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 ;

 }