hdu1839 二分最短路

题意:

      给你n个城市,m条双向边,每条边有自己的长度和最大运输量,让你找到一条时间小于等于T的运输能力最大的那条路...

思路:

      刚开始以为是费用流呢,后来发现根本不是,因为根本不是在求最优和最优下的其他最优,其实这个题目可以二分最大运输量,每次都根据二分结果建图,比如对于当前的mid,枚举每一条边,如果当前边的流量大于等于mid那么就把当前边连接到图里,枚举玩之后跑最短路,看如果1到n的距离小于等于T则满足,如果满足 low = mod + 1,ans = mid,如果不满足则

up = mid - 1.......二分枚举,建图,找到答案.

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

#define N_node 10000 + 500
#define N_edge 100000 + 10000
#define inf 2000000000

using namespace std;

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

typedef struct
{
   int a ,b ,c ,d;
}EDGE;

STAR E[N_edge];
EDGE edge[N_edge];
int list[N_node] ,tot;
int s_x[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;
}

void SPFA(int s ,int n)
{
   for(int i = 0 ;i <= n ;i ++)
   s_x[i] = inf;
   s_x[s] = 0;
   int mark[N_node] = {0};
   mark[s] = 1;
   queue<int>q;
   q.push(s);
   while(!q.empty())
   {
      int tou ,xin;
      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)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
}

void Buid(int m ,int mid)
{
   memset(list ,0 ,sizeof(list));
   tot = 1;
   for(int i = 1 ;i <= m ;i ++)
   if(edge[i].c >= mid)
   {
      add(edge[i].a ,edge[i].b ,edge[i].d);
      add(edge[i].b ,edge[i].a ,edge[i].d);
   }
}

bool OK(int T ,int n)
{
   SPFA(1 ,n);
   return s_x[n] <= T;
}

int main ()
{
   int t ,n ,m ,T;
   int i ,a ,b ,c ,d;
   int max;
   scanf("%d" ,&t);
   while(t--)
   {
      scanf("%d %d %d" ,&n ,&m ,&T);
      max = -1;
      for(i = 1 ;i <= m ;i ++)
      {
         scanf("%d %d %d %d" ,&a ,&b ,&c ,&d);
         if(max < c) max = c;
         edge[i].a = a;
         edge[i].b = b;
         edge[i].c = c;
         edge[i].d = d;
      }
      int low ,mid ,up;
      low = 0;
      up = max;
      int ans = 0;
      while(low <= up)
      {
         mid = (low + up) >> 1;
         Buid(m ,mid);
         if(OK(T ,n))
         {
            low = mid + 1;
            ans = mid;
         }
         else
         up = mid - 1;
      }
      printf("%d\n" ,ans);
   }
   return 0;
}