hdu4292——Food（SAP+拆点）

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 3
1 1 1
1 1 1
YYN
NYY
YNY
YNY
YNY
YYN
YYN
NNY

Sample Output
3

这题原理上和poj3281很相似，我用刚学的SAP模板套了一遍，发现以下几个问题（不敢相信我居然能发现）：
n是所有的点的个数，理论上n=des+1就行；
点的个数MAXN和边的个数很玄学，以我的知识水平不知道它们之间应该有什么样的关系，总之一开始边数是点数的4倍，WA，改成32倍就过了。

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 11000#define Mod 10001using namespace std;struct E{    int to, frm, nxt, cap;}edge[MAXN<<5];int head[MAXN], e, n, m, src, des;int dep[MAXN], gap[MAXN]; //gap[x]=y:说明残留网络中dep[i]=x的个数为yvoid addedge(int u, int v, int c){    edge[e].frm = u;    edge[e].to = v;    edge[e].cap = c;    edge[e].nxt = head[u];    head[u] = e++;    edge[e].frm = v;    edge[e].to = u;    edge[e].cap = 0;    edge[e].nxt = head[v];    head[v] = e++;}int Q[MAXN];void BFS(int src, int des){    memset(dep, -1, sizeof(dep));    memset(gap, 0, sizeof(gap));    gap[0] = 1;   //说明此时有1个dep[i] = 0    int front = 0, rear = 0;    dep[des] = 0;    Q[rear++] = des;    int u, v;    while (front != rear)    {        u = Q[front++];        //cout<<u<<endl;        front = front%MAXN;        for (int i=head[u]; i!=-1; i=edge[i].nxt)        {            //cout<<i<<endl;            v = edge[i].to;            if (edge[i].cap != 0 || dep[v] != -1)                continue;            Q[rear++] = v;            rear = rear % MAXN;            ++gap[dep[v] = dep[u] + 1];  //求出各层次的数量        }    }}int S[MAXN],cur[MAXN];int SAP(){    int res = 0;    BFS(src, des);    int  top = 0;    memcpy(cur, head, sizeof(head));    int u = src, i;    while (dep[src] < n)   //n为结点的个数    {        if (u == des)        {            int temp = INF, inser = n;            for (i=0; i!=top; ++i)                if (temp > edge[S[i]].cap)                {                    temp = edge[S[i]].cap;                    inser = i;                }            for (i=0; i!=top; ++i)            {                edge[S[i]].cap -= temp;                edge[S[i]^1].cap += temp;            }            res += temp;            top = inser;            u = edge[S[top]].frm;        }        if (u != des && gap[dep[u] -1] == 0)//出现断层，无增广路            break;        for (i = cur[u]; i != -1; i = edge[i].nxt)//遍历与u相连的未遍历结点            if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1) //层序关系， 找到允许                break;        if (i != -1)//找到允许弧        {            cur[u] = i;            S[top++] = i;//加入路径栈            u = edge[i].to;//查找下一个结点        }        else   //无允许的路径，修改标号 当前点的标号比与之相连的点中最小的多1        {            int min = n;            for (i = head[u]; i != -1; i = edge[i].nxt) //找到与u相连的v中dep[v]最小的点            {                if (edge[i].cap == 0)                    continue;                if (min > dep[edge[i].to])                {                    min = dep[edge[i].to];                    cur[u] = i;          //最小标号就是最新的允许弧                }            }            --gap[dep[u]];          //dep[u] 的个数变化了 所以修改gap            ++gap[dep[u] = min + 1]; //将dep[u]设为min(dep[v]) + 1, 同时修改相应的gap[]            if (u != src) //该点非源点&&以u开始的允许弧不存在，退点                u = edge[S[--top]].frm;        }    }    return res;}int main(){    int N,D,F,w;    while(~scanf("%d%d%d",&N,&F,&D))    {        //1~F,F+1~F+N,F+N+1~F+N+N,F+N+N+1~F+N+N+D        src=0;        des=F+N+N+D+1;        n=des+1; //还有源点，汇点        e=0;        memset(head,-1,sizeof(head));        for(int i=1;i<=F;++i)        {            scanf("%d",&w);            addedge(src,i,w);        }        for(int i=1;i<=D;++i)        {            scanf("%d",&w);            addedge(F+N+N+i,des,w);        }        for(int i=1;i<=N;++i)        {            addedge(F+i,F+N+i,1);        }        string op;        for(int i=1;i<=N;++i)        {            cin>>op;            for(int j=0;j<F;++j)            {                if(op[j]=='Y')                {                    addedge(j+1,F+i,1);                }            }        }        for(int i=1;i<=N;++i)        {            cin>>op;            for(int j=0;j<D;++j)            {                if(op[j]=='Y')                {                    addedge(F+N+i,F+N+N+j+1,1);                }            }        }        int ans=SAP();        printf("%d\n",ans);    }    return 0;}