传送门
题解
同时分别限制流入和流出次数，所以把一个点拆成三个：入点in(x)、中间点mi(x)、出点ou(x)。
如果一个格子x在初始状态是黑点，则连(S, mi(x), 1, 0)
如果x在目标状态是黑点，则连(mi(x), T, 1, 0)
设x的交换次数限制是w
如果x在两种状态中颜色相同，则连(in(x), mi(x), w / 2, 0), (mi(x), ou(x), w / 2, 0)
如果x只在初始状态为黑色，则连(in(x), mi(x), w / 2, 0), (mi(x), ou(x), (w + 1) / 2, 0)
如果x只在目标状态为黑色，则连(in(x), mi(x), (w + 1) / 2, 0), (mi(x), ou(x), w / 2, 0)
每个点x和相邻的点y（八连通），连(ou(x), in(y), INF, 1)
然后最小费用最大流即可。
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
#define space putchar(' ')
#define enter putchar('
')
typedef long long ll;
using namespace std;
template <class T>
void read(T &x){
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-') op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op) x = -x;
}
template <class T>
void write(T x){
if(x < 0) putchar('-'), x = -x;
if(x >= 10) write(x / 10);
putchar('0' + x % 10);
}
const int N = 1234, M = 1000005, INF = 0x3f3f3f3f;
const int dx[] = {-1, 1, 0, 0, -1, -1, 1, 1};
const int dy[] = {0, 0, -1, 1, -1, 1, -1, 1};
int n, m, tot, maxflow, mincost, src = 1, des = 2, id[23][23][3], b1, b2;
int ecnt = 1, adj[N], pre[N], dis[N], nxt[M], go[M], cap[M], cost[M];
char st[23][23], ed[23][23], cp[23][23];
void _add(int u, int v, int w, int c){
go[++ecnt] = v;
nxt[ecnt] = adj[u];
adj[u] = ecnt;
cap[ecnt] = w;
cost[ecnt] = c;
}
void add(int u, int v, int w, int c){
_add(u, v, w, c);
_add(v, u, 0, -c);
}
bool spfa(){
queue <int> que;
static bool inq[N] = {0};
for(int i = 1; i <= tot; i++)
dis[i] = INF, pre[i] = 0;
dis[src] = 0, que.push(src), inq[src] = 1;
while(!que.empty()){
int u = que.front();
que.pop(), inq[u] = 0;
for(int e = adj[u], v; e; e = nxt[e]){
if(cap[e] && dis[u] + cost[e] < dis[v = go[e]]){
dis[v] = dis[u] + cost[e], pre[v] = e;
if(!inq[v]) que.push(v), inq[v] = 1;
}
}
}
return dis[des] < INF;
}
void mcmf(){
while(spfa()){
int delta = INF;
for(int e = pre[des]; e; e = pre[go[e ^ 1]])
delta = min(delta, cap[e]);
for(int e = pre[des]; e; e = pre[go[e ^ 1]])
cap[e] -= delta, cap[e ^ 1] += delta;
maxflow += delta;
mincost += delta * dis[des];
}
}
bool legal(int x, int y){
return x > 0 && y > 0 && x <= n && y <= m;
}
int main(){
read(n), read(m), tot = 3 * n * m + 2;
for(int i = 1, cnt = 2; i <= n; i++)
for(int j = 1; j <= m; j++)
id[i][j][0] = ++cnt, id[i][j][1] = ++cnt, id[i][j][2] = ++cnt;
for(int i = 1; i <= n; i++) scanf("%s", st[i] + 1);
for(int i = 1; i <= n; i++) scanf("%s", ed[i] + 1);
for(int i = 1; i <= n; i++) scanf("%s", cp[i] + 1);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++){
int w = cp[i][j] - '0';
if(st[i][j] == '1') b1++, add(src, id[i][j][0], 1, 0);
if(ed[i][j] == '1') b2++, add(id[i][j][0], des, 1, 0);
if(st[i][j] == ed[i][j]){
add(id[i][j][1], id[i][j][0], w / 2, 0);
add(id[i][j][0], id[i][j][2], w / 2, 0);
}
else if(st[i][j] == '1'){
add(id[i][j][1], id[i][j][0], w / 2, 0);
add(id[i][j][0], id[i][j][2], (w + 1) / 2, 0);
}
else if(ed[i][j] == '1'){
add(id[i][j][1], id[i][j][0], (w + 1) / 2, 0);
add(id[i][j][0], id[i][j][2], w / 2, 0);
}
for(int d = 0, x, y; d < 8; d++)
if(legal(x = i + dx[d], y = j + dy[d]))
add(id[i][j][2], id[x][y][1], INF, 1);
}
mcmf();
if(maxflow < max(b1, b2)) puts("-1");
else write(mincost), enter;
return 0;
}