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hdu4292

2019-11-06 07:12:46
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Food

Time Limit: 2000/1000 MS (java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3851    Accepted Submission(s): 1289PRoblem 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 31 1 11 1 1YYNNYYYNYYNYYNYYYNYYNNNY Sample Output
3
输入三个数,人数,食品数,和饮料数,每个人只能选择一种视频和饮料,问最多能让多少人得到最大的选择,前几天看了匈牙利算法但是没看看完,看这个匹配还以为是,结果原来是最大流,关键在图建的好,顺序无关,但是要让两种食品分别在他两边,那个数量就是流量,其他的全设置成1

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define MAXN 1000#define MAXM 200000+10#define INF 0x3f3f3f3fusing namespace std;struct Edge{    int from, to, cap, flow, next;};Edge edge[MAXM];int head[MAXN], edgenum;int dist[MAXN], cur[MAXN];bool vis[MAXN];int N, F, D;int source, sink;//超级源点 超级汇点void init(){    edgenum = 0;    memset(head, -1, sizeof(head));}void addEdge(int u, int v, int w){    Edge E1 = {u, v, w, 0, head[u]};    edge[edgenum] = E1;    head[u] = edgenum++;    Edge E2 = {v, u, 0, 0, head[v]};    edge[edgenum] = E2;    head[v] = edgenum++;}void getMap(){    int a;    source = 0, sink = 2*N + F + D + 1;    //人拆点后 编号从1到2*N    for(int i = 1; i <= N; i++)//对人拆点        addEdge(i, i+N, 1);//只能选一种食物和一种饮料    //食物编号从2*N+1到2*N+F    for(int i = 1; i <= F; i++)    {        scanf("%d", &a);        addEdge(source, 2*N+i, a);//源点 到 食物建边    }    //饮料编号2*N+F+1到2*N+F+D    for(int i = 1; i <= D; i++)    {        scanf("%d", &a);        addEdge(2*N+F+i, sink, a);//饮料 到 汇点建边    }    char str[300];    for(int i = 1; i <= N; i++)    {        scanf("%s", str);        for(int j = 0; j < F; j++)        {            if(str[j] == 'Y')//第i个人 接受 第j+1种食物                addEdge(2*N+j+1, i, 1);        }    }    for(int i = 1; i <= N; i++)    {        scanf("%s", str);        for(int j = 0; j < D; j++)        {            if(str[j] == 'Y')//第i个人 接受 第j+1种饮料                addEdge(i+N, 2*N+F+j+1, 1);        }    }}/*bool BFS(int s, int t){    queue<int> Q;    memset(dist, -1, sizeof(dist));    memset(vis, false, sizeof(vis));    dist[s] = 0;    vis[s] = true;    Q.push(s);    while(!Q.empty())    {        int u = Q.front();        Q.pop();        for(int i = head[u]; i != -1; i = edge[i].next)        {            Edge E = edge[i];            if(!vis[E.to] && E.cap > E.flow)            {                dist[E.to] = dist[u] + 1;                if(E.to == t) return true;                vis[E.to] = true;                Q.push(E.to);            }        }    }    return false;}*/bool BFS(int s,int t){     queue <int>Q;//存东西用     memset(dist,-1,sizeof(dist));     memset(vis,false,sizeof(vis));     dist[s]=0;//分层,先给第一层个0:     vis[s]=true;     Q.push(s);     while(!Q.empty())   {   int u=Q.front();//从门口拿走一个数;        Q.pop();       for(int i=head[u];i!=-1;i=edge[i].next)    {  Edge a=edge[i];       if(!vis[a.to]&&a.cap>a.flow)       {   dist[a.to]=dist[u]+1;           if(a.to==t)return true;            vis[a.to]=true;             Q.push(a.to);       }    }   }    return false;}int DFS(int x, int a, int t){    if(x == t || a == 0) return a;    int flow = 0, f;    for(int &i = cur[x]; i != -1; i = edge[i].next)    {        Edge &E = edge[i];        if(dist[E.to] == dist[x] + 1 && (f = DFS(E.to, min(a, E.cap - E.flow), t)) > 0)        {            edge[i].flow += f;//为啥加的 都是反的            edge[i^1].flow -= f;            flow += f;            a -= f;            if(a == 0) break;        }    }    return flow;}//这样写有一个好处,就是如果想得到最大流的流向,//那么直接就把flow提出来就可以了int Maxflow(int s, int t){    int flow = 0;    while(BFS(s, t))    {        memcpy(cur, head, sizeof(head));        flow += DFS(s, INF, t);    }    return flow;}int main(){    while(scanf("%d%d%d", &N, &F, &D) != EOF)    {        init();        getMap();        printf("%d/n", Maxflow(source, sink));    }    return 0;}


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