Gym

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
", ans);
}