1.​​题目链接​​。给定一个网格图，求这个网格图从左上角(1,1)到右下角（n,m)的最大流。裸题。

#include<bits/stdc++.h>
#include<queue>
using namespace std;
typedef long long LL;
const int MAXM = 6000011;
const int MAXN = 1000011;
int inf;
int n, m;
int first[MAXN];
int ecnt;
int s, t;
int dis[MAXN];
int ans;
#pragma warning(disable:4996)
struct edge {
  int v, f;
  int next;
}e[MAXM];

void link(int x, int y, int z) 
{
  e[++ecnt].next = first[x]; first[x] = ecnt; e[ecnt].v = y; e[ecnt].f = z;
  e[++ecnt].next = first[y]; first[y] = ecnt; e[ecnt].v = x; e[ecnt].f = z;
}

 bool bfs() {
  memset(dis, 127 / 3, sizeof(dis));
  int cun = dis[t];
  queue<int>Q;
  while (!Q.empty()) Q.pop();

  dis[1] = 1;  Q.push(1);
  while (!Q.empty()) {
    int u = Q.front(); Q.pop();
    for (int i = first[u]; i; i = e[i].next) {
      if (e[i].f && dis[e[i].v] == cun) {
        dis[e[i].v] = dis[u] + 1; Q.push(e[i].v);
      }
    }
    if (dis[t] != cun) return true;
  }

  return false;
}

 int maxflow(int now, int remain) 
{
  if (remain == 0 || now == t) return remain;
  int flow = 0;
  for (int i = first[now]; i; i = e[i].next) {
    if (dis[e[i].v] == dis[now] + 1 && e[i].f) {
      int f = maxflow(e[i].v, min(remain, e[i].f));
      if (f) {
        e[i].f -= f; e[i ^ 1].f += f;
        flow += f; remain -= f;
        if (remain == 0) return flow;
      }
      else dis[e[i].v] = -1;
    }
  }
  return flow;
}

void solve() {
  s = 1, t = n * m; inf = 1;
  for (int i = 1; i <= 30; i++) inf *= 2;
  while (bfs()) 
  {
    ans += maxflow(s, inf);
  }
  printf("%d", ans);
}

int main()
{
  scanf("%d%d", &n, &m);
  int x; ecnt = 1;
  int nowx, nownex;
  ///横向加边
  for (int i = 1; i <= n; i++)
    for (int j = 1; j < m; j++) 
    {
      nowx = (i - 1)*m + j, nownex = nowx + 1;
      scanf("%d", &x);
      link(nowx, nownex, x);
    }

  for (int i = 1; i < n; i++)
    for (int j = 1; j <= m; j++) {
      nowx = (i - 1)*m + j, nownex = nowx + m;
      scanf("%d", &x);
      link(nowx, nownex, x);
    }

  for (int i = 1; i < n; i++)
    for (int j = 1; j < m; j++) 
    {
      nowx = (i - 1)*m + j, nownex = nowx + m + 1;
      scanf("%d", &x);
      link(nowx, nownex, x);
    }

  solve();
  return 0;
}