hdu 4292 Food
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
题目大意：给出客人数量，食物种类，饮品种类，每种食物和饮品的库存，还有每位顾客的要求。求出，最多能满足多少位客人的要求。
解题思路：跟poj 3281 Dining是一种类型的题目。 设置超级源点，连向所有的食物，容量为该食物的库存，设置超级汇点，使所有饮品指向超级汇点，容量为该饮品的库存。然后客户连接对应的食物和饮品，容量为1。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cstdlib>
using namespace std;
const int N = 20005;
const int M = 200005;
const int OF = 300;
const int INF = 0x3f3f3f3f;
const int FIN = 1999;
typedef long long ll;
int n, f, d, s, t;
int ec, head[N], first[N], que[N], lev[N];
int Next[M], to[M], v[M];
void addEdge(int a,int b,int c) {
to[ec] = b;
v[ec] = c;
Next[ec] = first[a];
first[a] = ec++;
to[ec] = a;
v[ec] = 0;
Next[ec] = first[b];
first[b] = ec++;
}
int BFS() {
int kid, now, f = 0, r = 1, i;
memset(lev, 0, sizeof(lev));
que[0] = s, lev[s] = 1;
while (f < r) {
now = que[f++];
for (i = first[now]; i != -1; i = Next[i]) {
kid = to[i];
if (!lev[kid] && v[i]) {
lev[kid] = lev[now] + 1;
if (kid == t) return 1;
que[r++] = kid;
}
}
}
return 0;
}
int DFS(int now, int sum) {
int kid, flow, rt = 0;
if (now == t) return sum;
for (int i = head[now]; i != -1 && rt < sum; i = Next[i]) {
head[now] = i;
kid = to[i];
if (lev[kid] == lev[now] + 1 && v[i]) {
flow = DFS(kid, min(sum - rt, v[i]));
if (flow) {
v[i] -= flow;
v[i^1] += flow;
rt += flow;
} else lev[kid] = -1;
}
}
return rt;
}
int dinic() {
int ans = 0;
while (BFS()) {
for (int i = 0; i <= t; i++) {
head[i] = first[i];
}
ans += DFS(s, INF);
}
return ans;
}
char c[205];
void input() {
int a;
for (int i = 1; i <= f; i++) {
scanf("%d", &a);
addEdge(s, i, a);
}
for (int i = 1; i <= d; i++) {
scanf("%d", &a);
addEdge(i + 3 * OF, t, a);
}
for (int i = 1; i <= n; i++) {
addEdge(i + OF, i + 2 * OF, 1);
}
getchar();
for (int i = 1; i <= n; i++) {
scanf("%s", c);
for (int j = 0; j < f; j++) {
if (c[j] == ‘Y‘) {
addEdge(j + 1, i + OF, 1);
}
}
}
for (int i = 1; i <= n; i++) {
scanf("%s", c);
for (int j = 0; j < d; j++) {
if (c[j] == ‘Y‘) {
addEdge(i + 2 * OF, (j + 1) + 3 * OF, 1);
}
}
}
}
int main() {
while (scanf("%d %d %d", &n, &f, &d) == 3) {
ec = 0;
memset(first, -1, sizeof(first));
s = 0, t = FIN;
input();
printf("%d\n", dinic());
}
return 0;
}版权声明：本文为博主原创文章，未经博主允许也可以转载，不过要注明出处哦。