UVa 1515 - Pool construction (最小割)

题目链接：https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=0&problem=4261&mosmsg=Submission+received+with+ID+26560200

首先将边缘的洞全部填成草，然后从源点向草连容量为 \(d\) 的边，表示将草变为洞需要 \(d\) 的代价，

从洞向汇点连容量为 \(f\) 的边，表示将洞变成草需要 \(f\) 的代价，

相邻的格子互相连容量为 \(b\) 的边，表示如果一个为草，一个为洞，建立栅栏需要 \(b\) 的代价

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 2555;
const int INF = 1000000007;

int T, n, m, ans;
int D, F, B;
int mat[101][101];

int h[maxn], cnt = 1;
struct E{
	int to, cap, next;
}e[1000100];
void add(int u, int v, int c){
	e[++cnt].to = v;
	e[cnt].cap = c;
	e[cnt].next = h[u];
	h[u] = cnt;
	
	e[++cnt].to = u;
	e[cnt].cap = 0;
	e[cnt].next = h[v];
	h[v] = cnt;
}

int s,t;
int vis[maxn], d[maxn], cur[maxn];

  bool BFS() {
    memset(vis, 0, sizeof(vis));
    queue<int> Q;
    Q.push(s);
    vis[s] = 1;
    d[s] = 0;
    while(!Q.empty()) {
      int u = Q.front(); Q.pop();
      for(int i = h[u]; i != -1 ; i = e[i].next) {
        if(!vis[e[i].to] && e[i].cap) {
          vis[e[i].to] = 1;
          d[e[i].to] = d[u] + 1;
          Q.push(e[i].to);
        }
      }
    }
    return vis[t];
  }

  int DFS(int x, int a) {
    if(x == t || a == 0) return a;
    int flow = 0, f;
    for(int &i = cur[x]; i != -1 ; i = e[i].next) {
      if(d[x] + 1 == d[e[i].to] && (f = DFS(e[i].to, min(a, e[i].cap))) > 0) {
        e[i].cap -= f;
        e[i^1].cap += f;
        flow += f;
        a -= f;
        if(a == 0) break;
      }
    }
    return flow;
  }

  int Maxflow() {
    int flow = 0;
    while(BFS()) {
	  memcpy(cur, h, sizeof(h));
      flow += DFS(s, INF);
    }
    return flow;
  }

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

inline int ID(int i, int j) { return (i-1)*m+j; }

int main(){
	scanf("%d", &T);
	while(T--){
		memset(h, -1, sizeof(h)); cnt = 1;
		ans = 0;
		
		scanf("%d%d", &m, &n);
		scanf("%d%d%d", &D, &F, &B);
		
		char ss[100];
		for(int i = 1 ; i <= n ; ++i){
			scanf("%s", ss + 1);
			for(int j = 1 ; j <= m ; ++j){
				if(ss[j] == '.') mat[i][j] = 1;
				else mat[i][j] = 0;
			}
		}
		
		for(int i = 1 ; i <= n ; ++i){
			for(int j = 1 ; j <= m ; ++j){
				if((i == 1 || i == n || j == 1 || j == m) && (mat[i][j] == 1)){	// 边缘的洞都填成草
					mat[i][j] = 0;
					ans += F;
				}
			}
		}
		
		s = n * m + 1; t = n * m + 2;
		for(int i = 1 ; i <= n ; ++i){
			for(int j = 1 ; j <= m ; ++j){
				if(!mat[i][j]){ // 源点 -> 草 
					int C;
					if(i == 1 || i == n || j == 1 || j == m) C = INF;
					else C = D;
					add(s, ID(i, j), C);
				} else{ // 洞 -> 汇点 
					add(ID(i, j), t, F);
				}
				
				// 相邻的边相互连接 
				
				if(i > 1) {
					add(ID(i,j), ID(i-1,j), B);
				} 
        		if(i < n) {
        			add(ID(i,j), ID(i+1,j), B);
				}
        		if(j > 1) {
        			add(ID(i,j), ID(i,j-1), B);
				}
        		if(j < m) {
        			add(ID(i,j), ID(i,j+1), B);
				}
			}
		}
		
		ans += Maxflow();
		
		printf("%d\n", ans);
	}
	return 0;
}