How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15228    Accepted Submission(s): 5351

Problem Description

TT and FF are ... friends. Uh... very very good friends -________-b

FF is a bad boy, he is always wooing TT to play the following game with him. This is a very humdrum game. To begin with, TT should write down a sequence of integers-_-!!(bored).

Then, FF can choose a continuous subsequence from it(for example the subsequence from the third to the fifth integer inclusively). After that, FF will ask TT what the sum of the subsequence he chose is. The next, TT will answer FF's question. Then, FF can redo this process. In the end, FF must work out the entire sequence of integers.

Boring~~Boring~~a very very boring game!!! TT doesn't want to play with FF at all. To punish FF, she often tells FF the wrong answers on purpose.

The bad boy is not a fool man. FF detects some answers are incompatible. Of course, these contradictions make it difficult to calculate the sequence.

However, TT is a nice and lovely girl. She doesn't have the heart to be hard on FF. To save time, she guarantees that the answers are all right if there is no logical mistakes indeed.

What's more, if FF finds an answer to be wrong, he will ignore it when judging next answers.

But there will be so many questions that poor FF can't make sure whether the current answer is right or wrong in a moment. So he decides to write a program to help him with this matter. The program will receive a series of questions from FF together with the answers FF has received from TT. The aim of this program is to find how many answers are wrong. Only by ignoring the wrong answers can FF work out the entire sequence of integers. Poor FF has no time to do this job. And now he is asking for your help~(Why asking trouble for himself~~Bad boy)

 

Input

Line 1: Two integers, N and M (1 <= N <= 200000, 1 <= M <= 40000). Means TT wrote N integers and FF asked her M questions.

Line 2..M+1: Line i+1 contains three integer: Ai, Bi and Si. Means TT answered FF that the sum from Ai to Bi is Si. It's guaranteed that 0 < Ai <= Bi <= N.

You can assume that any sum of subsequence is fit in 32-bit integer.

 

Output

A single line with a integer denotes how many answers are wrong.
 

Sample Input

10 5
1 10 100
7 10 28
1 3 32
4 6 41
6 6 1
 

Sample Output

1

Source

 
 #include <bits/stdc++.h>

 using namespace std;

 const int N = ;

 int fa[N], rk[N];

 void init(){
for(int i = ; i < N; i++){
fa[i] = i;
rk[i] = ;
}
} int getfa(int x){
if(x == fa[x])return x;
int pre = fa[x];
fa[x] = getfa(fa[x]);
rk[x] += rk[pre];
return fa[x];
} bool merge_(int a, int b, int val){
int af = getfa(a);
int bf = getfa(b);
if(af != bf){
fa[af] = bf;
rk[af] = rk[b]-rk[a]+val;
return true;
}else{
return rk[a]-rk[b] == val;
}
} int main(){
int n, m, a, b, val;
while(cin>>n>>m){
int ans = ;
init();
while(m--){
cin>>a>>b>>val;
a--;
if(!merge_(a, b, val))
ans++;
}
cout<<ans<<endl;
} return ;
}

HDU3038(KB5-D加权并查集)的更多相关文章

  1. hdu 3047 Zjnu Stadium(加权并查集)2009 Multi-University Training Contest 14

    题意: 有一个运动场,运动场的坐席是环形的,有1~300共300列座位,每列按有无限个座位计算T_T. 输入: 有多组输入样例,每组样例首行包含两个正整数n, m.分别表示共有n个人,m次操作. 接下 ...

  2. hdu 3635 Dragon Balls(加权并查集)2010 ACM-ICPC Multi-University Training Contest(19)

    这道题说,在很久很久以前,有一个故事.故事的名字叫龙珠.后来,龙珠不知道出了什么问题,从7个变成了n个. 在悟空所在的国家里有n个城市,每个城市有1个龙珠,第i个城市有第i个龙珠. 然后,每经过一段时 ...

  3. HDU 3407.Zjnu Stadium 加权并查集

    Zjnu Stadium Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  4. A Bug's Life(加权并查集)

    Description Background  Professor Hopper is researching the sexual behavior of a rare species of bug ...

  5. A Bug's Life(加权并查集)

    Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs ...

  6. P1196 银河英雄传说(加权并查集)

    P1196 银河英雄传说 题目描述 公元五八○一年,地球居民迁移至金牛座α第二行星,在那里发表银河联邦 创立宣言,同年改元为宇宙历元年,并开始向银河系深处拓展. 宇宙历七九九年,银河系的两大军事集团在 ...

  7. Zjnu Stadium(加权并查集)

    Zjnu Stadium Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

  8. 洛谷 P2024 [NOI2001]食物链(种类并查集,加权并查集)

    传送门 解题思路 加权并查集: 什么是加权并查集? 就是记录着每个节点到它的父亲的信息(权值等). 难点:在路径压缩和合并节点时把本节点到父亲的权值转化为到根节点的权值 怎么转化呢? 每道题都不一样Q ...

  9. UVALive 4487 Exclusive-OR 加权并查集神题

    已知有 x[0-(n-1)],但是不知道具体的值,题目给定的信息 只有 I P V,说明 Xp=V,或者 I P Q V,说明 Xp ^ Xq=v,然后要求回答每个询问,询问的是 某任意的序列值 Xp ...

  10. 牛客网-Beauty of Trees 【加权并查集】

    锟斤拷锟接o拷https://www.nowcoder.com/acm/contest/119/A锟斤拷源锟斤拷牛锟斤拷锟斤拷 锟斤拷目锟斤拷锟斤拷 It锟斤拷s universally acknow ...

随机推荐

  1. 用Ubuntu快速安装Jenkins

    一.安装操作系统,安装前准备. 1.操作系统:Ubuntu 18.04 (大家都知道Ubuntu的特点,在线安装,方便很多) 2.apt源.apt源在官网上面分很多种,每个版本的源不一样,如果是其他版 ...

  2. 分布式文件系统 / MQ / 鉴权(轮廓)

    FastDFS的轮廓   /  RabbitMQ的轮廓  /  JWT和RSA非对称加密的轮廓

  3. webpack严格模式!!!忽略

    1. babel5 babel: { options: { blacklist: ["useStrict"], // ... }, // ... } 2. babel6 修改.ba ...

  4. vi命令详解(转)

    vi编辑器是所有Unix及Linux系统下标准的编辑器,它的强大不逊色于任何最新的文本编辑器,这里只是简单地介绍一下它的用法和一小部分指令.由于对Unix及Linux系统的任何版本,vi编辑器是完全相 ...

  5. jQuery文档操作

    jQuery文档操作 1.jq文档结构 var $sup = $('.sup'); $sup.children(); // sup所有的子级们 $sup.parent(); // sup的父级(一个, ...

  6. 简介 - SAFe(Scaled Agile Framework,规模化敏捷框架)

    简介 SAFe(Scaled Agile Framework,规模化敏捷框架) SAFe:https://www.scaledagileframework.com/ Scaled Agile Fram ...

  7. 安卓Listview和Adapter数据设计

    ListView是一种用于垂直显示的列表控件,如果显示内容过多,则会自动出现垂直滚动条,每一行是一个View对象,在每一行上可以放置任何组件,Adapter适配器是数据和UI的桥梁,为数据显示提供了统 ...

  8. Kubernetes集群搭建之Etcd集群配置篇

    介绍 etcd 是一个分布式一致性k-v存储系统,可用于服务注册发现与共享配置,具有以下优点. 简单 : 相比于晦涩难懂的paxos算法,etcd基于相对简单且易实现的raft算法实现一致性,并通过g ...

  9. [原创]IIS提权工具-VBS提权脚本免杀生成器

    [原创]添加系统用户 VBS提权脚本随机加密生成器[K.8] 2011-05-05 02:42:53|  分类: 原创工具 VBS提权脚本随机加密生成器[K.8]  Author: QQ吻 QQ:39 ...

  10. VSCode typescript ctrl+shift+b can't be compiled error:TS5007

    环境: vscode:1.12.2 node 7.4.0 TypeScript:2.3.2 从svn 更新下来,别的电脑环境编译是没问题的,在我的电脑上编译失败并出现以下错误 error TS5007 ...