BZOJ 1221 软件开发(费用流)

容易看出这是显然的费用流模型。

把每天需要的餐巾数作为限制。需要将天数拆点，x’表示每天需要的餐巾，x’’表示每天用完的餐巾。所以加边 (s,x',INF,0),(x'',t,INF,0).

餐巾可以新买。所以需要加边(s,x'',INF,f)。

没用完餐巾可以留到下一天，所以加边(x',x+1',INF,0).

送往快洗店，加边(x',x+a+1'',INF,fa). 送往慢洗店，加边(x',x+b+1'',INF,fb).

跑一遍费用流即可。由于该图是一种特殊的结构，类二分图结构。用ZKW费用流可以快速出解。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=2005;
//Code begin...

struct Edge{
    int to, next, cap, flow, cost;
    Edge(int _to=0, int _next=0, int _cap=0, int _flow=0, int _cost=0):
    to(_to), next(_next), cap(_cap), flow(_flow), cost(_cost){}
}edge[50005];
struct ZKW_MinCostMaxFlow{
    int head[50005], tot, cur[50005], dis[50005], ss, tt, n, min_cost, max_flow;
    bool vis[50005];
    void init(){tot=0; mem(head,-1);}
    void addedge(int u, int v, int cap, int cost){
        edge[tot]=Edge(v,head[u],cap,0,cost);
        head[u]=tot++;
        edge[tot]=Edge(u,head[v],0,0,-cost);
        head[v]=tot++;
    }
    int aug(int u, int flow){
        if (u==tt) return flow;
        vis[u]=true;
        for (int i=cur[u]; i!=-1; i=edge[i].next) {
            int v=edge[i].to;
            if (edge[i].cap>edge[i].flow&&!vis[v]&&dis[u]==dis[v]+edge[i].cost) {
                int tmp=aug(v,min(flow,edge[i].cap-edge[i].flow));
                edge[i].flow+=tmp; edge[i^1].flow-=tmp; cur[u]=i;
                if (tmp) return tmp;
            }
        }
        return 0;
    }
    bool modify_label(){
        int d=INF;
        FO(u,0,n) if (vis[u]) for (int i=head[u]; i!=-1; i=edge[i].next) {
            int v=edge[i].to;
            if (edge[i].cap>edge[i].flow&&!vis[v]) d=min(d,dis[v]+edge[i].cost-dis[u]);
        }
        if (d==INF) return false;
        FO(i,0,n) if (vis[i]) vis[i]=false, dis[i]+=d;
        return true;
    }
    PII mincostmaxflow(int start, int end, int nn){
        ss=start, tt=end, n=nn; min_cost=max_flow=0;
        FO(i,0,n) dis[i]=0;
        while (1) {
            FO(i,0,n) cur[i]=head[i];
            while (1) {
                FO(i,0,n) vis[i]=false;
                int tmp=aug(ss,INF);
                if (tmp==0) break;
                max_flow+=tmp; min_cost+=tmp*dis[ss];
            }
            if (!modify_label()) break;
        }
        return mp(min_cost,max_flow);
    }
}solve;
int val[N];
int main ()
{
    int n,a,b,fa,fb,f;
    scanf("%d%d%d%d%d%d",&n,&a,&b,&f,&fa,&fb);
    solve.init();
    FOR(i,1,n) scanf("%d",val+i), solve.addedge(0,i,val[i],0), solve.addedge(0,i+n,INF,f), solve.addedge(i+n,2*n+1,val[i],0);
    FOR(i,1,n) {
        if (i<n) solve.addedge(i,i+1,INF,0);
        if (i<n-a) solve.addedge(i,i+n+a+1,INF,fa);
        if (i<n-b) solve.addedge(i,i+n+b+1,INF,fb);
    }
    printf("%d\n",solve.mincostmaxflow(0,2*n+1,2*n+2).first);
    return 0;
}