传送门
题目大意：求给出图的最小割（……题目就是这个意思）。
明眼一看，这就是一个裸的最大流，可是被数据范围吓住了，有106个点，然后又有3∗106条边，对于网络流可怜的O(n2m)的时间复杂度来说，简直是怪兽……但之前有同学告诉我，这道题就是最大流，我就硬着头皮上了一个ISAP，光荣超时，然后去找题解（想不出来了……）发现了两种，一种是把每一个面当做一个点，然后做最短路，即得最小割的容量，另一种竟然还是硬上网络流……黄学长用的dinic只用了800ms，而我之前的ISAP却用了15000+ms这不科学啊。然后我尝试着预标号，然后再ISAP，竟然只要500ms，太神了……
以后一定记住，要先预处理一遍！

代码：

/**************************************************************Problem: 1001User: geng4512Language: C++Result: AcceptedTime:552 msMemory:87896 kb
****************************************************************/#include<cstdio>
#include<cstring>
const int MAXN = 1001*1001;
inline int min(int a, int b) {return a < b ? a : b;}
int ecnt = -1;
struct node {int v, w, nxt;
}Edge[MAXN*6];
int Adj[MAXN], q[MAXN];
bool vis[MAXN];
void Addedge(int u, int v, int w) {node *t = &Edge[++ ecnt];t->v = v; t->w = w; t->nxt = Adj[u];Adj[u] = ecnt;t = &Edge[++ ecnt];t->v = u; t->w = w; t->nxt = Adj[v];Adj[v] = ecnt;
}
char c;
inline void GET(int &n) {n = 0;do c = getchar(); while(c > '9' || c < '0');while('0' <= c && c <= '9') {n=n*10+c-'0';c=getchar();}
}
int n, m, ans, S, T, d[MAXN], vd[MAXN], N;
int aug(int u, int augco) {int dmin = N-1, augc = augco, delta;if(u == T) return augco;for(int i = Adj[u]; ~i; i = Edge[i].nxt)if(Edge[i].w > 0) {if(d[Edge[i].v]+1 == d[u]) {delta = min(augc, Edge[i].w);delta = aug(Edge[i].v, delta);Edge[i].w -= delta;Edge[i^1].w += delta;augc -= delta;if(d[S] >= N) return augco - augc;if(!augc) return augco;}if(dmin > d[Edge[i].v]) dmin = d[Edge[i].v];}if(augco == augc) {-- vd[d[u]];if(!vd[d[u]]) d[S] = N;d[u] = ++ dmin;++ vd[dmin];}return augco - augc;}
void bfs() {int l = 0, r = 0, u;q[++ r] = T; vis[T] = 1; vd[0] ++;while(l < r) {u = q[++ l];for(int i = Adj[u]; ~i; i = Edge[i].nxt) {if(vis[Edge[i].v]) continue;d[Edge[i].v] = d[u] + 1;++ vd[d[u]+1];vis[Edge[i].v] = 1;q[++r] = Edge[i].v;}}
}
void sap() {bfs();while(d[S] < N)ans += aug(S, 0x7fffffff);
}
int main()
{int w;memset(Adj, -1, sizeof Adj);GET(n); GET(m);for(int i = 0; i < n; ++ i)for(int j = 0; j < m-1; ++ j) {GET(w);Addedge(i*m+j, i*m+j+1, w);}for(int i = 0; i < n-1; ++ i)for(int j = 0; j < m; ++ j) {GET(w);Addedge(i*m+j, (i+1)*m+j, w);}for(int i = 0; i < n-1; ++ i)for(int j = 0; j < m-1; ++ j) {GET(w);Addedge(i*m+j, (i+1)*m+j+1, w);}T = n*m-1; N = n*m;sap();printf("%d\n", ans);return 0;
}