C - Design the city Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Description Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere.…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

Design the city Time Limit: 1 Second      Memory Limit: 32768 KB Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason…

Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he wan…

Opening Portals Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 196E64-bit integer IO format: %I64d      Java class name: (Any)     Pavel plays a famous computer game. A player is responsible for a wh…

Checkposts Time Limit: 2000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 427C64-bit integer IO format: %I64d      Java class name: (Any)     Your city has n junctions. There are m one-way roads between the junction…

题目大概说给一棵树,每次询问三个点,问要把三个点连在一起的最少边权和是多少. 分几种情况..三个点LCA都相同,三个点有两对的LCA是某一点,三个点有两对的LCA各不相同...%……￥…… 画画图可以发现..虽然好像不太严谨..连接(a,b,c)三个点的最短边权和=(dist(a,b)+dist(b,c)+dist(a,c))/2,而dist(u,v)的计算可以预处理出各点到根的距离weight,dist(u,v)=weight(u)+weight(v)-2*weight(lca(u,v))..…

题意:给定n个点,下面n-1行 u , v ,dis 表示一条无向边和边权值,这里给了一颗无向树 下面m表示m个询问,问 u v n 三点最短距离 典型的LCA转RMQ #include<stdio.h> #include<string.h> #include<math.h> #define N 100000 #define INF 1<<29 #define Logo 17 using namespace std; inline int Min(int a…

题目要对每次询问将一个树形图的三个点连接,输出最短距离. 利用tarjan离线算法,算出每次询问的任意两个点的最短公共祖先,并在dfs过程中求出离根的距离.把每次询问的三个点两两求出最短距离,这样最终结果就是3个值一半. 其实开始我用的一种很挫的方法才AC的,具体思路就不说了,感觉很麻烦又不好写的样子.怎么没想到上面的简便方法呢. 初始代码: #include <iostream> #include <sstream> #include <cstdio> #includ…