2015
05-23

# Food

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.

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).

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).

4 3 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY

3

#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>

using namespace std;

// the maximum number of vertices
#define NN 802

// adjacency matrix (fill this up)
// If you fill adj[][] yourself, make sure to include both u->v and v->u.

// BFS stuff
int q[NN], prev[NN];

int dinic( int n, int s, int t )
{
int flow = 0;

while( true )
{
// find an augmenting path
memset( prev, -1, sizeof( prev ) );
int qf = 0, qb = 0;
prev[q[qb++] = s] = -2;
while( qb > qf && prev[t] == -1 )
for( int u = q[qf++], i = 0, v; i < deg[u]; i++ )
if( prev[v = adj[u][i]] == -1 && cap[u][v] )
prev[q[qb++] = v] = u;

// see if we're done
if( prev[t] == -1 ) break;

// try finding more paths
for( int z = 0; z < n; z++ ) if( cap[z][t] && prev[z] != -1 )
{
int bot = cap[z][t];
for( int v = z, u = prev[v]; u >= 0; v = u, u = prev[v] )
bot = min(bot, cap[u][v]);
if( !bot ) continue;

cap[z][t] -= bot;
cap[t][z] += bot;
for( int v = z, u = prev[v]; u >= 0; v = u, u = prev[v] )
{
cap[u][v] -= bot;
cap[v][u] += bot;
}
flow += bot;
}
}

return flow;
}

int n, f, d;

int main(void)
{
while(scanf("%d %d %d", &n, &f, &d) != -1)
{
memset(cap, 0, sizeof(cap));
int start = f + d + n * 2, end = f + d + 1 + n * 2;
int foodBase = n * 2;
int drinkBase = foodBase + f;

for(int i=0;i<n;i++) cap[i][n + i] = 1;
for(int i=0;i<f;i++)
{
int curFood;
scanf("%d", &curFood);
cap[start][foodBase + i] = curFood;
}
for(int i=0;i<d;i++)
{
int curDrink;
scanf("%d", &curDrink);
cap[drinkBase + i][end] = curDrink;
}

for(int i=0;i<n;i++)
{
char str[201];
scanf("%s", str);
for(int j=0;j<f;j++) if(str[j] == 'Y') cap[foodBase + j][i] = 1;
}

for(int i=0;i<n;i++)
{
char str[201];
scanf("%s", str);
for(int j=0;j<d;j++) if(str[j] == 'Y') cap[i + n][drinkBase + j] = 1;
}

memset( deg, 0, sizeof( deg ) );
for( int u = 0; u <= end; u++ )
for( int v = 0; v <= end; v++ ) if( cap[u][v] || cap[v][u] )
}