HDU 4292 Food（最小割，人数拆点）

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3542    Accepted Submission(s): 1191

Problem 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

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online 

题意：有N个人，F种食品，D种饮料，接下来一行表示F种食品的数量，再下一行D种饮料的数量，接下来的N行F列表示N个人对F种食品的可以接受情况：Y为可接受，N为不可接受。再接下来N行D列表示N个人对D种饮料可以接受的情况，Y与N和上面的代表相同。问最多能满足多少人，一人任意一种可接受的一个食品和一杯饮料。

解题：源点vs与每种食品建边，边权为当前食品的数量；每种饮料与汇点相连，边权为饮料的数量；N 个人点拆成一条有向边（两点先命名为左点与右点），边权为1 。每个人： 左点与能接受的食品种类点相连，边权为1；右点与能接受的饮料品种点相连，边权为1。最后跑一次最大流，最大流值就是最多能满足的人数。

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;#define captype intconst int MAXN = 1010;   //点的总数const int MAXM = 400010;    //边的总数const int INF = 1<<30;struct EDG{    int to,next;    captype cap;} edg[MAXM];int eid,head[MAXN];int gap[MAXN];  //每种距离(或可认为是高度)点的个数int dis[MAXN];  //每个点到终点eNode 的最短距离int cur[MAXN];  //cur[u] 表示从u点出发可流经 cur[u] 号边int pre[MAXN];void init(){    eid=0;    memset(head,-1,sizeof(head));}//有向边 三个参数，无向边4个参数void addEdg(int u,int v,captype c,captype rc=0){    edg[eid].to=v; edg[eid].next=head[u];    edg[eid].cap=c;  head[u]=eid++;    edg[eid].to=u; edg[eid].next=head[v];    edg[eid].cap=rc; head[v]=eid++;}captype maxFlow_sap(int sNode,int eNode, int n){//n是包括源点和汇点的总点个数，这个一定要注意    memset(gap,0,sizeof(gap));    memset(dis,0,sizeof(dis));    memcpy(cur,head,sizeof(head));    pre[sNode] = -1;    gap[0]=n;    captype ans=0;  //最大流    int u=sNode;    while(dis[sNode]<n){   //判断从sNode点有没有流向下一个相邻的点        if(u==eNode){   //找到一条可增流的路            captype Min=INF ;            int inser;            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to])    //从这条可增流的路找到最多可增的流量Min            if(Min>edg[i].cap){                Min=edg[i].cap;                inser=i;            }            for(int i=pre[u]; i!=-1; i=pre[edg[i^1].to]){                edg[i].cap-=Min;                edg[i^1].cap+=Min;  //可回流的边的流量            }            ans+=Min;            u=edg[inser^1].to;            continue;        }        bool flag = false;  //判断能否从u点出发可往相邻点流        int v;        for(int i=cur[u]; i!=-1; i=edg[i].next){            v=edg[i].to;            if(edg[i].cap>0 && dis[u]==dis[v]+1){                flag=true;                cur[u]=pre[v]=i;                break;            }        }        if(flag){            u=v;            continue;        }        //如果上面没有找到一个可流的相邻点，则改变出发点u的距离（也可认为是高度）为相邻可流点的最小距离+1        int Mind= n;        for(int i=head[u]; i!=-1; i=edg[i].next)        if(edg[i].cap>0 && Mind>dis[edg[i].to]){            Mind=dis[edg[i].to];            cur[u]=i;        }        gap[dis[u]]--;        if(gap[dis[u]]==0) return ans;  //当dis[u]这种距离的点没有了，也就不可能从源点出发找到一条增广流路径                                        //因为汇点到当前点的距离只有一种，那么从源点到汇点必然经过当前点，然而当前点又没能找到可流向的点，那么必然断流        dis[u]=Mind+1;//如果找到一个可流的相邻点，则距离为相邻点距离+1，如果找不到，则为n+1        gap[dis[u]]++;        if(u!=sNode) u=edg[pre[u]^1].to;  //退一条边    }    return ans;}int main(){    int N,D,F,vs , vt ;    char str[500];    while(scanf("%d%d%d",&N,&F,&D)>0){        init();        int num;        vs=0; vt=D+F+N*2+1;        for(int i=1; i<=F; i++){            scanf("%d",&num);            addEdg( vs , i , num);        }        for(int i=1; i<=D; i++){            scanf("%d",&num);            addEdg( i+F+N*2 , vt , num);        }        for(int i=1; i<=N; i++){            scanf("%s",str);            addEdg(i+F , i+F+N , 1);            for(int j=1; j<=F; j++)                if(str[j-1]=='Y')                  addEdg(j , i+F , 1);        }        for(int i=1; i<=N; i++){            scanf("%s",str);            for(int j=1; j<=D; j++)                if(str[j-1]=='Y')                  addEdg( i+F+N , j+F+N*2 , 1);        }        int ans = maxFlow_sap( vs , vt , vt+1);        printf("%d\n",ans);    }}