hdu4411 经典费用里建图

题意:

      给以一个无向图,0 - n,警察在0,他们有k个警队,要派一些警队去1--n个城市抓小偷,

问所有吧所有小偷全抓到然后在返回0的最小路径和是多少,当地i个城市被攻击的时候他会通知i-1城市,也就是说要么同时消灭他俩,要么消灭i-1在消灭i;

思路:

      经典的费用流建图 ,先来一遍floyd处理下图 ,首先虚拟出来源点s ,终点t ,超级源点ss,

建图:

       from       to         cost           flow

        ss         s          0              k      //限制最多k个警队

        s          i           map[0][i]   1     // 出发去

        i         i + n       -INF           1     //-INF 是为了保证每一个都得跑

        i + n     t          map[0][i]    1     // 回家

        i + n     j          map[i][j]     1     //i < j ,收拾完i再去城镇j收拾j

          s        t               0            1     //用这个边处理没有出动的警队

#include<stdio.h>
#include<string.h>
#include<queue>

#define N_node 210
#define N_edge 20000
#define INF 10000000

using namespace std;

typedef struct
{
   int from ,to ,cost ,flow ,next;
}STAR;

STAR E[N_edge];
int list[N_node] ,tot;
int mer[N_edge];
int s_x[N_node];
int map[N_node][N_node];

void add(int a, int b ,int c ,int d)
{
   E[++tot].from = a;
   E[tot].to = b;
   E[tot].cost = c;
   E[tot].flow = d;
   E[tot].next = list[a];
   list[a] = tot;

   E[++tot].from = b;
   E[tot].to = a;
   E[tot].cost = -c;
   E[tot].flow = 0;
   E[tot].next = list[b];
   list[b] = tot;
}

bool SPFA(int s, int t ,int n)
{
   int mark[N_node] = {0};
   for(int i = 0 ;i <= n ;i ++)
   s_x[i] = INF;
   mark[s] = 1 ,s_x[s] = 0;
   queue<int>q;
   q.push(s);
   memset(mer ,255 ,sizeof(mer));
   while(!q.empty())
   {
      int xin ,tou;
      tou = q.front();
      q.pop();
      mark[tou] = 0;
      for(int k = list[tou] ;k ;k = E[k].next)
      {
         xin = E[k].to;
         if(s_x[xin] > s_x[tou] + E[k].cost && E[k].flow)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            mer[xin] = k;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
   return mer[t] != -1;
}

int MCMF_SPFA(int s ,int t ,int n)
{
   int maxflow = 0,minflow;
   int mincost = 0;
   while(SPFA(s ,t ,n))
   {
      minflow = INF;
      for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
      {
         if(minflow > E[i].flow)
         minflow = E[i].flow;
      }
      for(int i = mer[t] ;i + 1 ;i = mer[E[i].from])
      {
         E[i].flow -= minflow;
         E[i^1].flow += minflow;
         mincost += minflow * E[i].cost;
      }
      maxflow += minflow;
   }
   return mincost;
}

int minn(int x ,int y)
{
   return x < y ? x : y;
}

void floyd(int n)
{
   int i ,j ,k;
   for(k = 0 ;k <= n ;k ++)
   for(i = 0 ;i <= n ;i ++)
   for(j = 0 ;j <= n ;j ++)
   map[i][j] = minn(map[i][j] ,map[i][k] + map[k][j]);
} 

int main ()
{
   int n ,m, i ,j ,K;
   int a ,b ,c;
   int ss ,s ,t;
   while(~scanf("%d %d %d" ,&n ,&m ,&K) && n + m + K)
   {
      ss = 0 ,s = n + n + 1 ,t = n + n + 1 + 1;
      for(i = 0 ;i <= n ;i ++)
      for(j = 0 ;j <= n ;j ++)
      {
         if(i == j) map[i][j] = 0;
         else map[i][j] = INF;
      }
      for(i = 1 ;i <= m ;i ++)
      {
         scanf("%d %d %d" ,&a ,&b ,&c);
         map[a][b] = map[b][a] = minn(map[a][b] ,c);
      }
      floyd(n);
      memset(list ,0 ,sizeof(list));
      tot = 1;
      for(i = 1 ;i <= n ;i ++)
      {
         add(s ,i ,map[0][i] ,1);
         add(i ,i + n ,-INF ,1);
         for(j = i + 1 ;j <= n ;j ++)
         add(i + n ,j ,map[i][j] ,1);
         add(i + n ,t ,map[i][0] ,1);
      }
      add(ss ,s ,0 ,K);
      add(s ,t ,0 ,K);
      int cost = MCMF_SPFA(ss ,t ,n + n + 1 + 1);
      printf("%d\n" ,cost + n * INF);
   }
   return 0;
}