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HDU 4292 Food

时间:2015-09-17 13:13:29      阅读:158      评论:0      收藏:0      [点我收藏+]

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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

 
解题:最大流,人拆点 边流为1
技术分享
 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 }
View Code

 

HDU 4292 Food

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原文地址:http://www.cnblogs.com/crackpotisback/p/4815822.html

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