分配问题

题目描述

题解：

裸的最小、大费用最大流。

其实和最小费用最大流一样，只是推进去时费用取反，输出也取反。

代码：

#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 105
#define ll long long
const int inf = 0x3f3f3f3f;
const ll Inf  = 0x3f3f3f3f3f3f3f3fll;
inline int rd()
{
    int f=1,c=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
    return f*c;
}
int n,S,T;
struct Mcmf
{
    int hed[2*N],cnt;
    Mcmf()
    {
        memset(hed,-1,sizeof(hed));
        cnt=-1;
    }
    struct EG
    {
        int to,nxt;
        ll w,c;
    }e[N*N*2];
    void ae(int f,int t,ll w,ll c)
    {
        e[++cnt].to = t;
        e[cnt].nxt = hed[f];
        e[cnt].w = w;
        e[cnt].c = c;
        hed[f] = cnt;
    }
    ll fl[N],dis[N];
    int fa[N],pre[N];
    bool vis[N];
    queue<int>q;
    bool spfa()
    {
        memset(dis,0x3f,sizeof(dis));
        dis[S] = 0,fl[S] = Inf,vis[S] = 1;
        q.push(S);
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            for(int j=hed[u];~j;j=e[j].nxt)
            {
                int to = e[j].to;
                if(e[j].w&&dis[to]>dis[u]+e[j].c)
                {
                    dis[to] = dis[u]+e[j].c;
                    fl[to] = min(fl[u],e[j].w);
                    pre[to] = j,fa[to] = u;
                    if(!vis[to])
                    {
                        vis[to] = 1;
                        q.push(to);
                    }
                }
            }
            vis[u] = 0;
        }
        return dis[T]!=Inf;
    }
    ll mcmf()
    {
        ll ret = 0;
        while(spfa())
        {
            ret+=fl[T]*dis[T];
            int u = T;
            while(u!=S)
            {
                e[pre[u]].w-=fl[T];
                e[pre[u]^1].w+=fl[T];
                u = fa[u];
            }
        }
        return ret;
    }
}f1,f2;
ll c[N][N];

int main()
{
    n = rd(),S = 2*n+1,T = 2*n+2;
    for(int i=1;i<=n;i++)
    {
        f1.ae(S,i,1,0);
        f1.ae(i,S,0,0);
        f1.ae(i+n,T,1,0);
        f1.ae(T,i+n,0,0);
        f2.ae(S,i,1,0);
        f2.ae(i,S,0,0);
        f2.ae(i+n,T,1,0);
        f2.ae(T,i+n,0,0);
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            c[i][j] = rd();
            f1.ae(i,j+n,1,c[i][j]);
            f1.ae(j+n,i,0,-c[i][j]);
            f2.ae(i,j+n,1,-c[i][j]);
            f2.ae(j+n,i,0,c[i][j]);
        }
    printf("%lld
%lld
",f1.mcmf(),-f2.mcmf());
    return 0;
}