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 Output3 Source2012 ACM/ICPC Asia Regional Chengdu Online
感觉和这题思路差不多：https://www.cnblogs.com/Fighting-sh/p/9818674.html
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#define maxn 200005
#define MAXN 200005
#define mem(a,b) memset(a,b,sizeof(a))
const int N=;
const int M=;
const int INF=0x3f3f3f3f;
using namespace std;
int n;
struct Edge{
int v,next;
int cap,flow;
}edge[MAXN*];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
int cnt=;//实际存储总边数
void isap_init()
{
cnt=;
memset(pre,-,sizeof(pre));
}
void isap_add(int u,int v,int w)//加边
{
edge[cnt].v=v;
edge[cnt].cap=w;
edge[cnt].flow=;
edge[cnt].next=pre[u];
pre[u]=cnt++;
}
void add(int u,int v,int w){
isap_add(u,v,w);
isap_add(v,u,);
}
bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
{
memset(dep,-,sizeof(dep));
memset(gap,,sizeof(gap));
gap[]=;
dep[t]=;
queue<int>q;
while(!q.empty())
q.pop();
q.push(t);//从汇点开始反向建层次图
while(!q.empty())
{
int u=q.front();
q.pop();
for(int i=pre[u];i!=-;i=edge[i].next)
{
int v=edge[i].v;
if(dep[v]==-&&edge[i^].cap>edge[i^].flow)//注意是从汇点反向bfs，但应该判断正向弧的余量
{
dep[v]=dep[u]+;
gap[dep[v]]++;
q.push(v);
//if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
//break;
}
}
}
return dep[s]!=-;
}
int isap(int s,int t)
{
if(!bfs(s,t))
return ;
memcpy(cur,pre,sizeof(pre));
//for(int i=1;i<=n;i++)
//cout<<"cur "<<cur[i]<<endl;
int u=s;
path[u]=-;
int ans=;
while(dep[s]<n)//迭代寻找增广路,n为节点数
{
if(u==t)
{
int f=INF;
for(int i=path[u];i!=-;i=path[edge[i^].v])//修改找到的增广路
f=min(f,edge[i].cap-edge[i].flow);
for(int i=path[u];i!=-;i=path[edge[i^].v])
{
edge[i].flow+=f;
edge[i^].flow-=f;
}
ans+=f;
u=s;
continue;
}
bool flag=false;
int v;
for(int i=cur[u];i!=-;i=edge[i].next)
{
v=edge[i].v;
if(dep[v]+==dep[u]&&edge[i].cap-edge[i].flow)
{
cur[u]=path[v]=i;//当前弧优化
flag=true;
break;
}
}
if(flag)
{
u=v;
continue;
}
int x=n;
if(!(--gap[dep[u]]))return ans;//gap优化
for(int i=pre[u];i!=-;i=edge[i].next)
{
if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
{
x=dep[edge[i].v];
cur[u]=i;//常数优化
}
}
dep[u]=x+;
gap[dep[u]]++;
if(u!=s)//当前点没有增广路则后退一个点
u=edge[path[u]^].v;
}
return ans;
} int a[maxn];
int F[maxn],D[maxn];
string mp[]; int main(){
std::ios::sync_with_stdio(false);
int m,s,t;
int f,d;
while(cin>>n>>f>>d){
for(int i=;i<=f;i++) cin>>F[i];
for(int i=;i<=d;i++) cin>>D[i];
for(int i=;i<=*n;i++) cin>>mp[i];
isap_init();
s=,t=n+n+f+d+;
for(int i=;i<=n;i++){
for(int j=;j<f;j++){
if(mp[i][j]=='Y'){
add(j+,f+i,);
}
}
}
for(int i=;i<=n;i++){
for(int j=;j<d;j++){
if(mp[i+n][j]=='Y'){
add(f+n+i,f+n+n+j+,);
}
}
}
for(int i=;i<=n;i++) add(f+i,f+n+i,);
for(int i=;i<=f;i++) add(s,i,F[i]);
for(int i=;i<=d;i++) add(f+n+n+i,t,D[i]);
n=n+n+f+d+;
cout<<isap(s,t)<<endl;
}
}
