hdu 4292 Food 最大流

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4292

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.

题意描述：你准备了F种食物（F<=200）和D种饮料（D<=200），每种食物和饮料都有一定的数量。有n个人（n<=200），每个人都有喜欢的食物和饮料并且只能吃自己喜欢的食物、喝喜欢的饮料，求出能够吃到喜欢的食物并喝到喜欢的饮料的最大人数。

算法分析：运用网络最大流求解即可。新增源点from和汇点to，from->食物（w为食物数量），饮料->to（w为饮料数量），对每个人 i 拆边 i 和 i+n ，喜欢的食物->i（w为1），i -> i+n（w为1），i+n->喜欢的饮料（w为1）。

  1 #include<iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<cstdlib>

  5 #include<cmath>

  6 #include<algorithm>

  7 #include<vector>

  8 #include<queue>

  9 #define inf 0x7fffffff

 10 using namespace std;

 11 const int maxn=200+10;

 12 const int M = 999999;

 13 

 14 int n,F,D;

 15 int from,to,d[maxn*10];

 16 struct node

 17 {

 18     int v,flow;

 19     int next;

 20 }edge[M*2];

 21 int head[maxn*10],edgenum;

 22 

 23 void add(int u,int v,int flow)

 24 {

 25     edge[edgenum].v=v ;edge[edgenum].flow=flow;

 26     edge[edgenum].next=head[u];

 27     head[u]=edgenum++;

 28 

 29     edge[edgenum].v=u ;edge[edgenum].flow=0;

 30     edge[edgenum].next=head[v];

 31     head[v]=edgenum++;

 32 }

 33 

 34 int bfs()

 35 {

 36     memset(d,0,sizeof(d));

 37     d[from]=1;

 38     queue<int> Q;

 39     Q.push(from);

 40     while (!Q.empty())

 41     {

 42         int u=Q.front() ;Q.pop() ;

 43         for (int i=head[u] ;i!=-1 ;i=edge[i].next)

 44         {

 45             int v=edge[i].v;

 46             if (!d[v] && edge[i].flow>0)

 47             {

 48                 d[v]=d[u]+1;

 49                 Q.push(v);

 50                 if (v==to) return 1;

 51             }

 52         }

 53     }

 54     return 0;

 55 }

 56 

 57 int dfs(int u,int flow)

 58 {

 59     if (u==to || flow==0) return flow;

 60     int cap=flow;

 61     for (int i=head[u] ;i!=-1 ;i=edge[i].next)

 62     {

 63         int v=edge[i].v;

 64         if (d[v]==d[u]+1 && edge[i].flow>0)

 65         {

 66             int x=dfs(v,min(cap,edge[i].flow));

 67             cap -= x;

 68             edge[i].flow -= x;

 69             edge[i^1].flow += x;

 70             if (cap==0) return flow;

 71         }

 72     }

 73     return flow-cap;

 74 }

 75 

 76 int dinic()

 77 {

 78     int sum=0;

 79     while (bfs()) sum += dfs(from,inf);

 80     return sum;

 81 }

 82 

 83 int main()

 84 {

 85     while (scanf("%d%d%d",&n,&F,&D)!=EOF)

 86     {

 87         memset(head,-1,sizeof(head));

 88         edgenum=0;

 89         from=F+2*n+D+1;

 90         to=from+1;

 91         int a;

 92         for (int i=1 ;i<=F ;i++)

 93         {

 94             scanf("%d",&a);

 95             add(from,i,a);

 96         }

 97         for (int i=1 ;i<=D ;i++)

 98         {

 99             scanf("%d",&a);

100             add(F+2*n+i,to,a);

101         }

102         char str[maxn];

103         memset(str,0,sizeof(str));

104         for (int i=1 ;i<=n ;i++)

105         {

106             scanf("%s",str+1);

107             for (int j=1 ;j<=F ;j++)

108             {

109                 if (str[j]=='Y')

110                     add(j,F+i,1);

111             }

112         }

113         for (int i=1 ;i<=n ;i++)

114         {

115             scanf("%s",str+1);

116             for (int j=1 ;j<=D ;j++)

117             {

118                 if (str[j]=='Y')

119                     add(F+n+i,F+2*n+j,1);

120             }

121         }

122         for (int i=1 ;i<=n ;i++)

123             add(F+i,F+n+i,1);

124         printf("%d\n",dinic());

125     }

126     return 0;

127 }