BZOJ 2406 二分+有上下界的网络流判定

思路：

求出每行的和  sum_row

每列的和   sum_line

二分最后的答案mid

S->i  流量[sum_row[i]-mid,sum_row[i]+mid]

i->n+j 流量[L,R]

n+j->T 流量 [sum_line[i]-mid,sum_line[i]+mid]

套用有上下界的网络流 判一下就好了..

这是道有上下界网络流的裸题

//By SiriusRen
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=,M=;
int n,m,a[N][N],sum_row[N],sum_line[N],L,R;
struct Dinic{
    int first[N*],next[M],v[M],w[M],vis[N*],tot,T,SS,TT,jy,du[N*],all;
    void Add(int x,int y,int z){w[tot]=z,v[tot]=y,next[tot]=first[x],first[x]=tot++;}
    void add(int x,int y,int z){Add(x,y,z),Add(y,x,);}
    bool tell(){
        memset(vis,-,sizeof(vis)),vis[SS]=;
        queue<int>q;q.push(SS);
        while(!q.empty()){
            int t=q.front();q.pop();
            for(int i=first[t];~i;i=next[i])
                if(vis[v[i]]==-&&w[i])
                    vis[v[i]]=vis[t]+,q.push(v[i]);
        }return vis[TT]!=-;
    }
    int zeng(int x,int y){
        if(x==TT)return y;
        int r=;
        for(int i=first[x];~i&&y>r;i=next[i])
            if(vis[v[i]]==vis[x]+&&w[i]){
                int t=zeng(v[i],min(y-r,w[i]));
                w[i]-=t,w[i^]+=t,r+=t;
            }
        if(!r)vis[x]=-;
        return r;
    }
    int flow(){
        int ans=;
        while(tell())while(jy=zeng(SS,0x3f3f3f3f))ans+=jy;
        return ans;
    }
    bool check(int x){
        memset(first,-,sizeof(first)),
        all=tot=,T=n+m+,SS=n+m+,TT=n+m+,
        memset(du,,sizeof(du));
        add(T,,0x3f3f3f3f);
        for(int i=;i<=n;i++)add(,i,*x),du[i]+=sum_row[i]-x,du[]-=sum_row[i]-x;
        for(int i=;i<=m;i++)add(i+n,T,*x),du[T]+=sum_line[i]-x,du[i+n]-=sum_line[i]-x;
        for(int i=;i<=n;i++)
            for(int j=;j<=m;j++)
                add(i,j+n,R-L),du[j+n]+=L,du[i]-=L;
        for(int i=;i<=T;i++){
            if(du[i]>)add(SS,i,du[i]),all+=du[i];
            else add(i,TT,-du[i]);
        }if(flow()>=all)return ;
        return ;
    }
}d;
int main(){
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++)
        for(int j=;j<=m;j++){
            scanf("%d",&a[i][j]);
            sum_row[i]+=a[i][j];
            sum_line[j]+=a[i][j];
        }
    scanf("%d%d",&L,&R);
    int l=,r=,ans=;
    while(l<=r){
        int mid=(l+r)>>;
        if(d.check(mid))r=mid-,ans=mid;
        else l=mid+;
    }printf("%d\n",ans);
}