网络流 最小生成树的最少割边数--How Many to Be Happy?

时间:2022-02-10 17:48:31

题意:https://blog.csdn.net/Ratina/article/details/95200594

思路:

首先我们知道最小生成树就是按长度枚举边,能连就连。

那么,如果这条边在最小生成树里,那我们只需要看比它短的边是不是已经使当前的u---v连通,如果连通最少需要切掉几条(边权为1跑最小割)。

所以我们对边排序,枚举边 重构图跑Dinic就行了。

  1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0);
  2 #include <cstdio>//sprintf islower isupper
  3 #include <cstdlib>//malloc  exit strcat itoa system("cls")
  4 #include <iostream>//pair
  5 #include <fstream>//freopen("C:\Users\13606\Desktop\Input.txt","r",stdin);
  6 #include <bitset>
  7 //#include <map>
  8 //#include<unordered_map>
  9 #include <vector>
 10 #include <stack>
 11 #include <set>
 12 #include <string.h>//strstr substr
 13 #include <string>
 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand() - 0~9;
 15 #include <cmath>
 16 #include <deque>
 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less
 18 #include <vector>//emplace_back
 19 //#include <math.h>
 20 #include <cassert>
 21 //#include <windows.h>//reverse(a,a len);// ~ ! ~ ! floor
 22 #include <algorithm>//sort   unique : sz=unique(b 1,b n 1)-(b 1); nth_element(first, nth, last, compare)
 23 using namespace std;//next_permutation(a 1,a 1 n);//prev_permutation
 24 //******************
 25 int abss(int a);
 26 int lowbit(int n);
 27 int Del_bit_1(int n);
 28 int maxx(int a,int b);
 29 int minn(int a,int b);
 30 double fabss(double a);
 31 void swapp(int &a,int &b);
 32 clock_t __STRAT,__END;
 33 double __TOTALTIME;
 34 void _MS(){__STRAT=clock();}
 35 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__STRAT)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;}
 36 //***********************
 37 #define rint register int
 38 #define fo(a,b,c) for(rint a=b;a<=c;  a)
 39 #define fr(a,b,c) for(rint a=b;a>=c;--a)
 40 #define mem(a,b) memset(a,b,sizeof(a))
 41 #define pr printf
 42 #define sc scanf
 43 #define ls rt<<1
 44 #define rs rt<<1|1
 45 typedef vector<int> VI;
 46 typedef long long ll;
 47 const double E=2.718281828;
 48 const double PI=acos(-1.0);
 49 //const ll INF=(1LL<<60);
 50 const int inf=(1<<30);
 51 const double ESP=1e-9;
 52 const int mod=(int)1e9 7;
 53 const int N=(int)1e3 10;
 54 const int M=(int)5e3 5;
 55 
 56 class DINIC
 57 {
 58 public:
 59 //    const int MAXN=10004,MAXWAY=100005;
 60     int n,way,max_flow,deep[N];
 61     int tot,head[N],cur[N];
 62     struct EDGE{
 63         int to,next;
 64         int dis;
 65     }edge[M];
 66     void Init(int n_)
 67     {
 68         tot=-1;//因为加反向边要^1,所以要从0开始;
 69         n=n_;
 70         max_flow=0;
 71         for(int i=0;i<=n_;  i)
 72             head[i]=-1;
 73     }
 74     void add(int from,int to,int V)
 75     {
 76         //正向
 77           tot;
 78         edge[tot].to=to;
 79         edge[tot].dis=V;
 80         edge[tot].next=head[from];
 81         head[from]=tot;
 82         //反向
 83         swap(from,to);
 84           tot;
 85         edge[tot].to=to;
 86         edge[tot].dis=V;
 87         edge[tot].next=head[from];
 88         head[from]=tot;
 89     }
 90     queue<int>q;
 91     bool bfs(int s,int t)
 92     {
 93         for(int i=1;i<=n;  i)
 94             deep[i]=inf;
 95         while(!q.empty())q.pop();
 96         for(int i=1;i<=n;  i)cur[i]=head[i];
 97         deep[s]=0;
 98         q.push(s);
 99 
100         while(!q.empty())
101         {
102             int now=q.front();q.pop();
103             for(int i=head[now];i!=-1;i=edge[i].next)
104             {
105                 if(deep[edge[i].to]==inf&&edge[i].dis)
106                 {
107                     deep[edge[i].to]=deep[now] 1;
108                     q.push(edge[i].to);
109                 }
110             }
111         }
112         return deep[t]<inf;
113     }
114     int dfs(int now,int t,int limit)
115     {
116         if(!limit||now==t)return limit;
117         int flow=0,f;
118         for(int i=cur[now];i!=-1;i=edge[i].next)
119         {
120             cur[now]=i;
121             if(deep[edge[i].to]==deep[now] 1&&(f=dfs(edge[i].to,t,min(limit,edge[i].dis))))
122             {
123                 flow =f;
124                 limit-=f;
125                 edge[i].dis-=f;
126                 edge[i^1].dis =f;
127                 if(!limit)break;
128             }
129         }
130         return flow;
131     }
132     void Dinic(int s,int t)
133     {
134         while(bfs(s,t))
135             max_flow =dfs(s,t,inf);
136     }
137 }G;
138 struct EDGE
139 {
140     int u,v;
141     int val;
142     friend bool operator<(EDGE a,EDGE b)
143     {
144         return a.val<b.val;
145     }
146 }edge[M];
147 
148 int main()
149 {
150     int n,m;
151     sc("%d%d",&n,&m);
152     for(int i=1;i<=m;  i)
153         sc("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].val);
154     sort(edge 1,edge 1 m);
155     int ans=0;
156     for(int i=1;i<=m;  i)
157     {
158         G.Init(n);
159         for(int j=1;j<i;  j)
160         {
161             if(edge[j].val<edge[i].val)
162                 G.add(edge[j].u,edge[j].v,1);
163         }
164         G.Dinic(edge[i].u,edge[i].v);
165         ans =G.max_flow;
166     }
167     pr("%dn",ans);
168     return 0;
169 }
170 
171 /**************************************************************************************/
172 
173 int maxx(int a,int b)
174 {
175     return a>b?a:b;
176 }
177 
178 void swapp(int &a,int &b)
179 {
180     a^=b^=a^=b;
181 }
182 
183 int lowbit(int n)
184 {
185     return n&(-n);
186 }
187 
188 int Del_bit_1(int n)
189 {
190     return n&(n-1);
191 }
192 
193 int abss(int a)
194 {
195     return a>0?a:-a;
196 }
197 
198 double fabss(double a)
199 {
200     return a>0?a:-a;
201 }
202 
203 int minn(int a,int b)
204 {
205     return a<b?a:b;
206 }