P4009 汽车加油行驶问题 费用流

 

题意:

 

 

 

题解：

• 如果无限油的话就是裸的分层图
• 最多有k滴油  那么可以建立分层图  每一滴油建立一层即可
• 然后按照题意模拟

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=500000+10,M=5e6+10;
ll maxflow,mincost;
int last[N],pre[N],dis[N],flow[N];
bool vis[N];
struct Edge{int next,to,flow,dis;}edge[M<<1];
int pos=1,head[N];
void init(){pos=1;memset(head,0,sizeof head);mincost=maxflow=0;}
queue <int> q;
void add(int from,int to,int flow,int dis)//flow流量 dis费用
{
    edge[++pos].next=head[from];edge[pos].flow=flow;edge[pos].dis=dis;edge[pos].to=to;head[from]=pos;
    edge[++pos].next=head[to];edge[pos].flow=0;edge[pos].dis=-dis;edge[pos].to=from;head[to]=pos;
}
bool spfa(int s,int t)
{
    memset(dis,0x3f,sizeof dis);
    memset(flow,0x3f,sizeof flow);
    memset(vis,0,sizeof vis);
    while (!q.empty()) q.pop();
    dis[s]=0; pre[t]=-1; q.push(s); vis[s]=1;
    int tot=0;
    while (!q.empty())
    {
        int now=q.front(); q.pop(); vis[now]=0;
        for (int i=head[now]; i; i=edge[i].next)
        {
            int to=edge[i].to;
            if  (edge[i].flow>0 && dis[to]>dis[now]+edge[i].dis)
            {
                dis[to]=edge[i].dis+dis[now];
                flow[to]=min(edge[i].flow,flow[now]);
                last[to]=i;
                pre[to]=now;
                if (!vis[to])
                {
                    q.push(to); vis[to]=1;
                }
            }
        }
    }
    return pre[t]!=-1;
}
void MCMF(int s,int t)
{
    while (spfa(s,t))
    {
        int now=t;
        maxflow+=flow[t];
        mincost+=flow[t]*dis[t];
        while (now!=s)
        {
            edge[last[now]].flow-=flow[t];//dis . flow
            edge[last[now]^1].flow+=flow[t];
            now=pre[now];
        }
    }
}
int n,m,A,B,C,K,s,t;
int id(int x,int y){return (x-1)*n+y;}

int main()
{
    cin>>n>>K>>A>>B>>C;
    int T=n*n;
    s=(K+2)*T;t=s+1;
    add(s,id(1,1)+K*T,1,0);
    for(int i=0;i<=K;i++)
        add(id(n,n)+T*i,t,1,0);
    int x;
    for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)
    {
        scanf("%d",&x);
        if(x==1)
        {
            for(int s=0;s<K;s++)add(id(i,j)+s*T,id(i,j)+K*T,1,A);
            if(i+1<=n)add(id(i,j)+K*T,id(i+1,j)+(K-1)*T,1,0);
            if(j+1<=n)add(id(i,j)+K*T,id(i,j+1)+(K-1)*T,1,0);
            if(i-1>=1)add(id(i,j)+K*T,id(i-1,j)+(K-1)*T,1,B);
            if(j-1>=1)add(id(i,j)+K*T,id(i,j-1)+(K-1)*T,1,B);
            continue;
        }
        for(int s=1;s<=K;s++)
        {
            if(i+1<=n)add(id(i,j)+s*T,id(i+1,j)+(s-1)*T,1,0);
            if(j+1<=n)add(id(i,j)+s*T,id(i,j+1)+(s-1)*T,1,0);
            if(i-1>=1)add(id(i,j)+s*T,id(i-1,j)+(s-1)*T,1,B);
            if(j-1>=1)add(id(i,j)+s*T,id(i,j-1)+(s-1)*T,1,B);
        }
        add(id(i,j),id(i,j)+T*K,1,A+C);
    }
    MCMF(s,t);
    cout<<mincost;
}

View Code