# 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 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.

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).

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

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#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=11000;
const int MAXM=405000;
const int INF=0x3f3f3f3f;
struct Node
{
int from,to,next;
int cap;
}edge[MAXM];
int tol;
int dep[MAXN];
int gap[MAXN];
int n;
void init()
{
tol=0;
}
{
edge[tol].from=u;
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].from=v;
edge[tol].to=u;
edge[tol].cap=0;
}

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;
{
int v=edge[i].to;
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;
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)//出现断层，无增广路
break;
for(i=cur[u];i!=-1;i=edge[i].next)
if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
break;
if(i!=-1)
{
cur[u]=i;
S[top++]=i;
u=edge[i].to;
}
else
{
int min=n;
{
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%d%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]);
}

for(i=F+2*N+1;i<=F+2*N+D;i++)
{
scanf("%d",&g[i][end]);
}

for(i=1;i<=N;i++)

for(i=1;i<=N;i++)
{
scanf("%s",&str);
for(int j=0;j<F;j++)
{
if(str[j]=='Y')
{
}
}
}
for(i=1;i<=N;i++)
{
scanf("%s",&str);
for(int j=0;j<D;j++)
{
if(str[j]=='Y')
{
}
}
}
start=0;
end=n+1;
n+=2;
printf("%d\n",SAP(start,end));
}
return 0;
}

posted on 2012-09-16 19:34  kuangbin  阅读(1082)  评论(0编辑  收藏  举报

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