P2045 方格取数加强版

思路

简单的拆点
拆成入点和出点，对应点之间连一条cap=1，cost=x和一条cap=INF，cost=0的边，然后相邻点的出点和其他点的入点连边，然后S有k的流量，然后跑最大费用最大流就好了

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
struct Edge{
    int u,v,cap,cost,flow;
};
const int MAXN = 10000;
const int INF = 0x3f3f3f3f;
vector<Edge> edges;
vector<int> G[MAXN];
void addedge(int u,int v,int cap,int cost){
    edges.push_back((Edge){u,v,cap,cost,0});
    edges.push_back((Edge){v,u,0,-cost,0});
    int cnt=edges.size();
    G[u].push_back(cnt-2);
    G[v].push_back(cnt-1);
}
int d[MAXN],a[MAXN],p[MAXN],vis[MAXN],s,t;
queue<int> q;
bool spfa(int &flow,int &cost){
    memset(d,0x3f,sizeof(d));
    memset(p,0,sizeof(p));
    q.push(s);
    vis[s]=true;
    a[s]=INF;
    d[s]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        vis[x]=false;
        for(int i=0;i<G[x].size();i++){
            Edge &e = edges[G[x][i]];
            if(e.cap>e.flow&&d[x]+e.cost<d[e.v]){
                d[e.v]=d[x]+e.cost;
                a[e.v]=min(a[x],e.cap-e.flow);
                p[e.v]=G[x][i];
                if(!vis[e.v]){
                    vis[e.v]=true;
                    q.push(e.v);
                }
            }
        }
    }
    if(d[t]==INF)
        return false;
    flow+=a[t];
    cost+=d[t]*a[t];
    for(int i=t;i!=s;i=edges[p[i]].u){
        edges[p[i]].flow+=a[t];
        edges[p[i]^1].flow-=a[t];
    }
    return true;
}
void MCMF(int &flow,int &cost){
    flow=0,cost=0;
    while(spfa(flow,cost));
}
int n,k;
inline int id(int x,int y){
    return (x-1)*n+y;
}
int main(){
    scanf("%d %d",&n,&k);
    s=MAXN-2;
    t=MAXN-3;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            int x;
            scanf("%d",&x);
            addedge(id(i,j),id(i,j)+n*n,1,-x);
            addedge(id(i,j),id(i,j)+n*n,k-1,0); 
        }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++){
            if(i<n)//xia
                addedge(id(i,j)+n*n,id(i+1,j),INF,0);
            if(j<n)
                addedge(id(i,j)+n*n,id(i,j+1),INF,0);
        }
    addedge(s,id(1,1),k,0);
    addedge(id(n,n)+n*n,t,INF,0);
    int cost,flow;
    MCMF(flow,cost);
    printf("%d\n",-cost);
    return 0;
}