洛谷$P3749$ [六省联考2017] 寿司餐厅 网络流

正解:网络流

解题报告:

传送门$QwQ$

这道题好烦昂,,,就给了好多变量,,,但仔细读一遍题还是能$get$的所以我就不再提取一遍题目大意辣$QwQ$?

显然考虑建两排点,一排收益一排支出然后最小流呗?

考虑连边?寿司和编号之间连$inf$嘛,编号和$T$连$m\cdot x^{2}$嘛,然后关于这个$c\cdot x$显然可以归结到每个寿司上不用通过编号背锅嘛$QwQ$.最后关于那个区间的考虑强制性就说如果选了$[l,r]$就还强制选$[l,r-1]$和$[l+1,r]$所以连个$inf$边嘛,然后就欧克了?仔细想下发现似乎很正确的样子于是就做完辣$QwQ$

最后总结下怎么建边趴$QwQ$,就首先建两排点,然后每个区间$[l,r]$作一个点,每个编号作一个点,然后收益的放一边支出的放一遍,分别和$ST$连收益/支出,然后每个寿司向编号连$inf$边,每个区间向相应的必选区间连$inf$边

跑个最小割,做完辣$QwQQQQQ$!

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

const int N=+,M=+,inf=1e8;
int n,m,dep[N],head[N],cur[N],S,T,ed_cnt=-,as;
bool vis[N];
struct ed{int to,nxt,wei;}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 l,ri r){return (l-)*n+r+;}
il void ad(ri x,ri y,ri z)
{edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],};head[x]=ed_cnt;}
il bool bfs()
{
    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;
    while(!Q.empty())
    {
        ri nw=Q.front();Q.pop();
        e(i,nw)if(w(i) && !dep[t(i)]){dep[t(i)]=dep[nw]+,Q.push(t(i));if(t(i)==T)return ;}
    }
    return ;
}
il int dfs(ri nw,ri flow)
{
    if(nw==T || !flow)return flow;ri ret=;
    for(ri &i=cur[nw];~i;i=n(i))
        if(w(i) && dep[t(i)]==dep[nw]+)
        {ri tmp=dfs(t(i),min(flow,w(i)));ret+=tmp,w(i)-=tmp;w(i^)+=tmp,flow-=tmp;}
    return ret;
}
il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}

int main()
{
    //freopen("3749.in","r",stdin);freopen("3749.out","w",stdout);
    n=read();m=read();S=;T=n*n++;memset(head,-,sizeof(head));
    rp(i,,n){ri tmp=read(),id=nam(i,i);if(!vis[tmp])ad(T,tmp,m*tmp*tmp),vis[tmp]=;ad(tmp,id,inf);ad(T,id,tmp);}
    rp(i,,n)
        rp(j,i,n)
        {
            ri tmp=read(),id1=nam(i,j),id2=nam(i,j-),id3=nam(i+,j);
            if(i!=j)ad(id2,id1,inf),ad(id3,id1,inf);
            if(tmp>)ad(id1,S,tmp),as+=tmp;else ad(T,id1,-tmp);
        }
//    printf("as=%d\n",as);
    printf("%d\n",as-dinic());
    return ;
}