poj3204Ikki's Story I - Road Reconstruction（最大流求割边）

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最大流=最小割  这题是求割边集 dinic求出残余网络 两边dfs分别以源点d找到可达点 再以汇点进行d找到可达汇点的点  

如果u,v为割边 那么s->u可达 v->t可达 并且为饱和边

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <queue>

  5 #include <algorithm>

  6 using namespace std;

  7 const int INF = 50000;

  8 const int N = 505;

  9 #define M 100005

 10 struct node

 11 {

 12     int u,v,next;

 13     int w;

 14 }edge[M],p[M];

 15 int head[N],t,vis[N],pp[N],dis[N];

 16 int st,en;

 17 bool vis1[N],vis2[N];

 18 void init()

 19 {

 20     t=0;

 21     memset(head,-1,sizeof(head));

 22 }

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

 24 {

 25     edge[t].u = u;

 26     edge[t].v = v;

 27     edge[t].w = w;

 28     edge[t].next = head[u];

 29     head[u] = t++;

 30     edge[t].u = v;

 31     edge[t].v = u;

 32     edge[t].w = 0;

 33     edge[t].next = head[v];

 34     head[v] = t++;

 35 }

 36 int bfs()

 37 {

 38     int i,u;

 39     int w;

 40     memset(dis,0,sizeof(dis));

 41     queue<int>q;

 42     q.push(st);

 43     dis[st]=  1;

 44     while(!q.empty())

 45     {

 46         u = q.front();

 47         q.pop();

 48         for(i = head[u] ; i != -1 ; i = edge[i].next)

 49         {

 50             int v = edge[i].v;

 51             w = edge[i].w;

 52             if(!dis[v]&&w>0)

 53             {

 54                 dis[v] = dis[u]+1;

 55                 q.push(v);

 56             }

 57         }

 58     }

 59     if(dis[en]>0) return 1;

 60     return 0;

 61 }

 62 int dfs(int u,int te)

 63 {

 64     int i;

 65     int s;

 66     if(u==en) return te;

 67     int tmp = te;

 68     for(i = head[u] ; i != -1 ; i = edge[i].next)

 69     {

 70         int v = edge[i].v;

 71         int w = edge[i].w;

 72         if(w>0&&dis[v]==dis[u]+1&&(s=dfs(v,min(te,w))))

 73         {

 74             edge[i].w-=s;

 75             edge[i^1].w+=s;

 76             tmp-=s;

 77         }

 78     }

 79     return te-tmp;

 80 }

 81 int dinic()

 82 {

 83     int flow = 0;

 84     while(bfs())

 85     {

 86         flow+=dfs(st,INF);

 87     }

 88     return flow;

 89 }

 90 void dfs1(int u)

 91 {

 92     int i;

 93     for(i = head[u] ; i != -1 ; i= edge[i].next)

 94     if(!vis1[edge[i].v]&&edge[i].w>0)

 95     {

 96         vis1[edge[i].v] = 1;

 97         dfs1(edge[i].v);

 98     }

 99 }

100 void dfs2(int u)

101 {

102     int i;

103     for(i = head[u] ; i != -1 ; i= edge[i].next)

104     {

105         int v =  edge[i].v;

106         //cout<<v<<" "<<edge[i].w<<endl;

107         if(!vis2[edge[i].v]&&edge[i^1].w>0)

108         {

109         vis2[edge[i].v] = 1;

110         dfs2(edge[i].v);

111         }

112     }

113 }

114 int main()

115 {

116     int n,m,i;

117     while(scanf("%d%d",&n,&m)!=EOF)

118     {

119         init();

120         memset(vis1,0,sizeof(vis1));

121         memset(vis2,0,sizeof(vis2));

122         for(i = 1 ;i <= m; i++)

123         {

124             int u,v,c;

125             scanf("%d%d%d",&u,&v,&c);

126             u++,v++;

127             add(u,v,c);

128         }

129         st = 1,en = n;

130         int kk = dinic();

131         //cout<<kk<<endl;

132         vis1[st] = 1;

133         dfs1(st);

134         vis2[en] = 1;

135         dfs2(en);

136         int ans = 0;

137         dinic();

138         for(i = 0; i < t; i+=2)

139         {

140             int u = edge[i].u;

141             int v = edge[i].v;

142             int w = edge[i].w;

143             //cout<<u<<" "<<v<<" "<<vis1[u]<<" "<<vis2[v]<<endl;

144             if(vis1[u]&&vis2[v]&&w<=0)

145             ans++;

146         }

147         cout<<ans<<endl;

148     }

149     return 0;

150 }

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