HDU 4292 Food （网络流，最大流）
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
Http
HDU:https://vjudge.net/problem/HDU-4292
Hit
网络流，最大流
题目大意
现在提供若干种饮料、食物，每种都有一定的数量。给定N个人对每一种食物和饮料的喜好，若给某人提供了其喜欢的食物和饮料，则称这个人是开心的。现在求能使最多的人开心的数量。
解决思路
我们按照源点-食物-人-饮料-汇点的顺序建图。对于每一个食物，从源点连容量为其数量的的边，而对于每一个人，连接他和他所有喜欢的食物，容量都是1,。因为给一个人只提供一份食物和一份饮料，所以我们把人拆点，中间连容量为1的边。而对于饮料则是类似的，从人到他喜欢的饮料连容量为1的边，从饮料到汇点连容量为其数量的边。这样跑最大流即可。
关于最大流，这里使用Dinic，可以参考这篇文章
代码
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxN=1001;
const int maxM=maxN*maxN*2;
const int inf=2147483647;
class Edge
{
public:
int u,v,flow;
};
int N,F,D;//点的编号安排：0汇点，[1,N]人1，[N+1,N*2]人2，[N*2+1,N*2+F]食物，[N*2+F+1,N*2+F+D]饮料，N*2+F+D+1汇点
int cnt=-1;
char str[maxN];
int Head[maxN];
int Next[maxM];
Edge E[maxM];
int depth[maxN];
int cur[maxN];
int Q[maxN];
void Add_Edge(int u,int v,int flow);
bool bfs();
int dfs(int u,int flow);
int main()
{
while (cin>>N>>F>>D)//多组数据
{
cnt=-1;//先清空
memset(Head,-1,sizeof(Head));
for (int i=1;i<=F;i++)//读入食物的数量
{
int maxflow;
scanf("%d",&maxflow);
Add_Edge(0,N*2+i,maxflow);//连接源点和食物
}
for (int i=1;i<=D;i++)//读入饮料的数量
{
int maxflow;
scanf("%d",&maxflow);
Add_Edge(N*2+F+i,N*2+F+D+1,maxflow);//连接饮料和汇点
}
for (int i=1;i<=N;i++)
{
scanf("%s",str);
for (int j=0;j<F;j++)
if (str[j]=='Y')
Add_Edge(N*2+j+1,i,1);//连接人与食物
}
for (int i=1;i<=N;i++)
{
scanf("%s",str);
for (int j=0;j<D;j++)
if (str[j]=='Y')
Add_Edge(N+i,N*2+F+j+1,1);//连接人与饮料
}
for (int i=1;i<=N;i++)
Add_Edge(i,N+i,1);//连接人1与人2，即拆开的两个点
int Ans=0;//求解最大流
while (bfs())
{
for (int i=0;i<=2*N+F+D+1;i++)
cur[i]=Head[i];
while (int di=dfs(0,inf))
Ans+=di;
}
cout<<Ans<<endl;
}
return 0;
}
void Add_Edge(int u,int v,int flow)
{
cnt++;
Next[cnt]=Head[u];
Head[u]=cnt;
E[cnt].u=u;
E[cnt].v=v;
E[cnt].flow=flow;
cnt++;
Next[cnt]=Head[v];
Head[v]=cnt;
E[cnt].u=v;
E[cnt].v=u;
E[cnt].flow=0;
}
bool bfs()
{
memset(depth,-1,sizeof(depth));
int h=1,t=0;
depth[0]=1;
Q[1]=0;
do
{
t++;
int u=Q[t];
for (int i=Head[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==-1)&&(E[i].flow>0))
{
depth[v]=depth[u]+1;
h++;
Q[h]=v;
}
}
}
while (t!=h);
if (depth[N*2+F+D+1]==-1)
return 0;
return 1;
}
int dfs(int u,int flow)
{
if (u==N*2+F+D+1)
return flow;
for (int i=Head[u];i!=-1;i=Next[i])
{
int v=E[i].v;
if ((depth[v]==depth[u]+1)&&(E[i].flow>0))
{
int di=dfs(v,min(flow,E[i].flow));
if (di>0)
{
E[i].flow-=di;
E[i^1].flow+=di;
return di;
}
}
}
return 0;
}
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