[CQOI2012] 交换棋子 - 费用流

有一个n行m列的黑白棋盘，你每次可以交换两个相邻格子（相邻是指有公共边或公共顶点）中的棋子，最终达到目标状态。要求第i行第j列的格子只能参与mi,j次交换。

Solution

一个点拆三份，入点，主点，出点

入点向主点连边，主点向出点连边，设该点允许的交换次数为 \(x\) ，根据以下规则确定

• 若为初态点，则入边限流 \(x/2\)，出边限流 \((x+1)/2\)

• 若为末态点，则入边限流 \((x+1)/2\)，出边限流 \(x/2\)

• 否则，入边限流 \(x/2\)，出边限流 \((x+1)/2\)

\(S \to\) 初态点，末态点 \(\to T\)，容量 \(1\)，费用 \(0\)

八连通相互连边，容量 \(\infty\)，费用 \(1\)

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define reset(x) memset(x,0,sizeof x)
#define reset3f(x) memset(x,0x3f,sizeof x)
const int N = 1005;

namespace flow {
const int N = 100005;
const int M = 1000005;
const int inf = 1e+12;
struct Edge {
    int p, c, w, nxt = -1;
} e[N];
int s, t, tans, ans, cost, ind, bus[N], qhead = 0, qtail = -1, qu[M],vis[N], dist[N];

void graph_link(int p, int q, int c, int w) {
    e[ind].p = q;
    e[ind].c = c;
    e[ind].w = w;
    e[ind].nxt = bus[p];
    bus[p] = ind;
    ++ind;
}
void make(int p, int q, int c, int w) {
    graph_link(p, q, c, w);
    graph_link(q, p, 0, -w);
}
int dinic_spfa() {
    qhead = 0;
    qtail = -1;
    memset(vis, 0x00, sizeof vis);
    memset(dist, 0x3f, sizeof dist);
    vis[s] = 1;
    dist[s] = 0;
    qu[++qtail] = s;
    while (qtail >= qhead) {
        int p = qu[qhead++];
        vis[p] = 0;
        for (int i = bus[p]; i != -1; i = e[i].nxt)
            if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0) {
                dist[e[i].p] = dist[p] + e[i].w;
                if (vis[e[i].p] == 0)
                    vis[e[i].p] = 1, qu[++qtail] = e[i].p;
            }
    }
    return dist[t] < inf;
}
int dinic_dfs(int p, int lim) {
    if (p == t)
        return lim;
    vis[p] = 1;
    int ret = 0;
    for (int i = bus[p]; i != -1; i = e[i].nxt) {
        int q = e[i].p;
        if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0) {
            int res = dinic_dfs(q, min(lim, e[i].c));
            cost += res * e[i].w;
            e[i].c -= res;
            e[i ^ 1].c += res;
            ret += res;
            lim -= res;
            if (lim == 0)
                break;
        }
    }
    return ret;
}
void solve(int _s,int _t) {
    s=_s; t=_t;
    while (dinic_spfa()) {
        memset(vis, 0x00, sizeof vis);
        ans += dinic_dfs(s, inf);
    }
}
void init() {
    memset(bus, 0xff, sizeof bus);
}
}

int n,m;
char a[N][N],b[N][N],c[N][N];

int idIn(int i,int j) {
    return i*m-m+j+2;
}

int idMid(int i,int j) {
    return 2ll + n*m + i*m-m+j;
}

int idOut(int i,int j) {
    return 2ll + 2*n*m + i*m-m+j;
}

int check(int i,int j) {
    if(i && j && i<=n && j<=m) return 1;
    return 0;
}

signed main() {
    flow::init();
	cin>>n>>m;
	for(int i=1;i<=n;i++) cin>>a[i]+1;
	for(int i=1;i<=n;i++) cin>>b[i]+1;
	for(int i=1;i<=n;i++) cin>>c[i]+1;
	int cnt=0,tot=0;
	for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
            if(a[i][j]=='1') cnt++;
            if(b[i][j]=='1') tot++;
        }
	}
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=m;j++) {
            if(a[i][j]=='1') flow::make(1,idMid(i,j),1,0);
            if(b[i][j]=='1') flow::make(idMid(i,j),2,1,0);
            int x=c[i][j]-'0';
            if(a[i][j]=='1' && b[i][j]=='0') {
                flow::make(idIn(i,j),idMid(i,j),x/2,0);
                flow::make(idMid(i,j),idOut(i,j),(x+1)/2,0);
            }
            else if(a[i][j]=='0' && b[i][j]=='1') {
                flow::make(idIn(i,j),idMid(i,j),(x+1)/2,0);
                flow::make(idMid(i,j),idOut(i,j),x/2,0);
            }
            else {
                flow::make(idIn(i,j),idMid(i,j),x/2,0);
                flow::make(idMid(i,j),idOut(i,j),(x+1)/2,0);
            }
            for(int k=i-1;k<=i+1;k++) {
                for(int l=j-1;l<=j+1;l++) {
                    if(i!=k || j!=l) {
                        if(check(k,l)) flow::make(idOut(i,j),idIn(k,l),99,1);
                    }
                }
            }
        }
    }
    flow::solve(1,2);
    if(flow::ans==cnt && cnt==tot) cout<<flow::cost;
    else cout<<-1;
}