hdu2435最大流最小割

2435  There is a war


题意:
      给你一个有向图,其中可以有一条边是无敌的,这条边可以是图中的边,也可以是自己任意加上去的图中没有的边,这条无敌的边不可以摧毁,让1和n无法连通的最大摧毁费用,就是1到n的最小割中的最大的那个,这个题卡了好几天,一开始是各种方法各种wa,后来无意中发现自己犯了个sb错误,结果改正后以前的各种方法各种ac,比赛要是碰到这样的事估计就跪了...


思路:
     首先能确定的就是题目要求咱们就最小割(最大流 = 最小割),但关键是有那么一条无坚不摧的nb道路,所以一开始的想法肯定是暴力枚举N*N的边,直接TLE出翔了,那么就优化,记得以前的一道题目 给你一个图求删除其中一条边最短路中最大的那个,答案是只枚举最短路上的边就可以了, 这个题目也是类似,只要枚举最小割后两个集合的点组成的边就行了,因为假如点a和点b是一个集合的,那么把边ab变成无敌的没有意思,最小割的值不会改变,,那么怎么吧分成两个集合呢,两种方法,一个是深搜,这个方法容易理解,先跑一遍最大流,然后从点1开始深搜,如果当前点走过或者没有流量了(跑完一遍最大流后的流量),直接continue,这样被mark的点就是左集合的点,剩下的就是右集合的点,还有一种方法就是直接看DINIC后的deep数组,如果不等于-1就是左集合的,否则的就是右集合的,这个我结论是网上的,我还不知道为什么,分成两个集合后就可以枚举两个集合的点建枚举的边了,这块也有两个方法,一个就是之前不是跑一边最大流了吗,加上当前枚举边,直接在残余网络上跑,取得最大的max最后输出一开始那个最大流maxflow+max,(记得每次跑之前都还原成第一次跑完的残余网路),第二种方法就是直接重新建边,一开始的时候吧m条边记录下来,每次枚举都重新建图,然后加上枚举的边跑,最后输出的是最大流中最大的那个maxflow.下面是三种方法的代码..




深搜找源集和汇集,在残余网络上跑 15ms AC

#include<stdio.h>

#include<string.h>

#include<queue>

#define N_node 120
#define N_edge 22000

#define inf 1000000000

using namespace std;

typedef struct
{

   int to ,next ,cost;

}STAR;

typedef struct

{

   int x ,t;

}DEP;



STAR E[N_edge] ,E_[N_edge];

DEP xin ,tou;

int list[N_node] ,list1[N_node] ,tot;

int list2[N_node];

int deep[N_node];

int mks[N_node] ,mks_;

int mkh[N_node] ,mkh_;

int mark[N_node];



void add(int a ,int b ,int c)

{

   E[++tot].to = b;

   E[tot].cost = c;

   E[tot].next = list[a];

   list[a] = tot;

   

   E[++tot].to = a;

   E[tot].cost = 0;

   E[tot].next = list[b];

   list[b] = tot;

}



int minn(int a ,int b)

{

   return a < b ? a : b;

}



bool BFS_DEEP(int s ,int t ,int n)

{

    memset(deep ,255 ,sizeof(deep));

    deep[s] = 0;

    xin.x = s;

    xin.t = 0;

    queue<DEP>q;

    q.push(xin);

    while(!q.empty())

    {

      tou = q.front();

      q.pop();

      for(int k = list[tou.x] ;k ;k = E[k].next)

      {

         xin.x = E[k].to;

         xin.t = tou.t + 1;

         if(deep[xin.x] != -1 || !E[k].cost)

         continue;

         deep[xin.x] = xin.t;

         q.push(xin);

      }

   }

   for(int i = 0 ;i <= n ;i ++)

   list1[i] = list[i];

   return deep[t] != -1;

}



int DFS_MAX_FLOW(int s ,int t ,int flow)

{

   if(s == t) return flow;

   int nowflow = 0;

   for(int k = list1[s] ;k ;k = E[k].next)

   {

      list1[s] = k;

      int to = E[k].to;

      int c = E[k].cost;

      if(deep[to] != deep[s] + 1||!E[k].cost)

      continue;

      int tmp = DFS_MAX_FLOW(to ,t ,minn(c ,flow - nowflow));

      nowflow += tmp;

      E[k].cost -= tmp;

      E[k^1].cost += tmp;

      if(nowflow == flow)

      break;

   }

   if(!nowflow)

   deep[s] = 0;

   return nowflow;

}



int DINIC(int s ,int t ,int n)

{

   int ans = 0;

   while(BFS_DEEP(s ,t ,n))

   {

      ans += DFS_MAX_FLOW(s ,t ,inf);

   }

   return ans;

}



void DFS(int s)

{

   for(int k = list[s] ;k ;k = E[k].next)

   {

      int to = E[k].to;

      if(mark[to] || !E[k].cost)

      continue;

      mark[to] = 1;

      DFS(to);

   }

   return ;

}



int main ()

{

   int n ,m ,i ,j ,t;

   int a ,b ,c;

   scanf("%d" ,&t);

   while(t--)

   {

      memset(list ,0 ,sizeof(list));

      tot = 1;

      scanf("%d %d" ,&n ,&m);

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

      {

         scanf("%d %d %d" ,&a ,&b ,&c);

         add(a ,b ,c);

      }

      int ans = DINIC(1 ,n ,n);

      mks_ = mkh_ = 0;

      memset(mark ,0 ,sizeof(mark));

      mark[1] = 1;

      DFS(1);

      for(i = 2 ;i < n ;i ++)

      if(mark[i]) mks[++mks_] = i;

      else mkh[++mkh_] = i;

      

      for(i = 1 ;i <= tot ;i ++)

      E_[i] = E[i];

      int mktot = tot;

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

      list2[i] = list[i];

      

      int max = 0;

      for(i = 1 ;i <= mks_ ;i ++)

      for(j = 1 ;j <= mkh_ ;j ++)

      {

         a = mks[i] ,b = mkh[j];

         for(int k = 1 ;k <= mktot ;k ++)

         E[k] = E_[k];

         memset(list ,0 ,sizeof(list));

         for(int k = 1 ;k <= n ;k ++)

         list[k] = list2[k];

         tot = mktot;

         add(a ,b ,inf);

         int tmp = DINIC(1 ,n ,n);

         if(max < tmp) max = tmp;

      }

      printf("%d\n" ,ans + max);

   }

   return 0;

}

         

         

      

      

根据deep数组找源集和汇集,在残余网络上跑 31ms AC


#include<stdio.h>

#include<string.h>

#include<queue>



#define N_node 120

#define N_edge 22000

#define inf 1000000000



using namespace std;



typedef struct

{

   int to ,next ,cost;

}STAR;

typedef struct

{

   int x ,t;

}DEP;



STAR E[N_edge] ,E_[N_edge];

DEP xin ,tou;

int list[N_node] ,list1[N_node] ,tot;

int list2[N_node];

int deep[N_node];

int mks[N_node] ,mks_;

int mkh[N_node] ,mkh_;



void add(int a ,int b ,int c)

{

   E[++tot].to = b;

   E[tot].cost = c;

   E[tot].next = list[a];

   list[a] = tot;

   

   E[++tot].to = a;

   E[tot].cost = 0;

   E[tot].next = list[b];

   list[b] = tot;

}



int minn(int a ,int b)

{

   return a < b ? a : b;

}



bool BFS_DEEP(int s ,int t ,int n)

{

    memset(deep ,255 ,sizeof(deep));

    deep[s] = 0;

    xin.x = s;

    xin.t = 0;

    queue<DEP>q;

    q.push(xin);

    while(!q.empty())

    {

      tou = q.front();

      q.pop();

      for(int k = list[tou.x] ;k ;k = E[k].next)

      {

         xin.x = E[k].to;

         xin.t = tou.t + 1;

         if(deep[xin.x] != -1 || !E[k].cost)

         continue;

         deep[xin.x] = xin.t;

         q.push(xin);

      }

   }

   for(int i = 0 ;i <= n ;i ++)

   list1[i] = list[i];

   return deep[t] != -1;

}



int DFS_MAX_FLOW(int s ,int t ,int flow)

{

   if(s == t) return flow;

   int nowflow = 0;

   for(int k = list1[s] ;k ;k = E[k].next)

   {

      list1[s] = k;

      int to = E[k].to;

      int c = E[k].cost;

      if(deep[to] != deep[s] + 1||!E[k].cost)

      continue;

      int tmp = DFS_MAX_FLOW(to ,t ,minn(c ,flow - nowflow));

      nowflow += tmp;

      E[k].cost -= tmp;

      E[k^1].cost += tmp;

      if(nowflow == flow)

      break;

   }

   if(!nowflow)

   deep[s] = 0;

   return nowflow;

}



int DINIC(int s ,int t ,int n)

{

   int ans = 0;

   while(BFS_DEEP(s ,t ,n))

   {

      ans += DFS_MAX_FLOW(s ,t ,inf);

   }

   return ans;

}



int main ()

{

   int n ,m ,i ,j ,t;

   int a ,b ,c;

   scanf("%d" ,&t);

   while(t--)

   {

      memset(list ,0 ,sizeof(list));

      tot = 1;

      scanf("%d %d" ,&n ,&m);

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

      {

         scanf("%d %d %d" ,&a ,&b ,&c);

         add(a ,b ,c);

      }

      int ans = DINIC(1 ,n ,n);

      mks_ = mkh_ = 0;

      for(i = 2 ;i < n ;i ++)

      if(deep[i] != -1) mks[++mks_] = i;

      else mkh[++mkh_] = i;

      

      for(i = 1 ;i <= tot ;i ++)

      E_[i] = E[i];

      int mktot = tot;

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

      list2[i] = list[i];

      

      int max = 0;

      for(i = 1 ;i <= mks_ ;i ++)

      for(j = 1 ;j <= mkh_ ;j ++)

      {

         a = mks[i] ,b = mkh[j];

         for(int k = 1 ;k <= mktot ;k ++)

         E[k] = E_[k];

         memset(list ,0 ,sizeof(list));

         for(int k = 1 ;k <= n ;k ++)

         list[k] = list2[k];

         tot = mktot;

         add(a ,b ,inf);

         int tmp = DINIC(1 ,n ,n);

         if(max < tmp) max = tmp;

      }

      printf("%d\n" ,ans + max);

   }

   return 0;

}

         
直接重新建图,深搜找源集和汇集(容易理解) 15msAC

#include<stdio.h>

#include<string.h>

#include<queue>



#define N_node 120

#define N_edge 22000

#define inf 1000000000



using namespace std;



typedef struct

{

   int to ,next ,cost;

}STAR;

typedef struct

{

   int x ,t;

}DEP;



typedef struct

{

   int a ,b ,c;

}EDGE;



STAR E[N_edge];

EDGE edge[N_edge];

DEP xin ,tou;

int list[N_node] ,list1[N_node] ,tot;

int deep[N_node];

int mks[N_node] ,mks_;

int mkh[N_node] ,mkh_;

int mark[N_node];



void add(int a ,int b ,int c)

{

   E[++tot].to = b;

   E[tot].cost = c;

   E[tot].next = list[a];

   list[a] = tot;

   

   E[++tot].to = a;

   E[tot].cost = 0;

   E[tot].next = list[b];

   list[b] = tot;

}



int minn(int a ,int b)

{

   return a < b ? a : b;

}



bool BFS_DEEP(int s ,int t ,int n)

{

    memset(deep ,255 ,sizeof(deep));

    deep[s] = 0;

    xin.x = s;

    xin.t = 0;

    queue<DEP>q;

    q.push(xin);

    while(!q.empty())

    {

      tou = q.front();

      q.pop();

      for(int k = list[tou.x] ;k ;k = E[k].next)

      {

         xin.x = E[k].to;

         xin.t = tou.t + 1;

         if(deep[xin.x] != -1 || !E[k].cost)

         continue;

         deep[xin.x] = xin.t;

         q.push(xin);

      }

   }

   for(int i = 0 ;i <= n ;i ++)

   list1[i] = list[i];

   return deep[t] != -1;

}



int DFS_MAX_FLOW(int s ,int t ,int flow)

{

   if(s == t) return flow;

   int nowflow = 0;

   for(int k = list1[s] ;k ;k = E[k].next)

   {

      list1[s] = k;

      int to = E[k].to;

      int c = E[k].cost;

      if(deep[to] != deep[s] + 1||!E[k].cost)

      continue;

      int tmp = DFS_MAX_FLOW(to ,t ,minn(c ,flow - nowflow));

      nowflow += tmp;

      E[k].cost -= tmp;

      E[k^1].cost += tmp;

      if(nowflow == flow)

      break;

   }

   if(!nowflow)

   deep[s] = 0;

   return nowflow;

}



int DINIC(int s ,int t ,int n)

{

   int ans = 0;

   while(BFS_DEEP(s ,t ,n))

   {

      ans += DFS_MAX_FLOW(s ,t ,inf);

   }

   return ans;

}



void DFS(int s)

{

   for(int k = list[s] ;k ;k = E[k].next)

   {

      int to = E[k].to;

      if(mark[to] || !E[k].cost)

      continue;

      mark[to] = 1;

      DFS(to);

   }

   return ;

}



int main ()

{

   int n ,m ,i ,j ,t;

   int a ,b ,c;

   scanf("%d" ,&t);

   while(t--)

   {

      memset(list ,0 ,sizeof(list));

      tot = 1;

      scanf("%d %d" ,&n ,&m);

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

      {

         scanf("%d %d %d" ,&a ,&b ,&c);

         add(a ,b ,c);

         edge[i].a = a ,edge[i].b = b ,edge[i].c = c;

      }

      int ans = DINIC(1 ,n ,n);

      mks_ = mkh_ = 0;

      memset(mark ,0 ,sizeof(mark));

      mark[1] = 1;

      DFS(1);

      for(i = 2 ;i < n ;i ++)

      if(mark[i]) mks[++mks_] = i;

      else mkh[++mkh_] = i;

      for(i = 1 ;i <= mks_ ;i ++)

      for(j = 1 ;j <= mkh_ ;j ++)

      {

         a = mks[i] ,b = mkh[j];

         memset(list ,0 ,sizeof(list));

         tot = 1;

         for(int k = 1 ;k <= m ;k ++)

         add(edge[k].a ,edge[k].b ,edge[k].c);

         add(a ,b ,inf);

         int tmp = DINIC(1 ,n ,n);

         if(ans < tmp) ans = tmp;

      }

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

   }

   return 0;

}