[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。