一、題目描述
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
二、分析
一個的最小費用最大流問題，詳細算法：最小費用最大流。
三、代碼
1
#include<iostream>
2#include<queue>
3using namespace std;
4int n, m, kind;
5int s, t;
6int order[51][51];
7int store[51][51];
8int cost[51][51][51];
9int c[102][102];
10int f[102][102];
11int b[102][102];
12int p[102];
13int d[102];
14bool visit[102]; //表示spfa中點是否在隊列中
15void spfa() //求Gf的最短路
16
{
17
queue<int> q;
18 memset(visit, 0, sizeof(visit));
19 q.push(s);
20 visit[s] = true;
21 while(!q.empty())
22
{
23 int u = q.front();
24 visit[u] = false;
25 q.pop();
26 for(int v=0; v<=n+m+1; v++)
27 if(c[u][v] > f[u][v] && d[v] > d[u] + b[u][v])
28 {
29 d[v] = d[u] + b[u][v];
30 p[v] = u;
31 if(!visit[v])
32 {
33 q.push(v);
34 visit[v] = true;
35
}
36 }
37 }
38
}
39void mcmf()
40{
41 while(1)
42 {
43 memset(p, -1, sizeof(p));
44 for(int i=1; i<=n+m+1; i++)
45 d[i] = 100000;
46 d[s] = 0;
47 spfa();
48 if(p[t] == -1) //表示已無增廣路
49 break;
50 int minf = INT_MAX;
51 int it = t;
52 while(p[it] != -1)
53 {
54 minf = min(minf, c[p[it]][it] - f[p[it]][it]);
55 it = p[it];
56 }
57 it = t;
58 while(p[it] != -1)
59 {
60 f[p[it]][it] += minf;
61 f[it][p[it]] = -f[p[it]][it];
62 it = p[it];
63 }
64 }
65}
66int main()
67{
68 while(1)
69 {
70 scanf("%d%d%d", &n, &m, &kind);
71 if(n==0 && m==0 && kind==0)
72 break;
73 for(int i=1; i<=n; i++)
74 for(int j=1; j<=kind; j++)
75 scanf("%d", &order[i][j]);
76 for(int i=1; i<=m; i++)
77 for(int j=1; j<=kind; j++)
78 scanf("%d", &store[i][j]);
79 for(int i=1; i<=kind; i++)
80 for(int j=1; j<=n; j++)
81 for(int k=1; k<=m; k++)
82 scanf("%d", &cost[i][k][j]);
83 s = 0; t = m+n+1;
84 int res = 0;
85 bool flag = true;
86 for(int i=1; i<=kind; i++)
87 {
88 memset(c, 0, sizeof(c));
89 for(int j=1; j<=m; j++)
90 c[s][j] = store[j][i];
91 for(int j=1; j<=m; j++)
92 for(int k=1; k<=n; k++)
93 c[j][k+m] = store[j][i];
94 for(int j=1; j<=n; j++)
95 c[j+m][t] = order[j][i];
96 memset(b, 0, sizeof(b));
97 for(int j=1; j<=m; j++)
98 for(int k=1; k<=n; k++)
99 {
100 b[j][k+m] = cost[i][j][k];
101 b[k+m][j] = -b[j][k+m]; //負費用，表示回流會減小費用
102 }
103 memset(f, 0, sizeof(f));
104 mcmf();
105 for(int j=1; j<=n; j++)
106 if(c[j+m][t] != f[j+m][t])
107 {
108 flag = false;
109 break;
110 }
111 if(!flag) break;
112 for(int j=1; j<=m; j++)
113 for(int k=1; k<=n; k++)
114 res += f[j][m+k] * b[j][m+k];
115 }
116 if(flag)
117 printf("%d\n", res);
118 else
119 printf("-1\n");
120 }
121}
posted on 2009-06-30 22:09
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