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最小费用最大流..乱搞即可
------------------------------------------------------------------------------
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define rep( i, n ) for( int i = 0; i < n; ++i )
#define Rep( i, n ) for( int i = 1; i <= n; ++i )
#define clr( x, c ) memset( x, c, sizeof( x ) )
using namespace std;
const int maxn = 50 + 5;
struct edge {
int to, cap, cost;
edge *next, *rev;
};
edge *head[ maxn ], *p[ maxn ];
edge* pt;
edge EDGE[ maxn << 3 ];
void init() {
pt = EDGE;
clr( head, 0 );
}
inline void add( int u, int v, int d, int w ) {
pt->to = v;
pt->cap = d;
pt->cost = w;
pt->next = head[ u ];
head[ u ] = pt++;
}
inline void add_edge( int u, int v, int d, int w ) {
add( u, v, d, w );
add( v, u, 0, -w );
head[ u ]->rev = head[ v ];
head[ v ]->rev = head[ u ];
}
int d[ maxn ], a[ maxn ];
bool inQ[ maxn ];
int minCost( int S, int T ) {
const int inf = 0x3f3f3f3f;
int flow = 0, cost = 0;
for( ; ; ) {
clr( inQ, 0 );
clr( d, inf );
d[ S ] = 0;
inQ[ S ] = 1;
a[ S ] = inf;
queue< int > Q;
Q.push( S );
while( !Q.empty() ) {
int x = Q.front(); Q.pop();
inQ[ x ] = 0;
for( edge* e = head[ x ]; e; e = e->next )
if( e->cap > 0 && d[ e->to ] > d[ x ] + e->cost ) {
int to = e->to;
d[ to ] = d[ x ] + e->cost;
a[ to ] = min( a[ x ], e->cap );
p[ to ] = e;
if( !inQ[ to ] )
Q.push( to ), inQ[ to ] = 1;
}
}
if( d[ T ] == inf ) break;
flow += d[ T ];
cost += d[ T ] * a[ T ];
int x = T;
while( x != S) {
p[ x ]->cap -= a[ T ];
p[ x ]->rev->cap += a[ T ];
x = p[ x ]->rev->to;
}
}
return cost;
}
const int INF = 0x7fffffff;
int main() {
freopen( "test.in", "r", stdin );
int n, m, V;
cin >> n >> m >> V;
int s = 0, t = n + 1;
init();
Rep( i, n ) {
int x;
scanf( "%d", &x );
add_edge( i, t, x, 0 );
}
Rep( i, n ) {
int x;
scanf( "%d", &x );
add_edge( s, i, INF, x );
}
Rep( i, n - 1 )
add_edge( i, i + 1, V, m );
cout << minCost( s, t ) << "\n";
return 0;
}
------------------------------------------------------------------------------
2424: [HAOI2010]订货
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 603 Solved: 394
[Submit][Status][Discuss]Description
某公司估计市场在第i个月对某产品的需求量为Ui，已知在第i月该产品的订货单价为di，上个月月底未销完的单位产品要付存贮费用m，假定第一月月初的库存量为零，第n月月底的库存量也为零，问如何安排这n个月订购计划，才能使成本最低？每月月初订购，订购后产品立即到货，进库并供应市场，于当月被售掉则不必付存贮费。假设仓库容量为S。
Input
第1行：n, m, S (0<=n<=50, 0<=m<=10, 0<=S<=10000)
第2行：U1 , U2 , ... , Ui , ... , Un (0<=Ui<=10000)
第3行：d1 , d2 , ..., di , ... , dn (0<=di<=100)
Output
Sample Input
3 1 1000
2 4 8
1 2 4
Sample Output
34
HINT
Source
BZOJ 2424: [HAOI2010]订货(最小费用最大流)
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原文地址：http://www.cnblogs.com/JSZX11556/p/4524112.html
