luogu P4542 [ZJOI2011]营救皮卡丘 网络流 拆点

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=1e6+10;
const int inf=0x3f3f3f3f;
bool vis[maxn];
int n,m,K,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 d[210][210];
int head[maxn],num_edge;
queue <int> q;
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,0x7f,sizeof(cost));
    memset(flow,0x7f,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+=flow[t]*cost[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];
        }
    }
}
signed main()
{
    memset(head,-1,sizeof(head));
    num_edge=-1;
    cin>>n>>m>>K;
    s=n*2+2,t=s+1;
    for(int i=0; i<=n; i++)
        for(int j=0; j<=n; j++)
            if(i!=j)
                d[i][j]=inf;
    for(int i=1; i<=m; i++)
    {
        int x,y,z;
        cin>>x>>y>>z;
        d[x][y]=d[y][x]=min(d[x][y],z);
    }
    for(int k = 0; k <= n; ++k)
        for(int i = 0; i <= n; ++i)
            for(int j = 0; j <= n; ++j)
                if(k < max(i, j) && d[i][j] > d[i][k] + d[k][j])
                    d[i][j] = d[i][k] + d[k][j];
    //拆点
    //源点向0的入点连流量k费用0的边，表示最多经过0 K次
    //源点向其余每个点的入点连流量1费用0的边
    for(int i = 0; i <= n; ++i)
    {
        add(s, i, !i ? K : 1, 0);
        add(i + n + 1, t, 1, 0);//每个点都要被到达一次,0号不用被破坏
        for(int j = i + 1; j <= n; ++j)
            //每个i的入点向j(j>i)连流量1费用d[i][j]的边
            add(i, j + n + 1, 1, d[i][j]);
    }
    MCMF();
    cout<<mincost<<endl;
    return 0;
}