Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3658 Accepted Submission(s): 1246
Problem Description
You, a part-time dining service worker in your college’s dining hall, are now confused with a new problem: serve as many 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 could 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 many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would just go away, refusing any service.
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 could 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 many people as possible. So you must decide a plan where some people are served while requirements of the rest of them are unmet. You should notice that, when one’s requirement is unmet, he/she would 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 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 whether people i would accept drink j. “Y” for yes and “N” for no.
Please process until EOF (End Of File).
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 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 whether people i would 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 Input
4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY
Sample Output
3
Source
题解：
建图：S-F-人-人-D-T
代码
//зїеп:1085422276 #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef __int64 ll; using namespace std; const int inf = 10000000; inline ll read() { ll x=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9') { if(ch=='-')f=-1; ch=getchar(); } while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar(); } return x*f; } //******************************* namespace NetFlow { const int MAXN=100000,MAXM=1000000,inf=1e9; struct Edge { int v,c,f,nx; Edge() {} Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {} } E[MAXM]; int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz; void init(int _n) { N=_n,sz=0; memset(G,-1,sizeof(G[0])*N); } void link(int u,int v,int c) { E[sz]=Edge(v,c,0,G[u]); G[u]=sz++; E[sz]=Edge(u,0,0,G[v]); G[v]=sz++; } int ISAP(int S,int T) {//S -> T int maxflow=0,aug=inf,flag=false,u,v; for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0; for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false) { for (int &it=cur[u];~it;it=E[it].nx) { if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+1) { if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f; pre[v]=u,u=v; flag=true; if (u==T) { for (maxflow+=aug;u!=S;) { E[cur[u=pre[u]]].f+=aug; E[cur[u]^1].f-=aug; } aug=inf; } break; } } if (flag) continue; int mx=N; for (int it=G[u];~it;it=E[it].nx) { if (E[it].c>E[it].f&&dis[E[it].v]<mx) { mx=dis[E[it].v]; cur[u]=it; } } if ((--gap[dis[u]])==0) break; ++gap[dis[u]=mx+1]; u=pre[u]; } return maxflow; } bool bfs(int S,int T) { static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N); dis[S]=0; Q[0]=S; for (int h=0,t=1,u,v,it;h<t;++h) { for (u=Q[h],it=G[u];~it;it=E[it].nx) { if (dis[v=E[it].v]==-1&&E[it].c>E[it].f) { dis[v]=dis[u]+1; Q[t++]=v; } } } return dis[T]!=-1; } int dfs(int u,int T,int low) { if (u==T) return low; int ret=0,tmp,v; for (int &it=cur[u];~it&&ret<low;it=E[it].nx) { if (dis[v=E[it].v]==dis[u]+1&&E[it].c>E[it].f) { if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f))) { ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp; } } } if (!ret) dis[u]=-1; return ret; } int dinic(int S,int T) { int maxflow=0,tmp; while (bfs(S,T)) { memcpy(cur,G,sizeof(G[0])*N); while (tmp=dfs(S,T,inf)) maxflow+=tmp; } return maxflow; } } using namespace NetFlow; int main() { int n,f,d; int fs,ds; while(scanf("%d%d%d",&n,&f,&d)!=EOF){ init(900); for(int i=1;i<=f;i++) { fs=read(); link(801,i,fs); } for(int i=1;i<=d;i++) { ds=read(); link(i+600,802,ds); } char ch[201]; for(int i=1;i<=n;i++) { scanf("%s",ch); for(int j=0;j<f;j++) { if(ch[j]=='Y'){ link(j+1,200+i,1); } } } for(int i=1;i<=n;i++) { scanf("%s",ch); for(int j=0;j<d;j++) { if(ch[j]=='Y'){ link(400+i,j+1+600,1); } } }// for(int i=1;i<=n;i++)link(200+i,400+i,1); printf("%d ",ISAP(801,802));//cout<<1111<<endl; } return 0; }