Food
Time Limit: 1000ms
Memory Limit: 32768KB
This problem will be judged on HDU. Original ID: 4292
64-bit integer IO format: %I64d Java class name: Main
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
解题：最大流，人拆点 边流为1
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int INF = 0x3f3f3f3f;
4 const int maxn = 1010;
5 struct arc {
6 int to,flow,next;
7 arc(int x = 0,int y = 0,int z = -1) {
8 to = x;
9 flow = y;
10 next = z;
11 }
12 } e[maxn*maxn];
13 int head[maxn],cur[maxn],d[maxn],tot,S,T;
14 void add(int u,int v,int flow) {
15 e[tot] = arc(v,flow,head[u]);
16 head[u] = tot++;
17 e[tot] = arc(u,0,head[v]);
18 head[v] = tot++;
19 }
20 queue<int>q;
21 bool bfs() {
22 while(!q.empty()) q.pop();
23 memset(d,-1,sizeof d);
24 q.push(S);
25 d[S] = 1;
26 while(!q.empty()) {
27 int u = q.front();
28 q.pop();
29 for(int i = head[u]; ~i; i = e[i].next) {
30 if(e[i].flow && d[e[i].to] == -1) {
31 d[e[i].to] = d[u] + 1;
32 q.push(e[i].to);
33 }
34 }
35 }
36 return d[T] > -1;
37 }
38 int dfs(int u,int low) {
39 if(u == T) return low;
40 int tmp = 0,a;
41 for(int &i = cur[u]; ~i; i = e[i].next) {
42 if(e[i].flow && d[u]+1==d[e[i].to]&&(a=dfs(e[i].to,min(low,e[i].flow)))) {
43 low -= a;
44 tmp += a;
45 e[i].flow -= a;
46 e[i^1].flow += a;
47 if(!low) break;
48 }
49 }
50 if(!tmp) d[u] = -1;
51 return tmp;
52 }
53 int dinic() {
54 int ret = 0;
55 while(bfs()) {
56 memcpy(cur,head,sizeof cur);
57 ret += dfs(S,INF);
58 }
59 return ret;
60 }
61 char str[maxn];
62 int main() {
63 int N,F,D,flow;
64 while(~scanf("%d%d%d",&N,&F,&D)) {
65 memset(head,-1,sizeof head);
66 S = tot = 0;
67 T = 1000;
68 for(int i = 1; i <= F; ++i) {
69 scanf("%d",&flow);
70 add(S,i,flow);
71 }
72 for(int i = 1; i <= D; ++i) {
73 scanf("%d",&flow);
74 add(F + i,T,flow);
75 }
76 for(int i = 1; i <= N; ++i) {
77 add(F + D + i*2-1,F + D + i*2,1);
78 scanf("%s",str);
79 for(int j = 0; str[j]; ++j)
80 if(str[j] == ‘Y‘) add(j+1, F + D + i*2 - 1,INF);
81 }
82 for(int i = 1; i <= N; ++i) {
83 scanf("%s",str);
84 for(int j = 0; str[j]; ++j)
85 if(str[j] == ‘Y‘) add(F + D + i*2,F + j + 1,INF);
86 }
87 printf("%d\n",dinic());
88 }
89 return 0;
90 }