hdu2962 二分 + spfa

题意:
      给你一个无向图,每条路径上都有自己的长度和最大承受高度,给你起点终点还有车的最大承装高度,问你高度最大的前提下路径最短是多少,求高度和路径.


思路:

     这种类型题目太多了,就是给你一些限制,然后让你在这个限制的前提下找到另一个最优,涉及到线性单调的一般都可以直接二分枚举a掉,这个也不例外,二分高度,重新建图,或者在跑最短路的时候限制,都可以,具体看代码就懂了..

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

#define N_node 1000 + 100
#define N_edge 1000000 + 500
#define INF 1800000000

using namespace std;

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

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

EDGE edge[N_edge];
STAR E[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; 
   int mark[N_node] = {0};
   s_x[s] = 0;
   mark[s] = 1;
   queue<int>q;
   q.push(s);
   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)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
   return ;                                    
}

bool ok(int mid ,int s ,int t ,int n ,int m)
{
   memset(list ,0 ,sizeof(list));
   tot = 1;
   for(int i = 1 ;i <= m ;i ++)
   if(edge[i].c == -1 || edge[i].c >= mid)
   {
      add(edge[i].a ,edge[i].b ,edge[i].d);
      add(edge[i].b ,edge[i].a ,edge[i].d);
   }
   spfa(s ,n);
   return s_x[t] != INF;
}
            
int main ()
{
   int n ,m ,i ,j;
   int s ,t ,w ,cas = 1;
   while(~scanf("%d %d" ,&n ,&m) && n + m)
   {
      if(cas != 1) puts("");
      for(i = 1 ;i <= m ;i ++)
      scanf("%d %d %d %d" ,&edge[i].a ,&edge[i].b ,&edge[i].c ,&edge[i].d);
      scanf("%d %d %d" ,&s ,&t ,&w);
      int up ,low ,mid;
      low = 0,up = w;
      int answ = -1 ,ansl;
      while(low <= up)
      {
         mid = (low + up) >> 1;
         if(ok(mid ,s ,t ,n ,m))
         {
            low = mid + 1;
            answ = mid;
            ansl = s_x[t];
         }
         else
         up = mid - 1;
      }
      printf("Case %d:
" ,cas ++);
      if(answ == -1)
      printf("cannot reach destination
");
      else
      {
         printf("maximum height = %d
" ,answ);
         printf("length of shortest route = %d
" ,ansl);
      }
   }
   return 0;
}