HDU 4292 Food

Foodtime limit:1000msmemory limit:32768kbthis problem'll be judged onHDU. Original id:4292
64-bit integer IO format: %i64d Java class name: Main You, a part-time dining service worker in your college ' s dining hall, is now confused with a new problem:serve as many p Eople as possible.
The issue comes up as people in your college is more and more difficult to serve with meal:they eat only some certain ki NDS of food and drink, and with requirement unsatisfied, go away directly.
You are prepared F (1 <= f <=) kinds of food and D (1 <= d <=) kinds of drink. Each of the kind of food or drink have certain amount, that's, how many people could the food or drink serve. Besides, you know there ' re N (1 <= N <= $) people and you too can tell people ' s personal preference for food and D Rink.
Back to your goal:to serve as many people as possible. So you must decide a plan where some people is served while requirements of the rest of them is unmet. You should notice this, when one's requirement is unmet, he/she would just go away, and refusing any service.
InputThere is 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 whethe R people I would 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 whethe R people I would accept drink J. "Y" for Yes and "N" for No.
Please process until EOF (End of File).
OutputFor each test case, maximum number of people to be satisfied. * Print a single line with one integer.
Sample Input

4 3 to 1 1 1YYNNYYYNYYNYYNYYYNYYNNNY

Sample Output

3

SourceACM/ICPC Asia Regional Chengdu Online Solution: Maximum flow, man split edge stream is 1

1#include <bits/stdc++.h>2 using namespacestd;3 Const intINF =0x3f3f3f3f;4 Const intMAXN =1010;5 structArc {6     intTo,flow,next;7Arcintx =0,inty =0,intz =-1) {8to =x;9Flow =y;TenNext =Z; One     } A} e[maxn*MAXN]; - inthead[maxn],cur[maxn],d[maxn],tot,s,t; - voidAddintUintVintflow) { theE[tot] =arc (V,flow,head[u]); -Head[u] = tot++; -E[tot] = arc (U,0, Head[v]); -HEAD[V] = tot++; + } -queue<int>Q; + BOOLBFs () { A      while(!q.empty ()) Q.pop (); atmemset (d,-1,sizeofd); - Q.push (S); -D[s] =1; -      while(!Q.empty ()) { -         intU =Q.front (); - Q.pop (); in          for(inti = Head[u]; ~i; i =E[i].next) { -             if(E[i].flow && d[e[i].to] = =-1) { toD[e[i].to] = D[u] +1; + Q.push (e[i].to); -             } the         } *     } $     returnD[t] >-1;Panax Notoginseng } - intDfsintUintLow ) { the     if(U = = T)returnLow ; +     intTMP =0, A; A      for(int&i = Cur[u]; ~i; i =E[i].next) { the         if(E[i].flow && d[u]+1==d[e[i].to]&& (a=Dfs (E[i].to,min (low,e[i].flow)))) { +Low-=A; -TMP + =A; $E[i].flow-=A; $e[i^1].flow + =A; -             if(!low) Break; -         } the     } -     if(!tmp) D[u] =-1;Wuyi     returntmp; the } - intDinic () { Wu     intRET =0; -      while(BFS ()) { Aboutmemcpy (Cur,head,sizeofcur); $RET + =DFS (s,inf); -     } -     returnret; - } A CharSTR[MAXN]; + intMain () { the     intN,f,d,flow; -      while(~SCANF ("%d%d%d",&n,&f,&D)) { $memset (head,-1,sizeofhead); theS = tot =0; theR +; the          for(inti =1; I <= F; ++i) { thescanf"%d",&flow); - Add (s,i,flow); in         } the          for(inti =1; I <= D; ++i) { thescanf"%d",&flow); AboutAdd (F +i,t,flow); the         } the          for(inti =1; I <= N; ++i) { theAdd (F + D + i*2-1, F + D + i*2,1); +scanf"%s", str); -              for(intj =0; STR[J]; ++j) the                 if(Str[j] = ='Y') Add (j+1, F + D + i*2-1, INF);Bayi         } the          for(inti =1; I <= N; ++i) { thescanf"%s", str); -              for(intj =0; STR[J]; ++j) -                 if(Str[j] = ='Y') Add (F + D + i*2, F + j +1, INF); the         } theprintf"%d\n", Dinic ()); the     } the     return 0; -}

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HDU 4292 Food