Choose the best route Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8224    Accepted Submission(s): 2706 Problem Description One day , Kiki wants to visit one of her friends. As she is liable…

题目链接: Choose the best route Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K (Java/Others) Problem Description   One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her frie…

此题中起点有1000个,边有20000条.用链式前向星建图,再枚举起点用SPFA的话,超时了.(按理说,两千万的复杂度应该没超吧.不过一般说计算机计算速度 1~10 千万次/秒.也许拿最烂的计算机来卡时间) 有一个技巧,加一个超级源点.也就是加一个点,使得该点连通所有的起点,并且边的权值为0.这个技巧应用蛮多的.网络流.最小树形图都有题目这样做. #include<iostream> #include<cstdio> #include<cstring> #include…

http://acm.hdu.edu.cn/showproblem.php?pid=2680 Choose the best route Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9602    Accepted Submission(s): 3111 Problem Description One day , Kiki wants…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2680 Choose the best route Description One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a ma…

Choose the best route Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 86   Accepted Submission(s) : 35 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description One day , Kiki…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2680 Choose the best route Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7201    Accepted Submission(s): 2350 Problem Description One day , Kiki…

Choose the best route Problem Description One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon as possible . Now give you a map of the city’s traffic route, and the stati…

第一问最短路.第二问,先把最短路的图建出来(边(u,v)满足d[s->u]+d[v->t]+d(u,v)==最短路径长度,就在图中,可以从源点和汇点分别跑一次最短路得到每个点到源点和汇点的最短路),然后跑一遍最大流就OK了. --------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #…

题意: 有n个城市,有m条路,给出每条路的出发和结束的城市及长度,求从第一个城市到最后一个城市的最短路.按格式输出. power oj 2443 题解: 标准dijk算法. #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; const int INF=0x3f3f3f3f; const int maxn=505; typ…