POJ 1149 网络流最大流 解题报告

PIGS

Description

Mirko works on a pig farm that consists of M locked pig-houses and Mirko can’t unlock any pighouse because he doesn’t have the keys. Customers come to the farm one after another. Each of them has keys to some pig-houses and wants to buy a certain number of pigs.
All data concerning customers planning to visit the farm on that particular day are available to Mirko early in the morning so that he can make a sales-plan in order to maximize the number of pigs sold.
More precisely, the procedure is as following: the customer arrives, opens all pig-houses to which he has the key, Mirko sells a certain number of pigs from all the unlocked pig-houses to him, and, if Mirko wants, he can redistribute the remaining pigs across the unlocked pig-houses.
An unlimited number of pigs can be placed in every pig-house.
Write a program that will find the maximum number of pigs that he can sell on that day.

Input

The first line of input contains two integers M and N, 1 <= M <= 1000, 1 <= N <= 100, number of pighouses and number of customers. Pig houses are numbered from 1 to M and customers are numbered from 1 to N.
The next line contains M integeres, for each pig-house initial number of pigs. The number of pigs in each pig-house is greater or equal to 0 and less or equal to 1000.
The next N lines contains records about the customers in the following form ( record about the i-th customer is written in the (i+2)-th line):
A K1 K2 … KA B It means that this customer has key to the pig-houses marked with the numbers K1, K2, …, KA (sorted nondecreasingly ) and that he wants to buy B pigs. Numbers A and B can be equal to 0.

Output

The first and only line of the output should contain the number of sold pigs.

Sample Input

3 3
3 1 10
2 1 2 2
2 1 3 3
1 2 6

Sample Output

7

【解题报告】(转载自网络流建模汇by Edelweiss )

代码如下:

#include<cstdio>  #include<cstring> #include<string>#include<algorithm>#include<map>#include<queue>  #include<vector>using namespace std;  #define inf 0x3f3f3f3f  #define maxv 10005#define maxe 100005 int nume=0,head[maxv],e[maxe][3];void inline adde(int i,int j,int c)  {      e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;      e[nume++][2]=c;     e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;      e[nume++][2]=0;  }  int ss,tt,n,m;  int vis[maxv],lev[maxv]; bool bfs()  {      for(int i=0;i<maxv;i++)      vis[i]=lev[i]=0;      queue<int>q;      q.push(ss);      vis[ss]=1;      while(!q.empty())      {          int cur=q.front();          q.pop();          for(int i=head[cur];i!=-1;i=e[i][1])          {              int v=e[i][0];              if(!vis[v]&&e[i][2]>0)              {                  lev[v]=lev[cur]+1;                  vis[v]=1;                  q.push(v);              }          }      }      return vis[tt];  }  int dfs(int u,int minf)  {      if(u==tt||minf==0)return minf;      int sumf=0,f;      for(int i=head[u];i!=-1&&minf;i=e[i][1])      {          int v=e[i][0];          if(lev[v]==lev[u]+1&&e[i][2]>0)          {              f=dfs(v,minf<e[i][2]?minf:e[i][2]);              e[i][2]-=f;e[i^1][2]+=f;              sumf+=f;minf-=f;          }      }      if(!sumf) lev[u]=-1;      return sumf;  }  int Dinic()  {      int sum=0;      while(bfs())sum+=dfs(ss,inf);      return sum;  }  vector<int>a[200];int pig[1005];int L[1005];int main(){    memset(head,-1,sizeof(head));    scanf("%d%d",&m,&n);    ss=0,tt=n+1;    for(int i=1;i<=m;++i) scanf("%d",&pig[i]);    for(int i=1;i<=n;++i)    {        int x,y;        scanf("%d",&x);        while(x--)        {            scanf("%d",&y);            a[i].push_back(y);        }        scanf("%d",&y);        adde(i,tt,y);    }    for(int i=1;i<=n;i++)    for(int j=0;j<a[i].size();j++)    {        int v=a[i][j];        if(!L[v])        {            L[v]=i;            adde(ss,i,pig[v]);        }        else         {            adde(L[v],i,inf);            L[v]=i;        }    }    printf("%d\n",Dinic());    return 0;}

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