思路和P3355 骑士共存问题基本一样
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#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int maxn = 200 * 200 + 10;
const int N = 200 * 200 + 10;
const int M = N << 5;
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
ll dis[maxn];
bool vis[maxn];
struct node {
int v;
ll w;
int to;
} edge[M * 2];
int pre[N], cnt, dep[N];
int S, T, head[N];
int n, m, q[N], cur[N];
void add(int u, int v, ll w) {
// cout << u << " " << v << " " << w << endl;
edge[cnt] = {v, w, head[u]};
head[u] = cnt++;
edge[cnt] = {u, 0, head[v]};
head[v] = cnt++;
}
bool bfs() {
for (int i = 0; i <= T; i++) dep[i] = 0;
dep[S] = 1;
int l = 0, r = 1;
q[r] = S;
while (l < r) {
int u = q[++l];
for (int i = head[u]; i != -1; i = edge[i].to) {
int v = edge[i].v;
if (!dep[v] && edge[i].w) dep[v] = dep[u] + 1, q[++r] = v;
}
}
return dep[T];
}
ll dfs(int u, ll mi) {
int res = 0;
if (mi == 0 || u == T) return mi;
for (int &i = cur[u]; i != -1; i = edge[i].to) {
int v = edge[i].v;
if (dep[u] + 1 == dep[v] && edge[i].w) {
ll minn = dfs(v, min(mi - res, edge[i].w));
edge[i].w -= minn;
edge[i ^ 1].w += minn;
res += minn;
if (res == mi) return res;
}
}
if (res == 0) dep[u] = 0;
return res;
}
ll dinic() {
ll res = 0;
while (bfs()) {
memcpy(cur, head, sizeof(head));
// cout << res << endl;
res += dfs(S, INF);
}
return res;
}
int id(int x, int y) {
// cout<<x<<" "<<y<<" "<<(m * (x - 1) + y)<<endl;
return (m * (x - 1) + y); }
int mp[205][205], f;
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
int main() {
// freopen("C:\\Users\\lenovo\\Downloads\\P2774_3.in", "r", stdin);
memset(head, -1, sizeof head);
cin >> n >> m;
ll ans = 0;
S = n * m + 1, T = S + 1;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> mp[i][j];
ans += mp[i][j];
}
}
// cout << "00";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
// cout << i << " " << j << endl;
int now = id(i, j);
if ((i + j) & 1) {
add(S, now, mp[i][j]);
} else {
add(now, T, mp[i][j]);
continue;
}
for (int k = 0; k < 4; k++) {
int nx = i + dx[k], ny = j + dy[k];
if (nx < 1 || nx > n || ny < 1 || ny > m) continue;
int to = id(nx, ny);
add(now, to, inf);
// printf("(%d,%d) :%d\n", i, j, id(i, j));
// printf("(%d,%d) :%d\n", nx, ny, id(nx, ny));
// add(now, id(nx, ny), INF);
}
}
}
// cout << "ok\n";
ll flow = dinic();
// cout << flow << endl;
ans -= flow;
cout << ans;
return 0;
}