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  • Minimum Cost(最小费用最大流,好题)

    Minimum Cost

    http://poj.org/problem?id=2516

    Time Limit: 4000MS   Memory Limit: 65536K
    Total Submissions: 19019   Accepted: 6716

    Description

    Dearboy, a goods victualer, now comes to a big problem, and he needs your help. In his sale area there are N shopkeepers (marked from 1 to N) which stocks goods from him.Dearboy has M supply places (marked from 1 to M), each provides K different kinds of goods (marked from 1 to K). Once shopkeepers order goods, Dearboy should arrange which supply place provide how much amount of goods to shopkeepers to cut down the total cost of transport. 

    It's known that the cost to transport one unit goods for different kinds from different supply places to different shopkeepers may be different. Given each supply places' storage of K kinds of goods, N shopkeepers' order of K kinds of goods and the cost to transport goods for different kinds from different supply places to different shopkeepers, you should tell how to arrange the goods supply to minimize the total cost of transport.

    Input

    The input consists of multiple test cases. The first line of each test case contains three integers N, M, K (0 < N, M, K < 50), which are described above. The next N lines give the shopkeepers' orders, with each line containing K integers (there integers are belong to [0, 3]), which represents the amount of goods each shopkeeper needs. The next M lines give the supply places' storage, with each line containing K integers (there integers are also belong to [0, 3]), which represents the amount of goods stored in that supply place. 

    Then come K integer matrices (each with the size N * M), the integer (this integer is belong to (0, 100)) at the i-th row, j-th column in the k-th matrix represents the cost to transport one unit of k-th goods from the j-th supply place to the i-th shopkeeper. 

    The input is terminated with three "0"s. This test case should not be processed.

    Output

    For each test case, if Dearboy can satisfy all the needs of all the shopkeepers, print in one line an integer, which is the minimum cost; otherwise just output "-1".

    Sample Input

    1 3 3   
    1 1 1
    0 1 1
    1 2 2
    1 0 1
    1 2 3
    1 1 1
    2 1 1
    
    1 1 1
    3
    2
    20
    
    0 0 0
    

    Sample Output

    4
    -1
    

    Source

     
     
     
      因为每种物品是独立的,所以可以把每种物品拆开来算,再判断最大流是否符合要求即可
     
      1 #include<iostream>
      2 #include<algorithm>
      3 #include<queue>
      4 #include<cstring>
      5 using namespace std;
      6 
      7 const int INF=0x3f3f3f3f;
      8 const int N=50005;
      9 const int M=500005;
     10 int top;
     11 int dist[N],pre[N];
     12 bool vis[N];
     13 int c[N];
     14 int maxflow;
     15 
     16 struct Vertex{
     17     int first;
     18 }V[N];
     19 struct Edge{
     20     int v,next;
     21     int cap,flow,cost;
     22 }E[M];
     23 
     24 void init(){
     25     memset(V,-1,sizeof(V));
     26     top=0;
     27     maxflow=0;
     28 }
     29 
     30 void add_edge(int u,int v,int c,int cost){
     31     E[top].v=v;
     32     E[top].cap=c;
     33     E[top].flow=0;
     34     E[top].cost=cost;
     35     E[top].next=V[u].first;
     36     V[u].first=top++;
     37 }
     38 
     39 void add(int u,int v,int c,int cost){
     40     add_edge(u,v,c,cost);
     41     add_edge(v,u,0,-cost);
     42 }
     43 
     44 bool SPFA(int s,int t,int n){
     45     int i,u,v;
     46     queue<int>qu;
     47     memset(vis,false,sizeof(vis));
     48     memset(c,0,sizeof(c));
     49     memset(pre,-1,sizeof(pre));
     50     for(i=1;i<=n;i++){
     51         dist[i]=INF;
     52     }
     53     vis[s]=true;
     54     c[s]++;
     55     dist[s]=0;
     56     qu.push(s);
     57     while(!qu.empty()){
     58         u=qu.front();
     59         qu.pop();
     60         vis[u]=false;
     61         for(i=V[u].first;~i;i=E[i].next){
     62             v=E[i].v;
     63             if(E[i].cap>E[i].flow&&dist[v]>dist[u]+E[i].cost){
     64                 dist[v]=dist[u]+E[i].cost;
     65                 pre[v]=i;
     66                 if(!vis[v]){
     67                     c[v]++;
     68                     qu.push(v);
     69                     vis[v]=true;
     70                     if(c[v]>n){
     71                         return false;
     72                     }
     73                 }
     74             }
     75         }
     76     }
     77     if(dist[t]==INF){
     78         return false;
     79     }
     80     return true;
     81 }
     82 
     83 int MCMF(int s,int t,int n){
     84     int d;
     85     int i,mincost;
     86     mincost=0;
     87     while(SPFA(s,t,n)){
     88         d=INF;
     89         for(i=pre[t];~i;i=pre[E[i^1].v]){
     90             d=min(d,E[i].cap-E[i].flow);
     91         }
     92         maxflow+=d;
     93         for(i=pre[t];~i;i=pre[E[i^1].v]){
     94             E[i].flow+=d;
     95             E[i^1].flow-=d;
     96         }
     97         mincost+=dist[t]*d;
     98     }
     99     return mincost;
    100 }
    101 
    102 int seller[105][105];
    103 int storage[105][105];
    104 int matrix[105][105][105];
    105 
    106 int main(){
    107     int n,m,k;
    108     int v,u,w,c;
    109     int s,t;
    110     while(~scanf("%d %d %d",&n,&m,&k)){
    111         if(!n&&!m&&!k) break;
    112         int sum=0;
    113         for(int i=1;i<=n;i++){
    114             for(int j=1;j<=k;j++){
    115                 scanf("%d",&seller[i][j]);
    116                 sum+=seller[i][j];
    117             }
    118         }
    119         for(int i=1;i<=m;i++){
    120             for(int j=1;j<=k;j++){
    121                 scanf("%d",&storage[i][j]);
    122             }
    123         }
    124         for(int i=1;i<=k;i++){
    125             for(int j=1;j<=n;j++){
    126                 for(int w=1;w<=m;w++){
    127                     scanf("%d",&matrix[i][j][w]);
    128                 }
    129             }
    130         }
    131         s=0,t=n+m+1;
    132         int ANS=0;
    133         int flow=0;
    134         for(int i=1;i<=k;i++){
    135             init();
    136             for(int j=1;j<=n;j++){
    137                 add(s,j,seller[j][i],0);
    138             }
    139             for(int j=1;j<=n;j++){
    140                 for(int w=1;w<=m;w++){
    141                     add(j,n+w,storage[w][i],matrix[i][j][w]);
    142                 }
    143             }
    144             for(int j=1;j<=m;j++){
    145                 add(n+j,t,storage[j][i],0);///INF
    146             }
    147             int ans=MCMF(s,t,t+1);
    148             ANS+=ans;
    149             flow+=maxflow;
    150         }
    151 
    152         if(flow==sum) printf("%d
    ",ANS);
    153         else printf("-1
    ");
    154     }
    155 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/9967915.html
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