[BeiJing2006]狼抓兔子

题面

一眼看就是最小割板子题，建图也很直观，注意每一条边建双向边其实不用建4条边，只要反向边的容量和正边相同就行。然后直接跑最大流板子就行。不过此题拿vector存图会MLE……而拿链前存图就能卡过去……场面一度十分尴尬。

这里发一个vector80分代码，各位改成链前就能AC了……

 #include<cstdio>

 #include<iostream>

 #include<cmath>

 #include<algorithm>

 #include<cstring>

 #include<cstdlib>

 #include<queue>

 #include<stack>

 #include<vector>

 #include<cctype>

 using namespace std;

 #define enter puts("")

 #define space putchar(' ')

 #define Mem(a) memset(a, 0, sizeof(a))

 #define rg register

 typedef long long ll;

 typedef double db;

 const int INF = 0x3f3f3f3f;

 const db eps = 1e-;

 const int maxn = 1e6 + ;

 inline ll read()

 {

     ll ans = ;

     char ch = getchar(), last = ' ';

     while(!isdigit(ch)) {last = ch; ch = getchar();}

     while(isdigit(ch))

     {

         ans = ans *  + ch - ''; ch = getchar();

     }

     if(last == '-') ans = -ans;

     return ans;

 }

 inline void write(ll x)

 {

     if(x < ) putchar('-'), x = -x;

     if(x >= ) write(x / );

     putchar(x %  + '');

 }

 int n, m;

 struct Edge

 {

     int from, to, cap, flow;

 };

 vector<Edge> edges;

 vector<int> G[maxn];

 void addedge(int from, int to, int w)

 {

     edges.push_back((Edge){from, to, w, });

     edges.push_back((Edge){to, from, w, });

     int sz = edges.size();

     G[from].push_back(sz - );

     G[to].push_back(sz - );

 }

 int dis[maxn];

 bool vis[maxn];

 bool bfs()

 {

     for(rg int i = ; i <= n * m; ++i) vis[i] = ;

     dis[] = ; vis[] = ;

     queue<int> q; q.push();

     while(!q.empty())

     {

         int now = q.front(); q.pop();

         for(rg int i = ; i < (int)G[now].size(); ++i)

         {

             Edge& e = edges[G[now][i]];

             if(!vis[e.to] && e.cap > e.flow)

             {

                 vis[e.to] = ;

                 dis[e.to] = dis[now] + ;

                 q.push(e.to);

             }

         }

     }

     return vis[n * m];

 }

 int cur[maxn];

 int dfs(const int& now, int a)

 {

     if(now == n * m || !a) return a;

     int flow = , f;

     for(int& i = cur[now]; i < (int)G[now].size(); ++i)

     {

         Edge& e = edges[G[now][i]];

         if(dis[e.to] == dis[now] +  && (f = dfs(e.to, min(a, e.cap - e.flow))) > )

         {

             e.flow += f;

             edges[G[now][i] ^ ].flow -= f;

             flow += f; a -= f;

             if(!a) break;

         }

     }

     return flow;

 }

 int maxflow()

 {

     int flow = ;

     while(bfs())

     {

         for(rg int i = ; i <= n * m; ++i) cur[i] = ;

         flow += dfs(, INF);

     }

     return flow;

 }

 int main()

 {

     n = read(); m = read();

     for(rg int i = ; i <= n; ++i)

         for(rg int j = ; j < m; ++j)

         {

             int x = read(), from = (i - ) * m + j, to = (i - ) * m + j + ;

             addedge(from, to, x);

         }

     for(rg int i = ; i < n; ++i)

         for(rg int j = ; j <= m; ++j)

         {

             int x = read(), from = (i - ) * m + j, to = i * m + j;

             addedge(from, to, x);

         }

     for(rg int i = ; i < n; ++i)

         for(rg int j = ; j < m; ++j)

         {

             int x = read(), from = (i - ) * m + j, to = i * m + j + ;

             addedge(from, to, x);

         }

     write(maxflow()); enter;

     return ;

 }

然后此题据说也可以用对偶图的知识解决。