Codeforces-1913E：Matrix Problem（最小费用最大流）

E. Matrix Problem
time limit per test 1 second
memory limit per test 256 megabytes
inputstandard input
outputstandard output
You are given a matrix a, consisting of $n$ rows by $m$ columns. Each element of the matrix is equal to $0$ or $1$.

You can perform the following operation any number of times (possibly zero): choose an element of the matrix and replace it with either $0$ or $1$.

You are also given two arrays $A$ and $B$ (of length $n$ and $m$ respectively). After you perform the operations, the matrix should satisfy the following conditions:

1. the number of ones in the $i$-th row of the matrix should be exactly $A_i$ for every $i\in [1,n]$.
2. the number of ones in the $j$-th column of the matrix should be exactly $B_j$ for every $j\in [1,m]$.

Calculate the minimum number of operations you have to perform.

Input
The first line contains two integers $n$ and $m$ $(2\le n,m\le 50)$.

Then $n$ lines follow. The $i$-th of them contains $m$ integers $a_{i,1},a_{i,2},\dots ,a_{i,m}(0\le a_{i,j}\le 1)$.

The next line contains $n$ integers $A_1,A_2,\dots ,A_n(0\le A_i\le m)$.

The next line contains $m$ integers $B_1,B_2,\dots ,B_m(0\le B_i\le n)$.

Output
Print one integer — the minimum number of operations you have to perform, or $-1$ if it is impossible.

Examples
input
3 3
0 0 0
0 0 0
0 0 0
1 1 1
1 1 1
output
3
input
3 3
1 1 1
1 1 1
1 1 1
3 2 1
1 2 3
output
3
input
2 2
0 0
0 0
1 2
0 1
output
-1

思路:最小费用最大流。从源点$S$建一条至$i$的容量为$A_i$费用为0的有向边，从$j$建一条至汇点$T$的容量为$B_j$费用为0的有向边。同时$i,j$之间建一条$i$至$j$且容量为1费用为$a_{i,j}⨁1$的有向边。

1. 当$a_{i,j}$为$1$时，费用为0，流量流过这条边不增加费用，映射到操作代表不改变$a_{i,j}$，$a_{i,j}=1$。
2. 当$a_{i,j}$为$0$时，费用为1，若有流量流过这条边，则增加费用，映射到操作代表将$a_{i,j}=0$变为$a_{i,j}=1$。

求出最小费用最大流$c,f$后，若答案存在则有$f=\sum A_i=\sum B_j$，其中有$c$次操作将$a_{i,j}=0$变为$a_{i,j}=1$，那么在原图中有$f-c$个$a_{i,j}=1$未发生变化，那么原图中多出来的$\sum a_{i,j}-(f-c)$个$1$就需要变为$0$。
所以最终操作次数为$c+\sum a_{i,j}-(f-c)=\sum a_{i,j}-f+2\times c$。

#include
#define fi first
#define se second
#define lson (k<<1)
#define rson (k<<1)+1
#define mid ((l+r)/2)
#define sz(x) int(x.size())
#define pii pair<ll,ll>
#define bit bitset<100000>
using namespace std;
const int MAX=2e6+10;
const int MOD=1e9+7;
const int INF=INT_MAX/2;
const double PI=acos(-1.0);
typedef long long ll;
struct edg{int from,to,cap,cost,rev;};
class MCMF
{
    struct node
    {
        int x,dis,flow;
        bool operator<(const node& q) const {return q.dis<dis;}
    };
private:
    int s,t;
    vector<int>h,d,v;
    vector<edg*>p;
    vector<vector<edg>>g;
private:
    void Spfa() { //worst O(n*m)
        fill(h.begin(),h.end(),INF);
        fill(v.begin(),v.end(),0);
        queue<int>q;
        q.push(s);
        h[s]=0;
        while(!q.empty())
        {
            int x=q.front();q.pop();
            v[x]=0;
            for(auto &e:g[x])
            {
                int y=e.to;
                if(!e.cap||h[y]<=h[x]+e.cost)continue;
                h[y]=h[x]+e.cost;
                if(v[y])continue;
                v[y]=1;
                q.push(y);
            }
        }
    }
    int Dijkstra(int flow) //O(m*log(n))
    {
        fill(v.begin(),v.end(),0);
        fill(d.begin(),d.end(),-1);
        fill(p.begin(),p.end(),nullptr);
        priority_queue<node> q;
        q.push({s,0,flow});
        d[s]=0;
        int ret=0;
        while(!q.empty())
        {
            auto cur=q.top();q.pop();
            int x=cur.x;
            if(d[x]!=cur.dis)continue;
            if(x==t)ret=cur.flow;
            for(auto &e:g[x])
            {
                int y=e.to;
                if(e.cap<=0)continue;
                if(d[y]!=-1&&d[y]<=d[x]+e.cost+h[x]-h[y])continue;
                d[y]=d[x]+e.cost+h[x]-h[y];
                p[y]=&e;
                q.push({y,d[y],min(cur.flow,e.cap)});
            }
        }
        return ret;
    }
public:
    explicit MCMF(int _n,int _s,int _t)
    {
        s=_s,t=_t;
        h.resize(_n);d.resize(_n);
        v.resize(_n);p.resize(_n);g.resize(_n);
    }
    void AddEdg(int from,int to,int cap,int cost) {
        g[from].push_back((edg){from,to,cap,cost,sz(g[to])});
        g[to].push_back((edg){to,from,0,-cost,sz(g[from])-1});
    }
    pii MinCostMaxFlow(int flow)
    {
//        Spfa();
        fill(h.begin(),h.end(),0);
        int mf=0,mc=0,f=0;
        while(f=Dijkstra(flow))
        {
            for(int i=0;i<sz(v);i++)h[i]+=d[i]; //update h
            for(int i=t;p[i];i=p[i]->from)      //update cap
            {
                p[i]->cap-=f;
                g[p[i]->to][p[i]->rev].cap+=f;
            }
            flow-=f;
            mf+=f;
            mc+=f*h[t];
        }
        return {mf,mc};
    }
};
int solve()
{
    int n,m,s=0,sa=0,sb=0;
    cin>>n>>m;
    MCMF t(n+m+2,0,n+m+1);
    for(int i=1;i<=n;i++)
    for(int j=1;j<=m;j++)
    {
        int x;
        scanf("%d",&x);
        t.AddEdg(i,j+n,1,x^1);
        s+=x;
    }
    for(int i=1;i<=n;i++)
    {
        int x;
        scanf("%d",&x);
        sa+=x;
        t.AddEdg(0,i,x,0);
    }
    for(int i=1;i<=m;i++)
    {
        int x;
        scanf("%d",&x);
        sb+=x;
        t.AddEdg(i+n,n+m+1,x,0);
    }
    int flow,cost;
    tie(flow,cost)=t.MinCostMaxFlow(n*m);
    if(flow!=sa||flow!=sb)return puts("-1");
    return printf("%d\n",s-flow+2*cost);
}
int main()
{
    int T=1;
    //cin>>T;
    while(T--)solve();
    return 0;
}