Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5945    Accepted Submission(s): 2010

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

 
建图与POJ3281一毛一样,边开少了狂T。。。
 //2017-08-24
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; const int N = ;
const int M = ;
const int INF = 0x3f3f3f3f;
int head[N], tot;
struct Edge{
int next, to, w;
}edge[M]; void add_edge(int u, int v, int w){
edge[tot].w = w;
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++; edge[tot].w = ;
edge[tot].to = u;
edge[tot].next = head[v];
head[v] = tot++;
} struct Dinic{
int level[N], S, T;
void init(int _S, int _T){
S = _S;
T = _T;
tot = ;
memset(head, -, sizeof(head));
}
bool bfs(){
queue<int> que;
memset(level, -, sizeof(level));
level[S] = ;
que.push(S);
while(!que.empty()){
int u = que.front();
que.pop();
for(int i = head[u]; i != -; i = edge[i].next){
int v = edge[i].to;
int w = edge[i].w;
if(level[v] == - && w > ){
level[v] = level[u]+;
que.push(v);
}
}
}
return level[T] != -;
}
int dfs(int u, int flow){
if(u == T)return flow;
int ans = , fw;
for(int i = head[u]; i != -; i = edge[i].next){
int v = edge[i].to, w = edge[i].w;
if(!w || level[v] != level[u]+)
continue;
fw = dfs(v, min(flow-ans, w));
ans += fw;
edge[i].w -= fw;
edge[i^].w += fw;
if(ans == flow)return ans;
}
if(ans == )level[u] = -;
return ans;
}
int maxflow(){
int flow = , f;
while(bfs())
while((f = dfs(S, INF)) > )
flow += f;
return flow;
}
}dinic; char str[N]; int main()
{
//std::ios::sync_with_stdio(false);
//freopen("inputH.txt", "r", stdin);
int n, f, d, w;
while(scanf("%d%d%d", &n, &f, &d) != EOF){
int s = , t = *n+f+d+;
dinic.init(s, t);
for(int i = ; i <= n; i++)
add_edge(i, n+i, );
for(int i = ; i <= f; i++){
scanf("%d", &w);
add_edge(s, *n+i, w);
}
for(int i = ; i <= d; i++){
scanf("%d", &w);
add_edge(*n+f+i, t, w);
}
for(int i = ; i <= n; i++){
scanf("%s", str);
for(int j = ; j < f; j++){
if(str[j] == 'Y')
add_edge(*n+j+, i, );
}
}
for(int i = ; i <= n; i++){
scanf("%s", str);
for(int j = ; j < d; j++){
if(str[j] == 'Y')
add_edge(n+i, *n+f+j+, );
}
}
printf("%d\n", dinic.maxflow());
}
return ;
}

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