luogu P4013 数字梯形问题 网络流24 最大费用最大流

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=100010;
bool vis[maxn];
int n,m,s,t,x,y,z,f,cost[maxn],pre[maxn],last[maxn],flow[maxn],maxflow,mincost;
//cost最小花费;pre每个点的前驱;last每个点的所连的前一条边;flow源点到此处的流量
//maxflow 最大流量
//mincost 最大流量的情况下的最小花费
struct Edge
{
    int to,next,flow,cost;//flow流量 cost花费
} edge[maxn];
int head[maxn],num_edge;
queue <int> q;
int mp[1001][1001],id[1001][1001],cnt;
const int inf=0x3f3f3f3f;
void add(int from,int to,int flow,int cost)
{
    edge[++num_edge].next=head[from];
    edge[num_edge].to=to;
    edge[num_edge].flow=flow;
    edge[num_edge].cost=cost;
    head[from]=num_edge;

    edge[++num_edge].next=head[to];
    edge[num_edge].to=from;
    edge[num_edge].flow=0;
    edge[num_edge].cost=-cost;
    head[to]=num_edge;
}
bool spfa(int s,int t)
{
    memset(cost,0x3f,sizeof(cost));
    memset(flow,0x3f,sizeof(flow));
    memset(vis,0,sizeof(vis));
    q.push(s);
    vis[s]=1;
    cost[s]=0;
    pre[t]=-1;
    while (!q.empty())
    {
        int now=q.front();
        q.pop();
        vis[now]=0;
        for (int i=head[now]; i!=-1; i=edge[i].next)
        {
            if (edge[i].flow>0 && cost[edge[i].to]>cost[now]+edge[i].cost)//正边
            {
                cost[edge[i].to]=cost[now]+edge[i].cost;
                pre[edge[i].to]=now;
                last[edge[i].to]=i;
                flow[edge[i].to]=min(flow[now],edge[i].flow);//
                if (!vis[edge[i].to])
                {
                    vis[edge[i].to]=1;
                    q.push(edge[i].to);
                }
            }
        }
    }
    return pre[t]!=-1;
}

void MCMF()
{
    while (spfa(s,t))
    {
        int now=t;
        maxflow+=flow[t];
        mincost+=cost[t]*flow[t];
        while (now!=s)
        {
            //从源点一直回溯到汇点
            edge[last[now]].flow-=flow[t];//flow和cost容易搞混
            edge[last[now]^1].flow+=flow[t];
            now=pre[now];
        }
    }
}
int main()
{
    memset(head,-1,sizeof(head));
    num_edge=1;
    cin>>n>>m;
    cnt=1;
    for(int i=1; i<=m; i++)
        for(int j=1; j<=n+i-1; j++)
        {
            cin>>mp[i][j];
            id[i][j]=++cnt;
        }
    s=0,t=1;
    int T=cnt;
    for(int i=1; i<=m; i++)
        for(int j=1; j<=n+i-1; j++)
        {
            add(id[i][j],id[i][j]+T,1,-mp[i][j]);//建反向的 
            if(i==m)
                continue;
            add(id[i][j]+T,id[i+1][j],1,0);
            add(id[i][j]+T,id[i+1][j+1],1,0);
        }
    for(int i=1; i<=n; i++)
        add(s,id[1][i],1,0);
    for(int i=1; i<=n+m-1; i++)
        add(id[m][i]+T,t,1,0);
    MCMF();
    cout<<-mincost<<endl;//输出负数 
    for(int i=2; i<=num_edge; i+=2)
        edge[i].flow+=edge[i^1].flow,edge[i^1].flow=0;
    maxflow=mincost=0;
    for(int i=1; i<=m; i++)
        for(int j=1; j<=n+i-1; j++)
        {
            add(id[i][j],id[i][j]+T,inf,-mp[i][j]);
        }
    for(int i=1; i<=n+m-1; i++)
        add(id[m][i]+T,t,inf,0);
    MCMF();
    cout<<-mincost<<endl;
    for(int i=2; i<=num_edge; i+=2)
        edge[i].flow+=edge[i^1].flow,edge[i^1].flow=0;
    maxflow=mincost=0;
    for(int i=1; i<=m-1; i++)
        for(int j=1; j<=n+i-1; j++)
        {
            add(id[i][j]+T,id[i+1][j],inf,0);
            add(id[i][j]+T,id[i+1][j+1],inf,0);
        }
    MCMF();
    cout<<-mincost<<endl;
    return 0;
}

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转载自www.cnblogs.com/QingyuYYYYY/p/13179510.html