Food

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8111    Accepted Submission(s): 2686

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

 

感覺和這題思路差不多：https://www.cnblogs.com/Fighting-sh/p/9818674.html

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cmath>
  5 #include<cstdio>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<vector>
  9 #include<set>
 10 #define maxn 200005
 11 #define MAXN 200005
 12 #define mem(a,b) memset(a,b,sizeof(a))
 13 const int N=200005;
 14 const int M=200005;
 15 const int INF=0x3f3f3f3f;
 16 using namespace std;
 17 int n;
 18 struct Edge{
 19     int v,next;
 20     int cap,flow;
 21 }edge[MAXN*20];//注意這裡要開的夠大。。不然WA在這裡真的想罵人。。問題是還不報RE。。
 22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
 23 int cnt=0;//實際儲存總邊數
 24 void isap_init()
 25 {
 26     cnt=0;
 27     memset(pre,-1,sizeof(pre));
 28 }
 29 void isap_add(int u,int v,int w)//加邊
 30 {
 31     edge[cnt].v=v;
 32     edge[cnt].cap=w;
 33     edge[cnt].flow=0;
 34     edge[cnt].next=pre[u];
 35     pre[u]=cnt++;
 36 }
 37 void add(int u,int v,int w){
 38     isap_add(u,v,w);
 39     isap_add(v,u,0);
 40 }
 41 bool bfs(int s,int t)//其實這個bfs可以融合到下面的迭代裡，但是好像是時間要長
 42 {
 43     memset(dep,-1,sizeof(dep));
 44     memset(gap,0,sizeof(gap));
 45     gap[0]=1;
 46     dep[t]=0;
 47     queue<int>q;
 48     while(!q.empty())
 49     q.pop();
 50     q.push(t);//從匯點開始反向建層次圖
 51     while(!q.empty())
 52     {
 53         int u=q.front();
 54         q.pop();
 55         for(int i=pre[u];i!=-1;i=edge[i].next)
 56         {
 57             int v=edge[i].v;
 58             if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是從匯點反向bfs，但應該判斷正向弧的餘量
 59             {
 60                 dep[v]=dep[u]+1;
 61                 gap[dep[v]]++;
 62                 q.push(v);
 63                 //if(v==sp)//感覺這兩句優化加了一般沒錯，但是有的題可能會錯，所以還是註釋出來，到時候視情況而定
 64                 //break;
 65             }
 66         }
 67     }
 68     return dep[s]!=-1;
 69 }
 70 int isap(int s,int t)
 71 {
 72     if(!bfs(s,t))
 73     return 0;
 74     memcpy(cur,pre,sizeof(pre));
 75     //for(int i=1;i<=n;i++)
 76     //cout<<"cur "<<cur[i]<<endl;
 77     int u=s;
 78     path[u]=-1;
 79     int ans=0;
 80     while(dep[s]<n)//迭代尋找增廣路,n為節點數
 81     {
 82         if(u==t)
 83         {
 84             int f=INF;
 85             for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增廣路
 86                 f=min(f,edge[i].cap-edge[i].flow);
 87             for(int i=path[u];i!=-1;i=path[edge[i^1].v])
 88             {
 89                 edge[i].flow+=f;
 90                 edge[i^1].flow-=f;
 91             }
 92             ans+=f;
 93             u=s;
 94             continue;
 95         }
 96         bool flag=false;
 97         int v;
 98         for(int i=cur[u];i!=-1;i=edge[i].next)
 99         {
100             v=edge[i].v;
101             if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
102             {
103                 cur[u]=path[v]=i;//當前弧優化
104                 flag=true;
105                 break;
106             }
107         }
108         if(flag)
109         {
110             u=v;
111             continue;
112         }
113         int x=n;
114         if(!(--gap[dep[u]]))return ans;//gap優化
115         for(int i=pre[u];i!=-1;i=edge[i].next)
116         {
117             if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
118             {
119                 x=dep[edge[i].v];
120                 cur[u]=i;//常數優化
121             }
122         }
123         dep[u]=x+1;
124         gap[dep[u]]++;
125         if(u!=s)//當前點沒有增廣路則後退一個點
126         u=edge[path[u]^1].v;
127      }
128      return ans;
129 }
130 
131 int a[maxn];
132 int F[maxn],D[maxn];
133 string mp[505];
134 
135 int main(){
136     std::ios::sync_with_stdio(false);
137     int m,s,t;
138     int f,d;
139     while(cin>>n>>f>>d){
140         for(int i=1;i<=f;i++) cin>>F[i];
141         for(int i=1;i<=d;i++) cin>>D[i];
142         for(int i=1;i<=2*n;i++) cin>>mp[i];
143         isap_init();
144         s=0,t=n+n+f+d+1;
145         for(int i=1;i<=n;i++){
146             for(int j=0;j<f;j++){
147                 if(mp[i][j]=='Y'){
148                     add(j+1,f+i,1);
149                 }
150             }
151         }
152         for(int i=1;i<=n;i++){
153             for(int j=0;j<d;j++){
154                 if(mp[i+n][j]=='Y'){
155                     add(f+n+i,f+n+n+j+1,1);
156                 }
157             }
158         }
159         for(int i=1;i<=n;i++) add(f+i,f+n+i,1);
160         for(int i=1;i<=f;i++) add(s,i,F[i]);
161         for(int i=1;i<=d;i++) add(f+n+n+i,t,D[i]);
162         n=n+n+f+d+2;
163         cout<<isap(s,t)<<endl;
164     }
165 }

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