poj2516

题意：有k种物品，m个供应商，n个收购商。每个供应商和收购商都需要一些种类的物品若干。每个供应商与每个收购商之间的对于不同物品的运费是不同的。求满足收购商要求的情况下，最小运费。

分析：最小费用最大流，最大流的前提下求最小费用。这题我们可以把k种物品分开计算，每次对一种物品进行最小费用最大流计算。如果不分开算会超时。对于每种物品，从源到供应商连接，容量为供应商的储存量，费用为0。采购商到汇连边，容量为需求量，费用为0。供应商到采购商连边，容量为无穷，费用为对应的运费。

最小费用最大流的计算流程大致是：每次找到一条根据费用来看的最短路，然后对这条最短路进行增加流量，直到所有路径流量都不能增加为止。

View Code

#include < iostream >
#include < cstdio >
#include < cstdlib >
#include < cstring >
using namespace std;

#define N 110
#define E 10000
#define inf 1000000000

#define typef int
// type of flow
#define typec int
// type of dis
const typef inff = 0x3f3f3f3f ;
// max of flow
const typec infc = 0x3f3f3f3f ;
// max of dis
int shop[N][N], supply[N][N];
struct network
{
int nv, ne, pnt[E], nxt[E];
int vis[N], que[N], head[N], pv[N], pe[N];
int sum;
typef flow, cap[E];
typec cost, dis[E], d[N];
void addedge( int u, int v, typef c, typec w)
{
pnt[ne] = v;
cap[ne] = c;
dis[ne] = + w;
nxt[ne] = head[u];
head[u] = (ne ++ );
pnt[ne] = u;
cap[ne] = 0 ;
dis[ne] = - w;
nxt[ne] = head[v];
head[v] = (ne ++ );
}
int mincost( int src, int sink)
{
int i, k, f, r;
typef mxf;
for (flow = 0 , cost = 0 ;;)
{
memset(pv, - 1 , sizeof (pv));
memset(vis, 0 , sizeof (vis));
for (i = 0 ; i < nv; ++ i)
d[i] = infc;
d[src] = 0 ;
pv[src] = src;
vis[src] = 1 ;
for (f = 0 , r = 1 , que[ 0 ] = src; r != f;)
{
i = que[f ++ ];
vis[i] = 0 ;
if (N == f)
f = 0 ;
for (k = head[i]; k != - 1 ; k = nxt[k])
if (cap[k] && dis[k] + d[i] < d[pnt[k]])
{
d[pnt[k]] = dis[k] + d[i];
if ( 0 == vis[pnt[k]])
{
vis[pnt[k]] = 1 ;
que[r ++ ] = pnt[k];
if (N == r)
r = 0 ;
}
pv[pnt[k]] = i;
pe[pnt[k]] = k;
}
}
if ( - 1 == pv[sink])
break ;
for (k = sink, mxf = inff; k != src; k = pv[k])
if (cap[pe[k]] < mxf)
mxf = cap[pe[k]];
flow += mxf;
cost += d[sink] * mxf;
for (k = sink; k != src; k = pv[k])
{
cap[pe[k]] -= mxf;
cap[pe[k] ^ 1 ] += mxf;
}
}
return cost;
}
void build( int n, int m, int o)
{
nv = n + m + 2 ;
ne = 0 ;
int src = 0 ;
int sink = 1 ;
sum = 0 ;
memset(head, - 1 , sizeof (head));
for ( int i = 0 ; i < n; i ++ )
{
addedge(i + 2 , sink, shop[i][o], 0 );
sum += shop[i][o];
}
for ( int i = 0 ; i < m; i ++ )
addedge(src, i + 2 + n, supply[i][o], 0 );
for ( int i = 0 ; i < n; i ++ )
for ( int j = 0 ; j < m; j ++ )
{
int a;
scanf( " %d " , & a);
addedge(j + 2 + n, i + 2 , inf, a);
}
}
};

int n, m, o;

void input()
{
for ( int i = 0 ; i < n; i ++ )
for ( int j = 0 ; j < o; j ++ )
scanf( " %d " , & shop[i][j]);
for ( int i = 0 ; i < m; i ++ )
for ( int j = 0 ; j < o; j ++ )
scanf( " %d " , & supply[i][j]);
}

int main()
{
// freopen("t.txt", "r", stdin);
while (scanf( " %d%d%d " , & n, & m, & o), n | m | o)
{
bool ok = true ;
int ans = 0 ;
input();
for ( int i = 0 ; i < o; i ++ )
{
network g;
g.build(n, m, i);
if (ok)
ans += g.mincost( 0 , 1 );
if (g.flow != g.sum)
ok = false ;
}
if (ok)
printf( " %d\n " , ans);
else
printf( " -1\n " );
}
return 0 ;
}