POJ 1273 (基础最大流) Drainage Ditches

虽然算法还没有理解透，但以及迫不及待地想要A道题了。

非常裸的最大流，试试lrj的模板练练手。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 using namespace std;
 
 const int maxn =  + ;
 const int INF = ;
 
 struct Edge
 {
     int from, to, cap, flow;
     Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
 };
 
 int n, m, M;
 vector<Edge> edges;
 vector<int> G[maxn];
 int a[maxn];    //到起点i的可改进量
 int p[maxn];    //最短路数上p的入弧编号
 
 void Init(int n)
 {
     for(int i = ; i < n; ++i) G[i].clear();
     edges.clear();
 }
 
 void AddEdge(int from, int to, int cap)
 {
     edges.push_back(Edge(from, to, cap, ));
     edges.push_back(Edge(to, from, , ));    //反向弧
     m = edges.size();
     G[from].push_back(m-);
     G[to].push_back(m-);
 }
 
 int MaxFlow(int s, int t)
 {
     int flow = ;
     for(;;)
     {
         memset(a, , sizeof(a));
         queue<int> Q;
         Q.push(s);
         a[s] = INF;
         while(!Q.empty())
         {
             int x = Q.front(); Q.pop();
             for(int i = ; i < G[x].size(); ++i)
             {
                 Edge& e = edges[G[x][i]];
                 if(!a[e.to] && e.cap > e.flow)
                 {
                     p[e.to] = G[x][i];
                     a[e.to] = min(a[x], e.cap-e.flow);
                     Q.push(e.to);
                 }
             }
             if(a[t]) break;
         }
         if(!a[t]) break;
         for(int u = t; u != s; u = edges[p[u]].from)
         {
             edges[p[u]].flow += a[t];
             edges[p[u]^].flow -= a[t];
         }
         flow += a[t];
     }
     return flow;
 }
 
 int main()
 {
     freopen("in.txt", "r", stdin);
 
     while(scanf("%d%d", &M, &n) == )
     {
         Init(n);
         int u, v, c;
         for(int i = ; i < M; ++i)
         {
             scanf("%d%d%d", &u, &v, &c);
             AddEdge(u-, v-, c);
         }
         printf("%d\n", MaxFlow(, n-));
     }
 
     return ;
 }

代码君