Drainage Ditches USACO 4.2 (最大流，BFS)

算法导论求最大流这一节还没看过，下午看了下，参照了大牛的代码模仿写了Ford-Fulkerson+BFS过了...

  1 /*
  2 
  3 ID: hubiao cave
  4 
  5 PROG: ditch
  6 
  7 LANG: C++
  8 
  9 */
 10 
 11 
 12 
 13 
 14 #include<iostream>
 15 
 16 #include<fstream>
 17 
 18 #include<cstring>
 19 #include<queue>
 20 #include<deque>
 21 
 22 using namespace std;
 23 
 24 
 25 
 26 int E[201][201];
 27 int m,n;
 28 bool visited[201];
 29 int prenumber;
 30 queue<pair<int,int> > intqueue;
 31 deque<pair<int,int> >intde;
 32 
 33 void BFS();
 34 void PreBFS();
 35 int main()
 36 
 37 {
 38 
 39 
 40     ifstream fin("ditch.in");
 41 
 42     ofstream fout("ditch.out");
 43 
 44     fin>>n>>m;
 45     for(int i=1;i<=n;i++)
 46     {
 47         int x,y,z;
 48         fin>>x>>y>>z;
 49         E[x][y]+=z;
 50     }
 51     int ans=0;
 52     for(;;)
 53     {
 54         PreBFS();
 55         BFS();
 56         if(prenumber==0)
 57             break;
 58         int min=2111111111;
 59         
 60         int pre=intde.back().second;
 61         do
 62         {
 63             while(intde.back().second!=pre)
 64                 intde.pop_back();
 65             pair<int,int>cur=intde.back();
 66             intde.pop_back();
 67             if(E[cur.first][cur.second]<min&&cur.first!=0)
 68                 min=E[cur.first][cur.second];
 69             if(cur.first!=0)
 70             intde.push_front(cur);
 71             pre=cur.first;
 72         }while(pre!=0);
 73         
 74         ans+=min;
 75         while(!intde.empty())
 76         {
 77             pair<int,int> pii=intde.front();
 78             intde.pop_front();
 79             E[pii.first][pii.second]-=min;
 80             E[pii.second][pii.first]+=min;
 81         }
 82     }
 83 
 84     fout<<ans<<endl;
 85     return 0;
 86 
 87 
 88 }
 89 
 90 void BFS()
 91 {
 92     intqueue.push(make_pair(0,1));
 93     while(!intqueue.empty())
 94     {
 95         pair<int,int> cur=intqueue.front();    
 96         if(visited[cur.second])
 97         {
 98             intqueue.pop();
 99             continue;
100         }
101         if(cur.second==m)
102         {
103             prenumber=cur.first;
104             intde.push_back(cur);
105             return;
106         }
107         visited[cur.second]=true;
108         intde.push_back(cur);
109         intqueue.pop();
110         for(int i=1;i<=m;i++)
111         {
112             if(E[cur.second][i]&&!visited[i])
113             {
114                 intqueue.push(make_pair(cur.second,i));
115             }
116         }
117     }
118 }
119 
120 void PreBFS()
121 {
122     while(!intqueue.empty())
123     {
124         intqueue.pop();
125     }
126     intde.clear();
127     prenumber=0;
128     memset(visited,0,sizeof(visited));
129 
130 }

 

posted @ 2013-11-11 22:50  cavehubiao  阅读(237)  评论(0)  编辑  收藏  举报