poj1815Friendship(最小割求割边）

链接

题意为去掉多少个顶点使图不连通，求顶点连通度问题。拆点，构造图，对于<u,v>可以变成<u2,v1> <v2,u1>容量为无穷，<u1,u2>容量为1.那么求出来的最大流（即最小割）就为所需要删除的顶点个数，需要字典序输出，从小到大枚举顶点，如果不加入当前点，最小割变小了的话 ，说明这个点是肯定要删除的。

 

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <cstring>

  4 #include <queue>

  5 #include <algorithm>

  6 using namespace std;

  7 #define INF 0x3f3f3f

  8 const int N = 415;

  9 #define M 160015

 10 struct node

 11 {

 12     int u,v,next;

 13     int w;

 14 } edge[M<<1];

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

 16 int o[N];

 17 int st,en;

 18 int x[N][N],f[N];

 19 void init()

 20 {

 21     t=0;

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

 23 }

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

 25 {

 26     edge[t].u = u;

 27     edge[t].v = v;

 28     edge[t].w = w;

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

 30     head[u] = t++;

 31     edge[t].u = v;

 32     edge[t].v = u;

 33     edge[t].w = 0;

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

 35     head[v] = t++;

 36 }

 37 int bfs()

 38 {

 39     int i,u;

 40     int w;

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

 42     queue<int>q;

 43     q.push(st);

 44     dis[st] = 0;

 45     while(!q.empty())

 46     {

 47         u = q.front();

 48         q.pop();

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

 50         {

 51             int v = edge[i].v;

 52             w = edge[i].w;

 53             if(dis[v]<0&&w>0)

 54             {

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

 56                 q.push(v);

 57             }

 58         }

 59     }

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

 61     return 0;

 62 }

 63 int dfs(int u,int te)

 64 {

 65     int i;

 66     int s;

 67     if(u==en) return 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             return s;

 77         }

 78     }

 79     dis[u] = -1;

 80     return 0;

 81 }

 82 int dinic()

 83 {

 84     int flow = 0;

 85     int res;

 86     while(bfs())

 87     {

 88         while(res = dfs(st,INF))

 89         flow+=res;

 90     }

 91     return flow;

 92 }

 93 int main()

 94 {

 95     int n,i,j;

 96     while(scanf("%d%d%d",&n,&st,&en)!=EOF)

 97     {

 98         init();

 99         // memset(x,0,sizeof(x));

100         memset(f,0,sizeof(f));

101         st+=n;

102         for(i = 1; i <= n ; i++)

103         {

104             for(j = 1; j <= n; j++)

105             {

106                 scanf("%d",&x[i][j]);

107                 if(i==j)

108                 {

109                     add(i,i+n,1);

110                 }

111                 else if(x[i][j])

112                 {

113                     add(i+n,j,INF);

114                 }

115             }

116         }

117         if(x[st-n][en])

118         {

119             puts("NO ANSWER!");

120             continue;

121         }

122         int ans = dinic();

123         int cnt = 0;

124         for(i = 1; i <= n ; i++)

125         {

126             if(ans==0) break;

127             if(i==st-n||i==en) continue;

128             f[i] = 1;

129             init();

130             for(j = 1; j <= n ; j++)

131             {

132                 if(f[j]) continue;

133                 for(int e = 1; e <= n ; e++)

134                 {

135                     if(f[e]) continue;

136                     if(j==e)

137                         add(j,j+n,1);

138                     else if(x[j][e])

139                     {

140                         add(j+n,e,INF);

141                     }

142                 }

143             }

144             int ts = dinic();

145             if(ts<ans)

146             {

147                 cnt++;

148                 ans = ts;

149             }

150             else f[i] = 0;

151         }

152         cout<<cnt<<endl;

153         for(j = 1; j <= n; j++)

154             if(f[j])

155                 printf("%d ",j);

156         printf("\n");

157     }

158     return 0;

159 }

View Code