luogu p4015 (minimum cost maximum flow)

Portal

Meaning of the questions:

There \ (m \) warehouse and \ (n-\) retailers, the \ (I \) warehouse to the section \ (J \) retailers takes \ (v [i] [j ] \) yuan. Now we need to supply the same amount of harvest and warehouse retailers, asking minimum cost and maximum cost.

analysis:

Quite the classic model of minimum cost maximum flow. Because you want to ensure the same supply and harvest, which represents the flow balance, so we can make a super source \ (sp \) with even a flow rate corresponding to the warehouse \ (a_i \) , cost \ (0 \) side, while allowing the corresponding retailers with super Meeting point \ (ep \) even a flow \ (b_i \) , cost \ (0 \) side. For warehouses and retailers, not even a traffic between warehouses and retailers we only need an infinite cost of \ (v [i] [j ] \) side.

For minimum cost, we only need to run a minimum cost maximum flow in the above figure.

As for the issue of a maximum cost, we only need to change the above graph warehouses and retailers even a flow of infinite, cost \ (- v [i] [ j] \) side, and finally in the new map while running on a minimum cost maximum flow, while the final cost of the minimum number that is the opposite answer.

Code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 505;
const int maxm = 20005;
int head[maxn],cnt=0;
int dis[maxn],vis[maxn],sp,ep,maxflow,cost;
const int INF=0x3f3f3f3f;
struct Node{
    int to,next,val,cost;
}q[maxm<<1];
void init(){
    memset(head,-1,sizeof(head));
    cnt=2;
    maxflow=cost=0;
}
void addedge(int from,int to,int val,int cost){
    q[cnt].to=to;
    q[cnt].next=head[from];
    q[cnt].val=val;
    q[cnt].cost=cost;
    head[from]=cnt++;
}
void add_edge(int from,int to,int val,int cost){
    addedge(from,to,val,cost);
    addedge(to,from,0,-cost);
}
bool spfa(){
    memset(vis,0,sizeof(vis));
    memset(dis,0x3f,sizeof(dis));
    dis[sp]=0;
    vis[sp]=1;
    queue<int>que;
    que.push(sp);
    while(!que.empty()) {
        int x = que.front();
        que.pop();
        vis[x]=0;
        for(int i=head[x];i!=-1;i=q[i].next){
            int to=q[i].to;
            if(dis[to]>dis[x]+q[i].cost&&q[i].val){
                dis[to]=dis[x]+q[i].cost;
                if(!vis[to]){
                    que.push(to);
                    vis[to]=1;
                }
            }
        }
    }
    return dis[ep]!=0x3f3f3f3f;
}
int dfs(int x,int flow){
    if(x==ep){
        vis[ep]=1;
        maxflow+=flow;
        return flow;
    }
    int used=0;
    vis[x]=1;
    for(int i=head[x];i!=-1;i=q[i].next){
        int to=q[i].to;
        if((vis[to]==0||to==ep)&&q[i].val!=0&&dis[to]==dis[x]+q[i].cost){
            int minflow=dfs(to,min(flow-used,q[i].val));
            if(minflow!=0){
                cost+=q[i].cost*minflow;
                q[i].val-=minflow;
                q[i^1].val+=minflow;
                used+=minflow;
            }
            if(used==flow) break;
        }
    }
    return used;
}
int mincostmaxflow(){
    while(spfa()){
        vis[ep]=1;
        while(vis[ep]){
            memset(vis,0,sizeof(vis));
            dfs(sp,INF);
        }
    }
    return maxflow;
}
int a[maxn],b[maxn],v[maxn][maxn];
int main()
{
    int n,m;
    init();
    scanf("%d%d",&n,&m);
    sp=n+m+1,ep=n+m+2;
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        add_edge(sp,i,a[i],0);
    }
    for(int i=1;i<=m;i++){
        scanf("%d",&b[i]);
        add_edge(i+n,ep,b[i],0);
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            scanf("%d",&v[i][j]);
            add_edge(i,j+n,INF,v[i][j]);
        }
    }
    mincostmaxflow();
    printf("%d\n",cost);
    init();
    for(int i=1;i<=n;i++){
        add_edge(sp,i,a[i],0);
    }
    for(int i=1;i<=m;i++){
        add_edge(i+n,ep,b[i],0);
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            add_edge(i,j+n,INF,-v[i][j]);
        }
    }
    mincostmaxflow();
    printf("%d\n",-cost);
    return 0;
}

Guess you like

Origin www.cnblogs.com/Chen-Jr/p/11348138.html