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
", ans);
}
return 0;
}