Appoint description:
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.
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.
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
把k个商品分开，一个一个对k求解，最后相加，具体看代码：
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200500
#define mod 1001
#define eps 1e-9
#define pi 3.1415926
#define mem(str) (memset(str,0,sizeof(str)))
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//*************************************************************************************
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge
{
int to, next, cap, flow, cost;
int x, y;
} edge[MAXM],HH[MAXN],MM[MAXN];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N, M;
char map[MAXN][MAXN];
void init()
{
N = MAXN;
tol = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v, int cap, int cost)//左端点，右端点，容量，花费
{
edge[tol]. to = v;
edge[tol]. cap = cap;
edge[tol]. cost = cost;
edge[tol]. flow = 0;
edge[tol]. next = head[u];
head[u] = tol++;
edge[tol]. to = u;
edge[tol]. cap = 0;
edge[tol]. cost = -cost;
edge[tol]. flow = 0;
edge[tol]. next = head[v];
head[v] = tol++;
}
bool spfa(int s, int t)
{
queue<int>q;
for(int i = 0; i < N; i++)
{
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i]. next)
{
int v = edge[i]. to;
if(edge[i]. cap > edge[i]. flow &&
dis[v] > dis[u] + edge[i]. cost )
{
dis[v] = dis[u] + edge[i]. cost;
pre[v] = i;
if(!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1) return false;
else return true;
}
//返回的是最大流， cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost)
{
int flow = 0;
cost = 0;
while(spfa(s,t))
{
int Min = INF;
for(int i = pre[t]; 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[t]; i != -1; i = pre[edge[i^1]. to])
{
edge[i]. flow += Min;
edge[i^1]. flow -= Min;
cost += edge[i]. cost * Min;
}
flow += Min;
}
return flow;
}
int sup[100][100];//供应数
int ned[100][100]; //需求数
int sum1[100],sum2[100];//判断是否供不应求
int cot[60][60][60];//花费
int main()
{
int n,m,k;
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
if(n==0&&m==0&&k==0)
return 0;
//初始化
mem(sup);
mem(ned);
mem(sum1);
mem(sum2);
mem(cot);
//添加边
for(int i=1;i<=n;i++)
{
for(int j=1;j<=k;j++)
{
cin >> ned[i][j];
sum1[j]+= ned[i][j];
}
}
for(int i=1;i<=m;i++)
{
for(int j=1;j<=k;j++)
{
cin >> sup[i][j];
sum2[j]+=sup[i][j];
}
}
for(int i=1;i<=k;i++)
{
for(int j=1;j<=n;j++)
{
for(int l=1;l<=m;l++)
{
cin >> cot[i][j][l];
}
}
}
//判断是否供不应求
int flag = 0;
for(int i=1;i<=k;i++)
{
if(sum1[i]>sum2[i])
{
flag = 1;
break;
}
}
if(flag)
{
cout << "-1" << endl;
continue;
}
ll re = 0;
for(int k1=1;k1<=k;k1++)
{
init();//注意
int beg = 0;//超级起点
int end = n+m+1 ;//超级汇点
// 超级起点连一条边到供应店，容量：供应数 花费：0
for(int i=1;i<=m;i++)
{
addedge(beg,i,sup[i][k1],0);
}
//供应点连到需求点， 容量：INF 花费：相应花费
for(int i=1;i<=m;i++)
{
for(int j=m+1;j<=n+m;j++)
addedge(i,j,INF,cot[k1][j-m][i]);
}
// 超级汇点连一条边到需求点，容量：需求数 花费：0
for(int i=1;i<=n;i++)
{
addedge(i+m,end,ned[i][k1],0);
}
//得到结果
int ans = 0;
minCostMaxflow(beg,end,ans);
//cout << ans<<endl;
re += ans;
}
cout << re <<endl;
}
}