POJ 3228 二分最大流

题意：

      给你N个位置，每个位置都有金矿数量和仓库数量，然后位置和位置之间的距离给了出来，最后问你吧所有的金矿都放到库里面走的路径 最长的最短 是多少？

思路：

     比较简单的一个题，直接二分答案，然后用最大流是否满流来判断二分方向，还有就是建图的时候不用拆点什么的，一开始建图想麻烦了，都快敲完了才反应过来，具体看代码。

#include<stdio.h>

#include<string.h>

#include<queue>

#define N_node 200  + 10

#define N_edge 90000

#define INF 1000000000

using namespace std;

typedef struct

{

   int to ,cost ,next;

}STAR;

typedef struct

{

   int x ,t;

}DEP;

typedef struct

{

   int a ,b ,c;

}EDGE;

EDGE edge[22000];

STAR E[N_edge];

DEP xin ,tou;

int list[N_node] ,list2[N_node] ,tot;

int deep[N_node];

int aaa[220] ,bbb[220];

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 = c;

     E[tot].next = list[b];

     list[b] = tot;

}

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

{

   memset(deep ,255 ,sizeof(deep));

   xin.x = s ,xin.t = 0;

   deep[xin.x] = xin.t;

   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 ++)

   list2[i] = list[i];

   return deep[t] != -1;

}

int minn(int x ,int y)

{

   return x < y ? x : y;

}

int DFS_Flow(int s ,int t ,int flow)

{

   if(s == t) return flow;

   int nowflow = 0;

   for(int k = list2[s] ;k ;k = E[k].next)

   {

       list2[s] = k;

       int to = E[k].to;

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

       continue;

       int tmp = DFS_Flow(to ,t ,minn(E[k].cost ,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_Flow(s ,t ,INF);

   }

   return Ans;

}

bool ok(int n ,int m ,int sum ,int mid)

{

     memset(list ,0 ,sizeof(list)) ,tot = 1;

     for(int i = 1 ;i <= n ;i ++)

     {

        add(0 ,i ,aaa[i]);

        add(i ,n + 1 ,bbb[i]);

     }

     for(int i = 1 ;i <= m ;i ++)

     if(edge[i].c <= mid)

     {

        add(edge[i].a ,edge[i].b ,INF);

     }

     int flow = DINIC(0 ,n + 1 ,n + 1);

     return  flow == sum;

}

    

int main ()

{

    int n ,m ,i ,sum;

    while(~scanf("%d" ,&n) && n)

    {

        for(sum = 0 ,i = 1 ;i <= n ;i ++)

        {

           scanf("%d" ,&aaa[i]);

           sum += aaa[i];

        }

        for(i = 1 ;i <= n ;i ++)

        scanf("%d" ,&bbb[i]);

        scanf("%d" ,&m);

        for(i = 1 ;i <= m ;i ++)

        scanf("%d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c);

        int low ,up ,mid ,Ans = -1;

        low = 0 ,up = 11000;

        while(low <= up)

        {

           mid = (low + up) >> 1;

           if(ok(n ,m ,sum ,mid))

           {

               Ans = mid;

               up = mid - 1;

           }

           else low = mid + 1;

        }

        Ans == -1 ? puts("No Solution"):printf("%d\n" ,Ans);

    }

    return 0;

}