Food
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7423 Accepted Submission(s): 2491
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
2012 ACM/ICPC Asia Regional Chengdu Online
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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAX_V=2005;
const int st=0;
const int en=2000;
const int INF=0x3f3f3f3f;
struct edge
{
int to,cap,rev; //用于表示边的结构体（终点，容量，反向边）
};
vector <edge> G[MAX_V]; //图的邻接表表示
int level[MAX_V]; //顶点到源点的距离标号
int iter[MAX_V]; //当前弧，在其之前的边已经没有用了
char t[222];
void add_edge(int from,int to,int cap)
{
G[from].push_back((edge){to,cap,G[to].size()});
G[to].push_back((edge){from,0,G[from].size()-1});
}
void bfs(int s) //通过bfs计算从源点出发的距离标号
{
memset(level,-1,sizeof(level));
queue <int> q;
level[s]=0;
q.push(s);
while(!q.empty())
{
int v=q.front();q.pop();
for(int i=0;i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&&level[e.to]<0)
{
level[e.to]=level[v]+1;
q.push(e.to);
}
}
}
}
int dfs(int v,int t,int f) //通过dfs寻找增广路
{
if(v==t) return f;
for(int &i=iter[v];i<G[v].size();i++)
{
edge &e=G[v][i];
if(e.cap>0&&level[v]<level[e.to])
{
int d=dfs(e.to,t,min(f,e.cap));
if(d>0)
{
e.cap-=d;
G[e.to][e.rev].cap+=d;
return d;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int flow=0;
for(;;)
{
bfs(s); //计算层次图
if(level[t]<0) return flow; //找不到s-t路径
memset(iter,0,sizeof(iter)); //初始化当前弧
int f;
while((f=dfs(s,t,INF))>0) //更新最大流
flow+=f;
}
return flow;
}
int main()
{
int n,f,d;
while(scanf("%d%d%d",&n,&f,&d)!=EOF)
{
for(int i=0;i<MAX_V;i++) G[i].clear();
for(int i=1;i<=f;i++)
{
int x;scanf("%d",&x);
add_edge(i,i+200,x); //每种food有几个
add_edge(st,i,INF); //起点到food
}
for(int i=1;i<=d;i++)
{
int x;scanf("%d",&x);
add_edge(i+800,i+1000,x); //每种饮料有几个
add_edge(i+1000,en,INF); //饮料到终点
}
for(int i=1;i<=n;i++) add_edge(i+400,i+600,1); //人自己连，注意是1
for(int i=1;i<=n;i++)
{
scanf("%s",t+1);
for(int j=1;j<=f;j++)
if(t[j]=='Y') add_edge(j+200,i+400,INF); //人和food
}
for(int i=1;i<=n;i++)
{
scanf("%s",t+1);
for(int j=1;j<=d;j++)
if(t[j]=='Y') add_edge(i+600,j+800,INF); //人到饮料
}
printf("%d\n",max_flow(st,en));
}
return 0;
}