好久没有写博客了。不过这个博客也没有多少人看
最近在写网络流，为了加深理解，来写一两篇题解。
对整个棋盘进行黑白染色以后可以发现，一次操作就是让二分图的两个点的值分别 (+1)。
这样，我们就可以对一个答案的合法性做出判断了。
对于每个白点，从 (S) 向它连 (ans - a[i][j]) 的边。黑点向 (T) 连 (ans - a[i][j]) 的边。每个白点向黑点建 (+infty) 的边。
如果满流就成立。
发现答案满足单调性，然后就开始非常开心地二分答案？
你会发现从 (S) 出来的容量和与流向 (T) 的容量和根本不一定等。
但是它们必须等。
于是我们设 (cnt_1) 为白点的数量，(sum_1) 为白点初始权值之和。(cnt_0, sum_0) 为黑点的。
那么必须有：
[ans cdot cnt_0 - sum_0 = ans cdot cnt_1 - sum_1
]
可以化成
[(cnt_o - cnt_1) cdot ans = sum_0 - sum_1
]
如果 (cnt_0 eq cnt_1) 可以直接把 (ans) 算出来验证一下就可以了。
如果 (cnt_o = cnt_1) 那么就是说 (sum_o = sum_1) 不满足就可以直接Skipped了。如果满足的话 (ans) 可以随便取值，那么就可以二分求出最小的了。
#include<bits/stdc++.h>
#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back
template<typename A, typename B> inline char SMAX(A &a, const B &b) {return a < b ? a = b , 1 : 0;}
template<typename A, typename B> inline char SMIN(A &a, const B &b) {return b < a ? a = b , 1 : 0;}
typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;
template<typename I>
inline void read(I &x) {
int f = 0, c;
while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
x = c & 15;
while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
f ? x = -x : 0;
}
const int MAXN = 40 + 7;
const int N = MAXN * MAXN;
const int M = MAXN * MAXN * 3;
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int INF_int = 0x3f3f3f3f;
int n, m, S, T, nod, cnt0, cnt1, hd, tl, allsize;
ll sum0, sum1, sum;
int a[MAXN][MAXN], id[MAXN][MAXN], col[MAXN][MAXN];
int q[N], cur[N], dis[N];
struct Edge {int to, ne; ll f;} g[M << 1]; int head[N], tot = 1;
inline void addedge(int x, int y, ll z) {g[++tot].to = y; g[tot].f = z; g[tot].ne = head[x]; head[x] = tot;}
inline void adde(int x, int y, ll z) {addedge(x, y, z); addedge(y, x, 0);}
inline bool bfs() {
memset(dis, 0x3f, allsize); memcpy(cur, head, allsize);
q[hd = 0, tl = 1] = S; dis[S] = 0;
while (hd < tl) {
int x = q[++hd];
for fec(i, x, y) if (g[i].f && dis[y] == INF_int) {
dis[y] = dis[x] + 1; q[++tl] = y;
// dbg("y = %d, T = %d
", y, T);
if (y == T) return 1;
}
}
return 0;
}
inline ll dfs(int x, ll a) {
// dbg("x = %d a = %lld
", x, a);
if (x == T || !a) return a;
ll flow = 0, f;
for (int &i = cur[x]; i; i = g[i].ne) {
int y = g[i].to;
if (dis[y] != dis[x] + 1) continue;
if (!(f = dfs(y, std::min(a, g[i].f)))) continue;
g[i].f -= f, g[i ^ 1].f += f;
flow += f, a -= f; if (!a) break;
}
if (!flow) dis[x] = INF_int;
return flow;
}
inline ll dinic() {
ll ans = 0;
while (bfs()) ans += dfs(S, INF);
// dbg("ans = %lld
", ans);
return ans;
}
inline bool build(ll mid) {
memset(head, 0, sizeof(head));
tot = 1; sum = 0;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) {
if (mid < a[i][j]) return 0;
if (col[i][j]) {
adde(S, id[i][j], mid - a[i][j]);
sum += mid - a[i][j];
for (int k = 0; k < 4; ++k) {
int px = i + dx[k], py = j + dy[k];
if (px < 1 || px > n || py < 1 || py > m) continue;
adde(id[i][j], id[px][py], INF);
}
} else adde(id[i][j], T, mid - a[i][j]);
}
return 1;
}
inline bool check(ll mid) {
if (!build(mid)) return 0;
return dinic() == sum;
}
inline void work() {
if (cnt0 == cnt1) {
if (sum0 != sum1) {
puts("-1");
return;
} else {
ll l = 0, r = INF;
while (l < r) {
ll mid = (l + r) >> 1;
if (check(mid)) r = mid;
else l = mid + 1;
}
// dbg("l = %d
", l);
// for (ll i = l - 10000; i <= l + 10000; ++i)
// dbg("When i = %lld, chk = %d
", i, (int)check(i));
if (l < INF) printf("%lld
", l * cnt0 - sum0);
else puts("-1");
}
} else {
ll dsum = sum0 - sum1;
int dcnt = cnt0 - cnt1;
if (((dsum < 0 && dcnt < 0) || (dsum > 0 && dcnt > 0)) && dsum % dcnt == 0) {
ll ans = dsum / dcnt;
if (check(ans)) printf("%lld
", ans * cnt0 - sum0);
else puts("-1");
}
}
}
inline void cls() {
sum0 = sum1 = 0;
cnt0 = cnt1 = 0;
}
inline void init() {
cls();
read(n), read(m);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
read(a[i][j]);
col[i][j] = (i + j) & 1;
if (col[i][j]) ++cnt1, sum1 += a[i][j];
else ++cnt0, sum0 += a[i][j];
if (j > 1) id[i][j] = id[i][j - 1] + 1;
else id[i][j] = id[i - 1][m] + 1;
}
}
S = id[n][m] + 1, T = nod = id[n][m] + 2;
allsize = (nod + 1) * sizeof(int);
}
int main() {
#ifdef hzhkk
freopen("hkk.in", "r", stdin);
#endif
int T; read(T);
while (T--) {
init();
work();
}
fclose(stdin);
}