题意

给你一个\(N×M\)的草地，有高地有低地。
收割机从低地走到高地或者从高地走到低地都要花费用\(A\)，你可以花费用\(B\)把一块高地变成低地，或者把一块低地变成高地。收割机每行每列都是必须要跑一趟的。

求最小花费。

解析

\(S\)向低地、高地向\(T\)建权为\(B\)的边，相邻的地之间建边权为\(A\)的边。

然后求最小割。

相同类型的地之间为什么也要建边呢？因为类型是可以改变的。

#include <bits/stdc++.h>
#define FOPI freopen("in.txt", "r", stdin)
#define FOPO freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 50 * 50 + 1000;
const int maxm = 1e5 + 100;struct Edge
{int to, next, cap, flow;
}edge[maxm];int tot;
int head[maxn];void init()
{tot = 2;memset(head, -1, sizeof(head));
}void build(int u, int v, int w, int rw = 0)
{edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;edge[tot].next = head[u]; head[u] = tot++;edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;edge[tot].next = head[v]; head[v] = tot++;
}int Q[maxn];
int dep[maxn], cur[maxn], sta[maxn];bool bfs(int s, int t, int n)
{int front = 0, tail = 0;memset(dep, -1, sizeof(dep[0]) * (n+1));dep[s] = 0;Q[tail++] = s;while(front < tail){int u = Q[front++];for (int i = head[u]; i != -1; i = edge[i].next){int v = edge[i].to;if (edge[i].cap > edge[i].flow && dep[v] == -1){dep[v] = dep[u] + 1;if (v == t) return true;Q[tail++] = v;}}}return false;
}LL dinic(int s, int t, int n)
{LL maxflow = 0;while(bfs(s, t, n)){for (int i = 0; i < n; i++) cur[i] = head[i];int u = s, tail = 0;while(cur[s] != -1){if (u == t){int tp = inf;for (int i = tail-1; i >= 0; i--)tp = min(tp, edge[sta[i]].cap - edge[sta[i]].flow);//if (tp >= inf) return -1;maxflow += tp;for (int i = tail-1; i >= 0; i--){edge[sta[i]].flow += tp;edge[sta[i]^1].flow -= tp;if (edge[sta[i]].cap - edge[sta[i]].flow == 0) tail = i;}u = edge[sta[tail]^1].to;}else if (cur[u] != -1 && edge[cur[u]].cap > edge[cur[u]].flow&& dep[u]+1 == dep[edge[cur[u]].to]){sta[tail++] = cur[u];u = edge[cur[u]].to;}else{while(u != s && cur[u] == -1) u = edge[sta[--tail]^1].to;cur[u] = edge[cur[u]].next;}}}return maxflow;
}int n, m, A, B;
int S, T;
char a[maxn][maxn];int id(int i, int j) { return (i-1)*m + j; }int main()
{//FOPI;init();scanf("%d%d%d%d", &n, &m, &A, &B);S = 0, T = n*m+1;for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++){scanf(" %c", &a[i][j]);if (a[i][j] == '.') build(S, id(i, j), B);else build(id(i, j), T, B);if (i > 1) build(id(i, j), id(i-1, j), A);if (i < n) build(id(i, j), id(i+1, j), A);if (j > 1) build(id(i, j), id(i, j-1), A);if (j < m) build(id(i, j), id(i, j+1), A);}LL ans = dinic(S, T, T+1);printf("%lld\n", ans);
}