HDU 4292 Food

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

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1010;
struct arc {
    int to,flow,next;
    arc(int x = 0,int y = 0,int z = -1) {
        to = x;
        flow = y;
        next = z;
    }
} e[maxn*maxn];
int head[maxn],cur[maxn],d[maxn],tot,S,T;
void add(int u,int v,int flow) {
    e[tot] = arc(v,flow,head[u]);
    head[u] = tot++;
    e[tot] = arc(u,0,head[v]);
    head[v] = tot++;
}
queue<int>q;
bool bfs() {
    while(!q.empty()) q.pop();
    memset(d,-1,sizeof d);
    q.push(S);
    d[S] = 1;
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        for(int i = head[u]; ~i; i = e[i].next) {
            if(e[i].flow && d[e[i].to] == -1) {
                d[e[i].to] = d[u] + 1;
                q.push(e[i].to);
            }
        }
    }
    return d[T] > -1;
}
int dfs(int u,int low) {
    if(u == T) return low;
    int tmp = 0,a;
    for(int &i = cur[u]; ~i; i = e[i].next) {
        if(e[i].flow && d[u]+1==d[e[i].to]&&(a=dfs(e[i].to,min(low,e[i].flow)))) {
            low -= a;
            tmp += a;
            e[i].flow -= a;
            e[i^1].flow += a;
            if(!low) break;
        }
    }
    if(!tmp) d[u] = -1;
    return tmp;
}
int dinic() {
    int ret = 0;
    while(bfs()) {
        memcpy(cur,head,sizeof cur);
        ret += dfs(S,INF);
    }
    return ret;
}
char str[maxn];
int main() {
    int N,F,D,flow;
    while(~scanf("%d%d%d",&N,&F,&D)) {
        memset(head,-1,sizeof head);
        S = tot = 0;
        T = 1000;
        for(int i = 1; i <= F; ++i) {
            scanf("%d",&flow);
            add(S,i,flow);
        }
        for(int i = 1; i <= D; ++i) {
            scanf("%d",&flow);
            add(F + i,T,flow);
        }
        for(int i = 1; i <= N; ++i) {
            add(F + D + i*2-1,F + D + i*2,1);
            scanf("%s",str);
            for(int j = 0; str[j]; ++j)
                if(str[j] == 'Y') add(j+1, F + D + i*2 - 1,INF);
        }
        for(int i = 1; i <= N; ++i) {
            scanf("%s",str);
            for(int j = 0; str[j]; ++j)
                if(str[j] == 'Y') add(F + D + i*2,F + j + 1,INF);
        }
        printf("%d\n",dinic());
    }
    return 0;
}