【BZOJ 1001】[BJOI2006]狼抓兔子（最大流）

题目链接
最大流裸题，没什么好说吧，恰好点数多，考验网络流的效率，正好练(Dinic)。

#include <cstdio>
#include <queue>
#include <cstring>
#define INF 2147483647
using namespace std;
const int MAXN = 1000010;
const int MAXM = 8000010;
inline int read(){
    int s = 0, w = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9'){ if(ch == '-') w = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); }
    return s * w;
}
struct Edge{
    int next, to, rest;
}e[MAXM];
int s, t, num = 1, n, m;
int head[MAXN];
inline void Add(int from, int to, int flow){
    e[++num] = (Edge){ head[from], to, flow }; head[from] = num;
    e[++num] = (Edge){ head[to], from, flow }; head[to] = num;
}
int level[MAXN], now, sum;
queue <int> q;
int re(){
	memset(level, 0, sizeof level);
    while(q.size()) q.pop();
    q.push(s); level[s] = 1;
    while(q.size()){
      now = q.front(); q.pop();
      for(int i = head[now]; i; i = e[i].next)
         if(e[i].rest && !level[e[i].to]){
           level[e[i].to] = level[now] + 1;
           q.push(e[i].to);
         }
    }
    return level[t];
}
int findflow(int u, int flow){
	if(!flow || u == t) return flow;
	int f = 0, t;
	for(int i = head[u]; i; i = e[i].next){
		if(e[i].rest && level[e[i].to] == level[u] + 1){
			f += (t = findflow(e[i].to, min(flow - f, e[i].rest)));
			e[i].rest -= t; e[i ^ 1].rest += t;
		}
	}
	if(!f) level[u] = 0;
	return f;
}
int dinic(){
    int ans = 0;
    while(re())
      ans += findflow(s, INF);
    return ans;
}
inline int id(int i, int j){
	return (i - 1) * m + j;
}
int main(){
    n = read(); m = read(); s = id(1, 1); t = id(n, m);
    for(int i = 1; i <= n; ++i)
    	for(int j = 1; j < m; ++j)
    	   Add(id(i, j), id(i, j + 1), read());
    for(int i = 1; i < n; ++i)
    	for(int j = 1; j <= m; ++j)
    	   Add(id(i, j), id(i + 1, j), read());
    for(int i = 1; i < n; ++i)
    	for(int j = 1; j < m; ++j)
    	   Add(id(i, j), id(i + 1, j + 1), read());
    printf("%d
", dinic());
    return 0;
}