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  • POJ 1149 PIGS

    PIGS
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 20892   Accepted: 9549

    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

    Source

     1 #include<iostream>
     2 #include<cstring>
     3 #include<cstdio>
     4 #include<queue>
     5 #include<vector>
     6 #define maxn 1050
     7 using namespace std;
     8 int dep[110],m,n,pig[maxn],house[maxn];
     9 int vis[110],map[110][110];
    10 #define INF 0x7fffffff
    11 bool BFS(){
    12     memset(dep,-1,sizeof(dep));
    13     queue<int>q;
    14     dep[0]=0;q.push(0);
    15     while(!q.empty()){
    16         int u=q.front();q.pop();
    17         for(int v=1;v<=n+1;v++){
    18             if(map[u][v]>0&&dep[v]==-1){
    19                 dep[v]=dep[u]+1;
    20                 if(v==n+1)return 1;
    21                 else q.push(v);
    22             }
    23         }
    24     }
    25     return 0;
    26 }
    27 int Dinic(){
    28     int maxf=0;vector<int>q;
    29     while(BFS()){
    30         memset(vis,0,sizeof(vis));
    31         q.push_back(0);vis[0]=1;
    32         while(!q.empty()){
    33             int p=q.back();
    34             if(p==n+1){
    35                 int minn,minx=0x7fffffff;
    36                 for(int i=1;i<q.size();i++){
    37                     int u=q[i-1],v=q[i];
    38                     if(map[u][v]<minx){
    39                         minx=map[u][v];
    40                         minn=u;
    41                     }
    42                 }
    43                 maxf+=minx;
    44                 for(int i=1;i<q.size();i++){
    45                     int u=q[i-1],v=q[i];
    46                     map[u][v]-=minx;map[v][u]+=minx;
    47                 }
    48                 while(!q.empty()&&q.back()!=minn){
    49                     vis[q.back()]=0;q.pop_back();
    50                 }
    51             }
    52             else {
    53                 int i;
    54                 for(i=0;i<=n+1;i++){
    55                     if(map[p][i]>0&&dep[i]==dep[p]+1
    56                         &&!vis[i]){
    57                         q.push_back(i);vis[i]=1;
    58                         break;
    59                     }
    60                 }
    61                 if(i>n+1)q.pop_back();
    62             }
    63         }
    64     }
    65     return maxf;
    66 }
    67 int main(){
    68     cin>>m>>n;
    69     for(int i=1;i<=m;i++)
    70       scanf("%d",pig+i);
    71     
    72     int num,k;
    73     for(int i=1;i<=n;i++){
    74         scanf("%d",&num);
    75         for(int j=0;j<num;j++){
    76             scanf("%d",&k);
    77             if(house[k]==0) map[0][i]+=pig[k];
    78             else map[house[k]][i]=INF;
    79             house[k]=i;
    80         }
    81         int tpig;
    82         scanf("%d",&tpig);
    83         map[i][n+1]=tpig;
    84     }
    85     cout<<Dinic()<<endl;
    86     return 0;
    87 }
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  • 原文地址:https://www.cnblogs.com/suishiguang/p/6485264.html
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