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Food (hdu 4292 网络流sap模板题)

时间:2015-02-17 17:43:42      阅读:229      评论:0      收藏:0      [点我收藏+]

标签:food   hdu 4292   

Food

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3102    Accepted Submission(s): 1034


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
 

Source
 

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题意:有N个人,准备了F种食物和D种饮料,每个人都有喜欢的食物和饮料,这些食物和饮料最多能满足多少人。

思路:网络流,添加超级源点和食物相连,边权为该食物的数量,添加超级汇点和饮料相连,边权为该种饮料的数量,将人拆点,边权为1,建图,s->食物->人->人->饮料->e。dinic超时,用sap。

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 1005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b)  for(i = a; i <= b; i++)
#define FRL(i,a,b)  for(i = a; i < b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define DBG         pf("Hi\n")
const int MAXM = 200010;
typedef long long ll;
using namespace std;

struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];

char str[222];
int D,N,F;
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}

//加边,单向图三个参数,双向图四个参数
void addedge(int u,int v,int w,int rw=0)
{
    edge[tol].to=v; edge[tol].cap=w; edge[tol].next=head[u];
    edge[tol].flow=0; head[u]=tol++;
    edge[tol].to=u; edge[tol].cap=rw; edge[tol].next=head[v];
    edge[tol].flow=0; head[v]=tol++;
}

//输入参数:起点,终点,点的总数
//点的编号没有影响,只要输入点的总数
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u=start;
    pre[u]=-1;
    gap[0]=N;
    int ans=0;
    while (dep[start]<N)
    {
        if (u==end)
        {
            int Min=INF;
            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
                if (Min>edge[i].cap-edge[i].flow)
                    Min=edge[i].cap-edge[i].flow;
            for (int i=pre[u];i!=-1;i=pre[edge[i^1].to])
            {
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
            }
            u=start;
            ans+=Min;
            continue;
        }
        bool flag=false;
        int v;
        for (int i=cur[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].to;
            if (edge[i].cap-edge[i].flow && dep[v]+1==dep[u])
            {
                flag=true;
                cur[u]=pre[v]=i;
                break;
            }
        }
        if (flag)
        {
            u=v;
            continue;
        }
        int Min=N;
        for (int i=head[u];i!=-1;i=edge[i].next)
            if (edge[i].cap-edge[i].flow && dep[edge[i].to]<Min)
            {
                Min=dep[edge[i].to];
                cur[u]=i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]]) return ans;
        dep[u]=Min+1;
        gap[dep[u]]++;
        if (u!=start) u=edge[pre[u]^1].to;
    }
    return ans;
}
int main()
{
    int i,j,x;
    while (~sfff(N,F,D))
    {
        init();
        int s=0,e=F+D+2*N+1;
        FRE(i,1,F)
        {
            sf(x);
            addedge(s,i,x);
        }
        FRE(i,1,D)
        {
            sf(x);
            addedge(F+2*N+i,e,x);
        }
        FRE(i,1,N)
        {
            scanf("%s",str);
            int len=strlen(str);
            FRL(j,0,len)
            {
                if (str[j]=='Y')
                    addedge(j+1,F+i,1);
            }
        }
        FRE(i,1,N)
        {
            scanf("%s",str);
            int len=strlen(str);
            FRL(j,0,len)
            {
                if (str[j]=='Y')
                    addedge(F+N+i,F+2*N+j+1,1);
            }
        }
        FRE(i,1,N)
            addedge(F+i,F+N+i,1);
        int ans=sap(s,e,e+1);
        printf("%d\n",ans);
    }
    return 0;
}



Food (hdu 4292 网络流sap模板题)

标签:food   hdu 4292   

原文地址:http://blog.csdn.net/u014422052/article/details/43866819

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