POJ 1986 Distance Queries LCA两点距离树
标题来源:POJ 1986 Distance Queries
意甲冠军:给你一棵树 q第二次查询 每次你问两个点之间的距离
思路:对于2点 u v dis(u,v) = dis(root,u) + dis(root,v) - 2*dis(roor,LCA(u,v)) 求近期公共祖先和dis数组
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 40010;
int first[maxn], head[maxn], cnt, sum;
struct edge
{
int u, v, w, next;
}e[maxn*2], qe[maxn], Q[maxn];
int ans[maxn];
int f[maxn], vis[maxn];
int d[maxn];
void AddEdge(int u, int v, int w)
{
e[cnt].u = u;
e[cnt].v = v;
e[cnt].w = w;
e[cnt].next = first[u];
first[u] = cnt++;
e[cnt].u = v;
e[cnt].v = u;
e[cnt].w = w;
e[cnt].next = first[v];
first[v] = cnt++;
} void AddEdge2(int u, int v, int w)
{
qe[sum].u = u;
qe[sum].v = v;
qe[sum].w = w;
qe[sum].next = head[u];
head[u] = sum++;
qe[sum].u = v;
qe[sum].v = u;
qe[sum].w = w;
qe[sum].next = head[v];
head[v] = sum++;
} int find(int x)
{
if(f[x] != x)
return f[x] = find(f[x]);
return f[x];
}
void LCA(int u, int k)
{
f[u] = u;
d[u] = k;
vis[u] = true;
for(int i = first[u]; i != -1; i = e[i].next)
{
int v = e[i].v;
if(vis[v])
continue;
LCA(v, k + e[i].w);
f[v] = u;
}
for(int i = head[u]; i != -1; i = qe[i].next)
{
int v = qe[i].v;
if(vis[v])
{
ans[qe[i].w] = find(v);
}
}
}
int main()
{
int n, m;
memset(first, -1, sizeof(first));
memset(head, -1, sizeof(head));
cnt = 0;
sum = 0;
scanf("%d %d", &n, &m);
for(int i = 0; i < m; i++)
{
int u, v, w;
char s[10];
scanf("%d %d %d %s", &u, &v, &w, s);
AddEdge(u, v, w);
}
int q;
scanf("%d", &q);
for(int i = 0; i < q; i++)
{
int u, v;
scanf("%d %d", &u, &v);
Q[i].u = u, Q[i].v = v;
AddEdge2(u, v, i);
AddEdge2(v, u, i); }
memset(vis, 0, sizeof(vis));
d[1] = 0;
LCA(1, 0);
for(int i = 0; i < q; i++)
{
int u = Q[i].u, v = Q[i].v;
printf("%d\n", d[u] + d[v] - 2*d[ans[i]]);
}
return 0;
}
版权声明:本文博客原创文章。博客,未经同意,不得转载。
POJ 1986 Distance Queries LCA两点距离树的更多相关文章
- POJ.1986 Distance Queries ( LCA 倍增 )
POJ.1986 Distance Queries ( LCA 倍增 ) 题意分析 给出一个N个点,M条边的信息(u,v,w),表示树上u-v有一条边,边权为w,接下来有k个询问,每个询问为(a,b) ...
- POJ 1986 - Distance Queries - [LCA模板题][Tarjan-LCA算法]
题目链接:http://poj.org/problem?id=1986 Description Farmer John's cows refused to run in his marathon si ...
- poj 1986 Distance Queries LCA
题目链接:http://poj.org/problem?id=1986 Farmer John's cows refused to run in his marathon since he chose ...
- POJ 1986 Distance Queries(LCA Tarjan法)
Distance Queries [题目链接]Distance Queries [题目类型]LCA Tarjan法 &题意: 输入n和m,表示n个点m条边,下面m行是边的信息,两端点和权,后面 ...
- POJ 1986 Distance Queries / UESTC 256 Distance Queries / CJOJ 1129 【USACO】距离咨询(最近公共祖先)
POJ 1986 Distance Queries / UESTC 256 Distance Queries / CJOJ 1129 [USACO]距离咨询(最近公共祖先) Description F ...
- POJ 1986 Distance Queries 【输入YY && LCA(Tarjan离线)】
任意门:http://poj.org/problem?id=1986 Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total ...
- POJ 1986 Distance Queries(Tarjan离线法求LCA)
Distance Queries Time Limit: 2000MS Memory Limit: 30000K Total Submissions: 12846 Accepted: 4552 ...
- poj 1986 Distance Queries(LCA)
Description Farmer John's cows refused to run in his marathon since he chose a path much too long fo ...
- poj 1986 Distance Queries(LCA:倍增/离线)
计算树上的路径长度.input要去查poj 1984. 任意建一棵树,利用树形结构,将问题转化为u,v,lca(u,v)三个点到根的距离.输出d[u]+d[v]-2*d[lca(u,v)]. 倍增求解 ...
随机推荐
- 魔兽争霸war3心得体会(三):UD内战
最近,经常匹配到UD内战.有输有赢,有的时候,自己双矿经济,人口优势巨大,却很遗憾地输掉比赛. 本文,简要分析下 对战过程. 前期狗流开局, 5只狗,一只出去骚扰,攻击商店,防止对方科技蜘蛛骚扰我.二 ...
- js课程 1-2 js概念
js课程 1-2 js概念 一.总结 一句话总结:js标签元素也是js对象,有属性和方法,方法就是事件,属性就是标签属性,可以直接调用. 1.js中如何获取标签对象? getElement获取的是标 ...
- sdo_geometry 转 st_geometry
CREATE OR REPLACE FUNCTION sdo2sde(geo SDO_GEOMETRY) RETURN st_geometry IS lx number; --类型 (点.线.面) c ...
- xv6 gdb
The "remote" target does not support "run". https://sourceware.org/gdb/onlinedoc ...
- 【最小树形图(奇怪的kruskal)】【SCOI 2012】【bzoj 2753】滑雪与时间胶囊
2753: [SCOI2012]滑雪与时间胶囊 Time Limit: 50 Sec Memory Limit: 128 MB Submit: 1621 Solved: 570 Description ...
- js进阶 11-3 jquery中css属性如何操作
js进阶 11-3 jquery中css属性如何操作 一.总结 一句话总结:通过css()方法 1.attr和css是有交叉的,比如width,两者中都可以设置,那么他们的区别是什么? 其实通俗一点 ...
- 卷积神经网络Lenet-5实现
卷积神经网络Lenet-5实现 原文地址:http://blog.csdn.net/hjimce/article/details/47323463 作者:hjimce 卷积神经网络算法是n年前就有的算 ...
- ZBar 是款桌面电脑用条形码/二维码扫描工具
ZBar 是款桌面电脑用条形码/二维码扫描工具 windows平台python 2.7环境编译安装zbar 最近一个项目需要识别二维码,找来找去找到了zbar和zxing,中间越过无数坑,总算基本 ...
- 【t077】宝物筛选
Time Limit: 1 second Memory Limit: 128 MB [问题描述] 小FF找到了王室的宝物室,里面堆满了无数价值连城的宝物--这下小FF可发财了.但是这里的宝物实在是太多 ...
- System.getProperty()获取系统的配置信息(系统变量)
原文地址:http://www.jsjtt.com/java/Javajichu/105.html 此处记录备用. 1. 通过System.getProperty()可以获取系统的配置信息,Syste ...