BZOJ 1001 狼抓兔子 平面图的最小割

题目链接：

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

题目大意：

见链接

思路：

求最小割，平面图的最小割等价于对偶图的最短路

直接建图求最短路即可，只是图比较难建。

 #include<bits/stdc++.h>
 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf
 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时
 #define Min(a, b) ((a) < (b) ? (a) : (b))
 #define Mem(a) memset(a, 0, sizeof(a))
 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1))
 #define MID(l, r) ((l) + ((r) - (l)) / 2)
 #define lson ((o)<<1)
 #define rson ((o)<<1|1)
 #pragma comment(linker, "/STACK:102400000,102400000")//栈外挂
 using namespace std;
 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;
 }

 typedef long long ll;
 const int maxn =  + ;
 const int mod = ;//const引用更快，宏定义也更快
 const int INF = 1e9;

 struct edge
 {
     int next;//指向下一个节点
     int u, v, w;
 };
 edge a[maxn];
 int head[maxn], node;//node记录节点的数目，head[i]记录连接着i的第一条边
 void addedge(int u, int v, int w)
 {
     a[node].u = u;
     a[node].v = v;
     a[node].w = w;
     a[node].next = head[u];
     head[u] = node++;
     a[node].u = v;
     a[node].v = u;
     a[node].w = w;
     a[node].next = head[v];
     head[v] = node++;
 }
 struct Heapnode
 {
     int d, u;//d为距离，u为起点
     Heapnode(){}
     Heapnode(int d, int u):d(d), u(u){}
     bool operator <(const Heapnode & a)const
     {
         return d > a.d;//这样优先队列先取出d小的
     }
 };
 ll n, m;
 bool v[maxn];//标记点是否加入集合
 int d[maxn];//起点s到各个点的最短路
 void init(int n)
 {
     node = ;
     memset(head, -, sizeof(head));
 }
 priority_queue<Heapnode>q;
 void dijkstra(ll s, ll n)//以s为起点
 {
     while(!q.empty())q.pop();
     for(int i = ; i <= n; i++)d[i] = INF;
     d[s] = ;
     memset(v, , sizeof(v));
     q.push(Heapnode(, s));
     while(!q.empty())
     {
         Heapnode now = q.top();
         q.pop();
         ll u = now.u;//当前起点
         if(v[u])continue;//如果已经加入集合，continue
         v[u] = ;
         for(int i = head[u]; i != -; i = a[i].next)
         {
             edge& e = a[i];//引用节省代码
             if(d[e.v] > d[u] + e.w)
             {
                 d[e.v] = d[u] + e.w;
                 q.push(Heapnode(d[e.v], e.v));
             }
         }
     }
 }

 int main()
 {
     ll n, m;
     while(scanf("%lld%lld", &n, &m) != EOF)
     {
         if(n ==  || m == )
         {
             ll ans = INF, x;
             for(int i = ; i < max(n, m); i++)
                 scanf("%lld", &x), ans = min(ans, x);
             printf("%lld\n", ans);
             continue;
         }
         ll x;
         ll t = , s = (n - ) * (m - ) *  + ;
         init(s);
         ll tmp = (m - ) * ;//每一行的数目
         for(int i = ; i <= n; i++)
             for(int j = ; j < m; j++)
             {
                 scanf("%lld", &x);
                 if(i == )addedge(j * , t, x);
                 else if(i == n)addedge(s, (n - ) * tmp +  * j - , x);
                 else addedge((i - ) * tmp +  * j, (i - ) * tmp +  * j - , x);
             }
         for(int i = ; i < n; i++)
             for(int j = ; j <= m; j++)
             {
                 scanf("%lld", &x);
                 if(j == )addedge(s, (i - ) * tmp + , x);
                 else if(j == m)addedge(i * tmp, t, x);
                 else addedge((i - ) * tmp +  * j - , (i - ) * tmp +  * j - , x);
             }
         for(int i = ; i < n; i++)
             for(int j = ; j < m; j++)
             {
                 scanf("%lld", &x);
                 addedge((i - ) * tmp +  * j - , (i - ) * tmp +  * j, x);
             }
         dijkstra(s, s);
         printf("%d\n", d[t]);
     }
     return ;
 }