Drainage Ditches(Dinic最大流)

http://poj.org/problem?id=1273

用Dinic求最大流的模板题，注意会有重边。

邻接矩阵建图

 1 #include<stdio.h>

 2 #include<string.h>

 3 #include<queue>

 4 #include<algorithm>

 5 using namespace std;

 6 const int maxn = 510;

 7 const int INF = 0x3f3f3f3f;

 8 int flow[maxn][maxn];//残量

 9 int m,n;

10 int dis[maxn];

11 int bfs()//按层数“建”图，就是对每层的点用dis标记它到源点的层数

12 {

13     queue<int>que;

14     memset(dis,-1,sizeof(dis));

15     while(!que.empty())

16         que.pop();

17     dis[1] = 0;

18     que.push(1);

19 

20     while(!que.empty())

21     {

22         int u = que.front();

23         que.pop();

24 

25         for(int i = 1; i <= n; i++)

26         {

27             if(flow[u][i] > 0 && dis[i] < 0)

28             {

29                 dis[i] = dis[u]+1;

30                 que.push(i);

31             }

32         }

33     }

34 

35     if(dis[n] > 0)

36         return 1;

37     else return 0;

38 }

39 //dfs寻找路径上的最小流量

40 int dfs(int s, int mf)

41 {

42     int a;

43     if(s == n)

44         return mf;

45     for(int i = 1; i <= n; i++)

46     {

47         if(dis[i] == dis[s] + 1 && flow[s][i] > 0 && (a = dfs(i,min(mf,flow[s][i]))))

48         {

49             flow[s][i] -= a;

50             flow[i][s] += a;

51             return a;

52         }

53     }

54     return 0;

55 }

56 

57 int main()

58 {

59     while(~scanf("%d %d",&m,&n))

60     {

61         int u,v,w;

62         memset(flow,0,sizeof(flow));

63         for(int i = 0; i < m; i++)

64         {

65             scanf("%d %d %d",&u,&v,&w);

66             flow[u][v] += w;

67         }

68         int ans = 0,res;

69         while(bfs())//bfs寻找源点到汇点是否有路

70         {

71             res = dfs(1,INF);

72             if(res > 0)

73                 ans += res;

74         }

75         printf("%d\n",ans);

76     }

77     return 0;

78 

79 }

View Code

邻接表建图

  1 #include<stdio.h>

  2 #include<string.h>

  3 #include<queue>

  4 #include<algorithm>

  5 using namespace std;

  6 const int maxn = 510;

  7 const int INF = 0x3f3f3f3f;

  8 int m,n,cnt;

  9 struct node

 10 {

 11     int u,v,w;

 12     int next;

 13 }edge[maxn];

 14 int p[maxn];

 15 int dis[maxn];

 16 void add(int u, int v, int w)

 17 {

 18 

 19     edge[cnt].u = u;

 20     edge[cnt].v = v;

 21     edge[cnt].w = w;

 22     edge[cnt].next = p[u];

 23     p[u] = cnt++;

 24 

 25     edge[cnt].u = v;

 26     edge[cnt].v = u;

 27     edge[cnt].w = 0;

 28     edge[cnt].next = p[v];

 29     p[v] = cnt++;

 30 }

 31 

 32 int bfs()

 33 {

 34     queue<int> que;

 35     while(!que.empty())

 36         que.pop();

 37     memset(dis,-1,sizeof(dis));

 38     dis[1] = 0;

 39     que.push(1);

 40 

 41     while(!que.empty())

 42     {

 43         int u = que.front();

 44         que.pop();

 45 

 46         for(int i = p[u]; i != -1; i = edge[i].next)

 47         {

 48             if(edge[i].w > 0 && dis[ edge[i].v ] < 0)

 49             {

 50                 dis[ edge[i].v ] = dis[u] + 1;

 51                 que.push(edge[i].v);

 52             }

 53         }

 54     }

 55     if(dis[n] > 0)

 56         return 1;

 57     else return 0;

 58 }

 59 

 60 int dfs(int s, int mf)

 61 {

 62     if(s == n)

 63         return mf;

 64     int a;

 65     int tf = 0;

 66     for(int i = p[s]; i != -1; i = edge[i].next)

 67     {

 68         int v = edge[i].v;

 69         int w = edge[i].w;

 70         if(dis[v] == dis[s] + 1 && w > 0 && (a = dfs(v,min(mf,w))))

 71         {

 72             edge[i].w -= a;

 73             edge[i^1].w += a;

 74             return a;

 75         }

 76     }

 77     if(!tf)//若找不到小流，从这个点到不了汇点，把这个点从分层图删去

 78         dis[s] = -1;

 79     return tf;

 80 }

 81 

 82 int main()

 83 {

 84     while(~scanf("%d %d",&m,&n))

 85     {

 86         int u,v,w;

 87         cnt = 0;

 88         memset(p,-1,sizeof(p));

 89         for(int i = 0; i < m; i++)

 90         {

 91             scanf("%d %d %d",&u,&v,&w);

 92             add(u,v,w);

 93         }

 94         int ans = 0;

 95         int res;

 96         while(bfs())

 97         {

 98             res = dfs(1,INF);

 99             if(res > 0)

100                 ans += res;

101         }

102         printf("%d\n",ans);

103     }

104 

105     return 0;

106 }

View Code