spoj freetour II
昨天吐槽还没A,今天就A了
有个变量开成了全局变量,应该携程局部变量
对于中间的solve我也不懂为什么是nlog2n,我不看题解也不会做
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 2000000000;
const int MAXN = 4e5+5;
int ans;
int co[MAXN];
int vis[MAXN];
struct Pode{
int to,nx, di;
Pode(int a=0,int b=0,int c=0):to(a), nx(b), di(c){}
}E[MAXN];
int head[MAXN], cot;
void add(int u, int v, int w) {
E[cot] = Pode(v,head[u],w); head[u] = cot++;
}
int N,K,M;
/***************WeightRoot************/
int all, num, center;
int dp[MAXN], nodes[MAXN];
void findRoot(int x,int pre) {
nodes[x] = 1; dp[x] = 0;
for(int i = head[x]; ~i; i = E[i].nx) {
int y = E[i].to; if(y == pre || vis[y]) continue;
findRoot(y,x);
nodes[x] += nodes[y];
dp[x] = max(dp[x], nodes[y]);
}
dp[x] = max(dp[x], all-nodes[x]);
if(dp[x] < num) {
num = dp[x]; center = x;
}
}
int getRoot(int root,int sn) {
num = INF; all = sn; center = root;
findRoot(root, -1);
return center;
}
/*************treecdq***************/
struct Node{
int dep, v, di;
}so[MAXN];
int cmp(Node a,Node b) {
return a.dep < b.dep;
}
int dep[MAXN]; // max dep (black)
int g[MAXN];
int mg[MAXN];
void getdep(int x,int pre) {
dep[x] = co[x]; int res = 0;
for(int i = head[x]; ~i; i = E[i].nx) {
int y = E[i].to; if(y == pre || vis[y]) continue;
getdep(y,x);
res = max(res, dep[y]);
}
dep[x] += res;
}
void getg(int x,int pre,int d,int c) {
g[c] = max(g[c], d);
for(int i = head[x]; ~i; i = E[i].nx) {
int y = E[i].to; if(y == pre || vis[y]) continue;
getg(y,x,d+E[i].di, c+co[y]);
}
}
void work(int x) {
vis[x] = 1; int tot = 0;
for(int i = head[x]; ~i; i = E[i].nx) {
int y = E[i].to; if(vis[y]) continue;
work(getRoot(y,nodes[y]));
}
for(int i = head[x]; ~i; i = E[i].nx) {
int y = E[i].to; if(vis[y]) continue;
getdep(y, x);
so[++tot].dep = dep[y]; so[tot].v = y; so[tot].di = E[i].di;
}
sort(so+1,so+tot+1,cmp);
// printf("%d:",x); for(int i = 1; i <= tot; ++i) printf("%d ",so[i].dep); printf("\n");
for(int i = 0; i <= so[tot].dep; ++i) mg[i] = -INF;
for(int i = 1; i <= tot; ++i) {
int t1 = so[i].dep; int t2 = so[i].v; int t3 = so[i].di;
for(int j = 0; j <= t1; ++j) g[j] = -INF;
getg(t2,x,t3,co[t2]);
if(i != 1) {
for(int j = 0; j <= K-co[x] && j <= t1; ++j) {
int tt = min(so[i-1].dep, K-co[x]-j);
if(mg[tt] == -INF) break;
if(g[j] != -INF) ans = max(ans, mg[tt]+g[j]);
}
}
for(int j = 0; j <= t1; ++j) {
mg[j] = max(g[j], mg[j]);
if(j) mg[j] = max(mg[j], mg[j-1]);
if(j+co[x] <= K) ans=max(ans, mg[j]);
}
}
vis[x] = 0;
}
int main(){
while(~scanf("%d %d %d",&N,&K,&M)) {
ans = 0;
memset(head,-1,sizeof(head)); cot = 0;
memset(co,0,sizeof(co));
memset(vis,0,sizeof(vis));
for(int i = 1; i <= M; ++i) {
int a; scanf("%d",&a);
co[a] ++;
}
for(int i = 1; i < N; ++i) {
int a,b,c; scanf("%d %d %d",&a,&b,&c);
add(a,b,c); add(b,a,c);
}
work(getRoot(1,N));
printf("%d\n",ans);
}
return 0;
}
spoj freetour II的更多相关文章
- hdu5977 Garden of Eden
都不好意思写题解了 跑了4000多ms 纪念下自己A的第二题 (我还有一道freetour II wa20多发没A...呜呜呜 #include<bits/stdc++.h> using ...
- SPOJ 1557. Can you answer these queries II 线段树
Can you answer these queries II Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 https://www.spoj.com/pr ...
- bzoj 2482: [Spoj GSS2] Can you answer these queries II 线段树
2482: [Spoj1557] Can you answer these queries II Time Limit: 20 Sec Memory Limit: 128 MBSubmit: 145 ...
- spoj gss2 : Can you answer these queries II 离线&&线段树
1557. Can you answer these queries II Problem code: GSS2 Being a completist and a simplist, kid Yang ...
- 【SPOJ】Longest Common Substring II (后缀自动机)
[SPOJ]Longest Common Substring II (后缀自动机) 题面 Vjudge 题意:求若干个串的最长公共子串 题解 对于某一个串构建\(SAM\) 每个串依次进行匹配 同时记 ...
- 【BZOJ2589】 Spoj 10707 Count on a tree II
BZOJ2589 Spoj 10707 Count on a tree II Solution 吐槽:这道题目简直...丧心病狂 如果没有强制在线不就是树上莫队入门题? 如果加了强制在线怎么做? 考虑 ...
- 【SPOJ】Count On A Tree II(树上莫队)
[SPOJ]Count On A Tree II(树上莫队) 题面 洛谷 Vjudge 洛谷上有翻译啦 题解 如果不在树上就是一个很裸很裸的莫队 现在在树上,就是一个很裸很裸的树上莫队啦. #incl ...
- spoj COT2 - Count on a tree II
COT2 - Count on a tree II http://www.spoj.com/problems/COT2/ #tree You are given a tree with N nodes ...
- SPOJ GSS2 - Can you answer these queries II(线段树 区间修改+区间查询)(后缀和)
GSS2 - Can you answer these queries II #tree Being a completist and a simplist, kid Yang Zhe cannot ...
随机推荐
- CenOS 上安装 Redis 服务器
1.构建 Redis 因为 Redis 官方没提供 RPM 安装包,所以需要编译源代码,则需要安装 GCC & MAKE. 终端输入: yum install gcc make 从官网下载 t ...
- POJ [P2631] Roads in the North
树的直径 树的直径求法: 任取一点u,找到树上距u最远的点s 找到树上距s点最远的点t,s->t的距离即为所求 #include <iostream> #include <cs ...
- bzoj 3864: Hero meet devil [dp套dp]
3864: Hero meet devil 题意: 给你一个只由AGCT组成的字符串S (|S| ≤ 15),对于每个0 ≤ .. ≤ |S|,问 有多少个只由AGCT组成的长度为m(1 ≤ m ≤ ...
- 51Nod 欢乐手速场1 A Pinball[DP 线段树]
Pinball xfause (命题人) 基准时间限制:1 秒 空间限制:262144 KB 分值: 20 Pinball的游戏界面由m+2行.n列组成.第一行在顶端.一个球会从第一行的某一列出发 ...
- 2018/2/5 ELK技术栈之ElasticSearch学习笔记
npm config set registry https://registry.npm.taobao.org npm config get registry 支持跨域访问http.cors.enab ...
- 正负样本比率失衡SMOTE
正负样本比率失衡SMOTE [TOC] 背景 这几天测试天池的优惠券预测数据在dnn上面会不会比集成树有较好的效果,但是正负样本差距太大,而处理这种情况的一般有欠抽样和过抽样,这里主要讲过抽样,过抽样 ...
- laravel5.4+vue+vux+element的环境搭配
最近因为项目的需要,需要搭配一个这样的环境.之前做过的东西没有这样用过,在网上找了半天不是过于简单就是根本行不通,自己踩了半天的坑,终于搭配成功. 首先下载laravel5.4,直接去官网一键安装包或 ...
- python爬虫登录
python3 urllib.request 网络请求操作 http://www.cnblogs.com/cocoajin/p/3679821.html python实现 爬取twitter用户姓名 ...
- Java开发API文档资源
<netty> http://netty.io/4.1/api/index.html < Spring FrameWork > 1 http://spring.io/ 2 ...
- 免费 Https 证书(Let's Encrypt)申请与配置
之前要申请免费的 https 证书操作步骤相当麻烦,今天看到有人在讨论,就搜索了一下.发现现在申请步骤简单多了. 1. 下载 certbot git clone https://github.com/ ...