edge [I]. Cap & mindis> Dis [v]) {cur [u] = I; mindis = Dis [v] ;}}  If (-- Gap [dis [u]) = 0 ) Break  ; Gap [dis [u] = Mindis + 1 ] ++ ; U = Pre [u];}  Return  Maxflow ;}  Int Num [ 1025  ];  Int A [ 11  ];  Int Bit [ 11  ];  Int Main (){  //  Freopen ("in.txt", "r", stdin );  //  Freopen ("out.txt", "W", stdout ); Bit [ 0 ] = 1  ;  For ( Int I = 1 ; I <= 10 ; I ++) bit [I] = bit [I- 1 ] * 2 ;  Int  N, m;  While (Scanf ( "  % D  " , & N, & M )! = EOF) {Init (); memset (Num,  0 , Sizeof  (Num ));  For ( Int I = 1 ; I <= N; I ++){  Int TMP = 0  ;  For ( Int J = 0 ; J <m; j ++) {scanf ( "  % D  " , & A [J]); TMP + = A [J] * Bit [J];} num [TMP] ++ ;}  Int Start = 0 , End =1024 + M + 1 , Nodenum = 2014 + M + 2  ;  For ( Int I = 0 ; I < 1024 ; I ++ ){  If (Num [I] = 0 ) Continue  ; Addedge (start, I +1  , Num [I]);  For ( Int J = 0 ; J < 10 ; J ++ )  If (I & Bit [J]) addedge (I + 1 , 1024 + J + 1  , INF );}  Int TMP;  For ( Int I = 1 ; I <= m; I ++ ) {Scanf (  "  % D  " ,& TMP); addedge (I + 1024  , End, TMP );}  If (SAP (START, end, nodenum) = N) printf ( "  Yes \ n "  );  Else Printf ( "  No \ n  "  );}  Return   0  ;}