洛谷$P2050\ [NOI2012]$美食节 网络流

正解:网络流

解题报告:

传送门$QwQ$

昂开始看到$jio$得,哇长得好像上一题嗷$QwQ$

然后仔细康康数据范围,发现,哇好像要几万个点,,,显然就$GG$了

但感$jio$思路方向好对的亚子?就考虑怎么优化趴$QwQ$

考虑到如果还没考虑好倒数第$i$道菜做啥就不用考虑倒数第$k+1$道?所以考虑动态加边,每次只有考虑完倒数第$k$道才做倒数第$k+1$道

所以动态加边就好$QwQ$?

就不详细港建图了$QwQ$和上题一样的嘛$QwQ$

$over$

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define lf double
#define gc getchar()
#define t(i) edge[i].to
#define w(i) edge[i].wei
#define fy(i) edge[i].fy
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define e(i,x) for(ri i=head[x];~i;i=edge[i].nxt)

const int N=1e5+,M=;
int head[N],ed_cnt=-,vis[N],S,T,fr_ed[N],fr_nod[N],dis[N],n,m,as,t[M][M],bl[N],tim[N],nod_cnt;
struct ed{int to,nxt,wei,fy;}edge[N<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il int nam(ri x,ri y){return (x-)*n+y;}
il void ad(ri x,ri y,ri z,ri p)
{edge[++ed_cnt]=(ed){x,head[y],z,p};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],,-p};head[x]=ed_cnt;}
il bool spfa()
{
    queue<int>Q;Q.push(S);memset(vis,,sizeof(vis));vis[S]=;memset(dis,,sizeof(dis));dis[S]=;
    while(!Q.empty())
    {
        ri nw=Q.front();Q.pop();vis[nw]=;
        e(i,nw)
            if(w(i) && fy(i)+dis[nw]<dis[t(i)])
            {dis[t(i)]=dis[nw]+fy(i),fr_ed[t(i)]=i,fr_nod[t(i)]=nw;if(!vis[t(i)])Q.push(t(i)),vis[t(i)]=;}
    }
    if(dis[T]==dis[T+])return ;
    ri flow=dis[T+];
    for(ri i=T;i!=S;i=fr_nod[i])flow=min(flow,w(fr_ed[i]));
    for(ri i=T;i!=S;i=fr_nod[i])w(fr_ed[i])-=flow,w(fr_ed[i]^)+=flow;as+=dis[T]*flow;
    return ;
}

int main()
{
    //freopen("2050.in","r",stdin);freopen("2050.out","w",stdout);
    memset(head,-,sizeof(head));n=read();m=read();S=,T=;
    rp(i,,n)ad(++nod_cnt,S,read(),);rp(i,,n)rp(j,,m)t[i][j]=read();
    rp(i,,m){tim[i]=;bl[++nod_cnt]=i;ad(T,nod_cnt,,);rp(j,,n)ad(nod_cnt,j,,t[j][i]);}
    while(spfa())
    {ri nw=bl[fr_nod[T]];++tim[nw];bl[++nod_cnt]=nw;ad(T,nod_cnt,,);rp(i,,n)ad(nod_cnt,i,,t[i][nw]*tim[nw]);}
    printf("%d\n",as);
    return ;
}