BZOJ4873 Shoi2017寿司餐厅（最小割）

　　选择了某个区间就必须选择其所有子区间，容易想到这是一个最大权闭合子图的模型。考虑将区间按长度分层，相邻层按包含关系连边，区间[i,j]的权值即d_i,j，其中最后一层表示长度为1的区间的同时也表示寿司本身，所以其权值减去x。这样建出原图，再用最大权闭合子图的方法重建就行了。于是m=0的情况就解决了。给最后一层的点连向寿司代号，m=1也就做完了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
#define M 6789
#define S 0
#define T 6666
#define inf 1000000010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,a[N],v[N][N],id[N][N],id2[],p[M],t=-,cnt,ans;
int d[M],cur[M],q[M];
struct data{int to,nxt,cap,flow;
}edge[M<<];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=,edge[t].flow=,p[y]=t;
}
bool bfs()
{
    memset(d,,sizeof(d));d[S]=;
    int head=,tail=;q[]=S;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-&&edge[i].flow<edge[i].cap)
        {
            q[++tail]=edge[i].to;
            d[edge[i].to]=d[x]+;
        }
    }while (head<tail);
    return ~d[T];
}
int work(int k,int f)
{
    if (k==T) return f;
    int used=;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
        edge[i].flow+=w,edge[i^].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==) d[k]=-;
    return used;
}
void dinic()
{
    while (bfs())
    {
        memcpy(cur,p,sizeof(p));
        ans-=work(S,inf);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4873.in","r",stdin);
    freopen("bzoj4873.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<=n;i++)
        for (int j=i;j<=n;j++)
        v[i][j]=read();
    memset(p,,sizeof(p));
    for (int k=n;k>=;k--)
        for (int i=;i<=n-k+;i++)
        {
            int j=i+k-;
            id[i][j]=++cnt;
            if (i>) addedge(id[i-][j],id[i][j],inf);
            if (j<n) addedge(id[i][j+],id[i][j],inf);
            int x=v[i][j];if (k==) x-=a[i];
            if (x>) ans+=x,addedge(S,id[i][j],x);
            else addedge(id[i][j],T,-x);
        }
    if (m==)
    for (int i=;i<=n;i++)
    {
        if (!id2[a[i]]) id2[a[i]]=++cnt,addedge(id2[a[i]],T,a[i]*a[i]);
        addedge(id[i][i],id2[a[i]],inf);
    }
    dinic();
    cout<<ans;
    return ;
}