hdu2962 二分 + spfa

来源：互联网发布：think in java 英文版编辑：程序博客网时间：2024/06/05 06:30

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 1800000000using 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:\n" ,cas ++);      if(answ == -1)      printf("cannot reach destination\n");      else      {         printf("maximum height = %d\n" ,answ);         printf("length of shortest route = %d\n" ,ansl);      }   }   return 0;}

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