Gym 101128F -- Landscaping

题目来源：http://codeforces.com/gym/101128/attachments

思路来自：https://blog.csdn.net/rain722/article/details/68928211

Ford-Fulkerson算法计算最大流。

所有低点与大汇点相连，左右高点与大源点相连，流量为B，在任意相邻两点之间连一条流量为A的边。建图之后跑图的最小割即为所求。

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e4+10;
const int maxm=4e4+10;
int dis[maxn],A,B;
struct edge{
    int to,next,cap,rev;
}e[maxm];
bool vis[maxn];
int head[maxn],n,m,cnt=0;
void ins(int s,int t,int c) {
    e[++cnt].to = t;
    e[cnt].next = head[s];
    head[s] = cnt;
    e[cnt].cap = c;
    e[cnt].rev = cnt + 1;
    e[++cnt].to = s;
    e[cnt].next = head[t];
    head[t] = cnt;
    e[cnt].cap = 0;
    e[cnt].rev = cnt - 1;
}
int dfs(int v,int t,int f) {
    if (v == t)return f;
    vis[v] = 1;
    for (int i = head[v]; i; i = e[i].next) {
        if (!vis[e[i].to] && e[i].cap > 0) {
            int d = dfs(e[i].to, t, min(f, e[i].cap));
            if (d > 0) {
                e[i].cap -= d;
                e[e[i].rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t) {
    int flow = 0;
    for (;;) {
        memset(vis, 0, sizeof(vis));
        int f = dfs(s, t, 1e9);
        if (f == 0)return flow;
        flow += f;
    }
}
int get_num(int i,int j) {
    return (i - 1) * m + j;
}
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};
char c[1001];
int main() {
    scanf("%d%d%d%d", &n, &m, &A, &B);
    memset(head, 0, sizeof(head));
    cnt = 0;
    int s = 0, t = n * m + 1;
    for (int i = 1; i <= n; ++i) {
        scanf("%s", c + 1);
        for (int j = 1; j <= m; ++j) {
            if (c[j] == '.')
                ins(get_num(i, j), t, B);
            else
                ins(s, get_num(i, j), B);
            for (int k = 0; k < 4; ++k) {
                if (i + dx[k] >= 1 && i + dx[k] <= n && j + dy[k] >= 1 && j + dy[k] <= m) {
                    ins(get_num(i, j), get_num(i + dx[k], j + dy[k]), A);
                }
            }
        }
    }
    printf("%d\n", max_flow(s, t));
    return 0;
}