interlinkage:

https://ac.nowcoder.com/acm/contest/847/F

description:

solution:

• 最大权闭合子图;
• 每个单元格看成一个正权点，每一行每一列分别看成一个负权点。各自的点权就是其对应获得或者是消耗的能量(获得为正，消耗为负);
• 每个单元格向所在行对应的点连inf边，所在列对应的点连inf边，表示选择这个单元格就必须选择其所在行和所在列;
• 对于每一个关联奖励，新建一个点权为k的点。新建点向四个对应的行，列节点连inf边;
• 源点向所有正权点连边，边权为正权点的权值。所有负权点向汇点连边，边权为负权点的权值的绝对值;
• 答案=正权点权值之和-最小割;

code:

#include<algorithm>
#include<cstring>
#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;

const int N=1e6+15;
const int inf=1e9+7;
int n,m,tot=1,S,T;
int head[N],cur[N],dep[N];
struct EDGE
{
    int to,nxt,cap;
}edge[N<<1];
inline int read()
{
    char ch=getchar();int s=0,f=1;
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9') {s=(s<<3)+(s<<1)+ch-'0';ch=getchar();}
    return s*f;
}
void add(int u,int v,int w)
{
    edge[++tot]=(EDGE){v,head[u],w};head[u]=tot;
    edge[++tot]=(EDGE){u,head[v],0};head[v]=tot;
}
int getid(int x,int y)
{
    return (x-1)*n+y+2*n;
}
queue <int> q;
int bfs()
{
    memset(dep,0,sizeof(dep));
    while (!q.empty()) q.pop();
    dep[S]=1;
    q.push(S);
    while (!q.empty())
    {
        int k=q.front();q.pop();
        for (int i=head[k];i;i=edge[i].nxt)
        {
            int y=edge[i].to;
            if (!dep[y]&&edge[i].cap)
            {
                dep[y]=dep[k]+1;
                q.push(y);
            }
        }
    }
    return dep[T];
}
int dfs(int x,int a)
{
    if (!a||x==T) return a;
    int f,flow=0;
    for (int &i=cur[x];i;i=edge[i].nxt)
    {
        int y=edge[i].to;
        if (dep[y]==dep[x]+1&&(f=dfs(y,min(edge[i].cap,a)))>0)
        {
            edge[i].cap-=f;
            edge[i^1].cap+=f;
            flow+=f;
            a-=f;
            if (!a) break;
        }
    }
    return flow;
}
int dinic()
{
    int ans=0;
    while (bfs())
    {
        memcpy(cur,head,sizeof(head));
        ans+=dfs(S,inf);
    }
    return ans;
}
int main()
{
    n=read();m=read();
    S=0;T=n*n+2*n+1;
    int sum=0;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=n;j++)
        {
            int c=read(),now=getid(i,j);
            sum+=c;
            add(S,now,c);
            add(now,i,inf);
            add(now,j+n,inf);
        }
    for (int i=1;i<=n;i++) add(i,T,read());
    for (int i=1;i<=n;i++) add(i+n,T,read());
    for (int i=1;i<=m;i++)
    {
        int i1=read(),j1=read(),i2=read(),j2=read(),k=read();
        sum+=k;
        add(S,T+i,k);
        add(T+i,i1,inf);add(T+i,j1+n,inf);
        add(T+i,i2,inf);add(T+i,j2+n,inf);
    }
    printf("%d\n",sum-dinic());
    return 0;
}