[BZOJ 2668] 交换棋子

Link:

BZOJ 2668 传送门

Solution:

重点在于对于每条转移路径：首尾算一次，中间节点算两次

可以一点拆三点，将原流量拆成入流量和出流量

但其实也可以就拆两点，分前后是否是一首尾点一普通点来确定是否有一条路径只占用1流量

Code:

#include <bits/stdc++.h>

using namespace std;
#define X first
#define Y second
typedef long long ll;
typedef pair<int,int> P;
const int MAXN=1e5+10,INF=1<<30;
int n,m,cnt1,cnt2;char dat[2][30][30],lmt[30][30];

namespace mcmf
{
    struct edge
    {int to,cap,cost,rev;};
    vector<edge> a[MAXN];
    int S,T,h[MAXN],dist[MAXN],preV[MAXN],preE[MAXN],maxf,minc;
    
    void add_edge(int from,int to,int cap,int cost)
    {
        a[from].push_back(edge{to,cap,cost,a[to].size()});
        a[to].push_back(edge{from,0,-cost,a[from].size()-1});
    }
    void min_cost_flow(int f)
    {
        while(f>0)
        {
            priority_queue<P,vector<P>,greater<P> > que;
            fill(dist,dist+T+1,INF);
            dist[S]=0;que.push(P(0,S));
            while(!que.empty())
            {
                P t=que.top();que.pop();
                int v=t.Y;if(dist[v]<t.X) continue;
                
                for(int i=0;i<a[v].size();i++)
                {
                    edge &e=a[v][i];
                    if(e.cap>0&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to])
                        dist[e.to]=dist[v]+e.cost+h[v]-h[e.to],
                        preV[e.to]=v,preE[e.to]=i,que.push(P(dist[e.to],e.to));
                }
            }
            if(dist[T]==INF) break;
            for(int i=1;i<=T;i++) h[i]+=dist[i];
            
            int d=f;
            for(int i=T;i!=S;i=preV[i])
                d=min(d,a[preV[i]][preE[i]].cap);
            f-=d;minc+=d*h[T];maxf+=d;
            for(int i=T;i!=S;i=preV[i])
            {
                edge &e=a[preV[i]][preE[i]];
                e.cap-=d,a[i][e.rev].cap+=d;
            }
        }
    }
}
using namespace mcmf;
int dx[]={1,1,1,-1,-1,-1,0,0};
int dy[]={1,-1,0,1,-1,0,1,-1};
int idx(int x,int y,int z){return (x-1)*m+y+n*m*z;}

int main()
{
    scanf("%d%d",&n,&m);
    S=0;T=2*n*m+1;
    for(int i=1;i<=n;i++) scanf("%s",dat[0][i]+1);
    for(int i=1;i<=n;i++) scanf("%s",dat[1][i]+1);
    for(int i=1;i<=n;i++) scanf("%s",lmt[i]+1);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            if(dat[0][i][j]=='1')
                cnt1++,add_edge(S,idx(i,j,0),1,0);
            if(dat[1][i][j]=='1')
                cnt2++,add_edge(idx(i,j,1),T,1,0);
            if(dat[0][i][j]==dat[1][i][j])
                add_edge(idx(i,j,0),idx(i,j,1),(lmt[i][j]-'0')/2+dat[0][i][j]-'0',0);
            else add_edge(idx(i,j,0),idx(i,j,1),(lmt[i][j]-'0'+1)/2,0);
            
            for(int k=0;k<8;k++)
            {
                int fx=i+dx[k],fy=j+dy[k];
                if(fx<1||fx>n||fy<1||fy>m) continue;
                add_edge(idx(i,j,1),idx(fx,fy,0),INF,1);
            }
        }
    
    min_cost_flow(INF);
    if(cnt1!=cnt2||maxf!=cnt1) puts("-1");
    else printf("%d",minc);
    return 0;
}