[Bzoj1001][BeiJing2006]狼抓兔子(网络流/对偶图)

题目链接：https://www.lydsy.com/JudgeOnline/problem.php?id=1001

看到大佬们都是对偶图过的，写了个最大流水过去了QAQ，网络流的无向图直接建双向边(不用建0边)，然后跑dinic，最基本的dinic会被卡，可以简单优化一下。

有空学了对偶图在补，(个_个)

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int maxn = 6e6 + ;
 const int inf = INT_MAX;
 struct node {
     int e, w, next;
 }edge[maxn];
 int head[maxn], len;
 int d[maxn];
 void init() {
     memset(head, -, sizeof(head));
     len = ;
 }
 void add(int s, int e, int w) {
     edge[len].e = e;
     edge[len].w = w;
     edge[len].next = head[s];
     head[s] = len++;
 }
 inline int read() {
     int x = , f = ; char ch = getchar();
     while (ch<'' || ch>'') { if (ch == '-')f = -; ch = getchar(); }
     while (ch >= ''&&ch <= '') { x = x *  + ch - ; ch = getchar(); }
     return x * f;
 }
 bool bfs(int s, int e) {
     queue<int>q;
     memset(d, , sizeof(d));
     d[s] = ;
     q.push(s);
     while (!q.empty()) {
         int x = q.front(); q.pop();
         for (int i = head[x]; i != -; i = edge[i].next) {
             int y = edge[i].e;
             if (edge[i].w && !d[y]) {
                 d[y] = d[x] + ;
                 q.push(y);
             }
         }
     }
     return d[e];
 }
 int dfs(int s, int e, int limit) {
     if (s == e)
         return limit;
     int add, ans = ;
     for (int i = head[s]; i != -; i = edge[i].next) {
         int y = edge[i].e;
         if (d[s] == d[y] -  && edge[i].w && (add = dfs(y, e, min(edge[i].w, limit)))) {
             edge[i].w -= add;
             edge[i ^ ].w += add;
             ans += add;
             limit -= add;
             if (!limit)
                 break;
         }
     }
     if (ans)
         return ans;
     d[s] = -;
     return ;
 }
 int dinic(int s, int e) {
     int ans = , f;
     while (bfs(s, e)) {
         while (f = dfs(s, e, inf))
             ans += f;
     }
     return ans;
 }
 int main() {
     int n, m, z;
     n = read(), m = read();
     init();
     for (int i = ; i <= n; i++)
         for (int j = ; j < m; j++) {
             z = read();
             add(m*(i - ) + j, m*(i - ) + j + , z);
             add(m*(i - ) + j + , m*(i - ) + j, z);
         }
     for (int i = ; i < n; i++)
         for (int j = ; j <= m; j++) {
             z = read();
             add(m*(i - ) + j, m*i + j, z);
             add(m*i + j, m*(i - ) + j, z);
         }
     for (int i = ; i < n; i++)
         for (int j = ; j < m; j++) {
             z = read();
             add(m*(i - ) + j, m*i + j + , z);
             add(m*i + j + , m*(i - ) + j, z);
         }
     printf("%d\n", dinic(, n*m));
     //system("pause");
 }