How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3404    Accepted Submission(s): 1310

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
 
 
 
传说中的带权并查集。实际上,可以输入的x,y,w可以理解成y+1比x大w,所以每次把x,y+1存起来,遇到sum[x]就表示x到根节点的和,根节点在后面。每次更新的方法就是用上一篇博文里的矩阵来更新。
#include <iostream>
#include <cstdio>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cmath>
#include <ctime>
using namespace std; const int SIZE = ;
int FATHER[SIZE],SUM[SIZE];
int N,M,ANS; void ini(void);
int find_father(int);
void unite(int,int,int);
int main(void)
{
int x,y,w; while(scanf("%d%d",&N,&M) != EOF)
{
ini();
while(M --)
{
scanf("%d%d%d",&x,&y,&w);
y ++;
unite(x,y,w);
}
printf("%d\n",ANS);
} return ;
} void ini(void)
{
ANS = ;
for(int i = ;i <= N + ;i ++)
{
FATHER[i] = i;
SUM[i] = ;
}
} int find_father(int n)
{
if(n == FATHER[n])
return n;
int temp = FATHER[n];
FATHER[n] = find_father(FATHER[n]);
SUM[n] = SUM[n] + SUM[temp];
return FATHER[n];
} void unite(int x,int y,int w)
{
int fx = find_father(x);
int fy = find_father(y); if(fx == fy)
{
if(SUM[x] - SUM[y] != w)
ANS ++;
return ;
} if(fx < fy)
{
FATHER[fx] = fy;
SUM[fx] = -SUM[x] + w + SUM[y];
}
else
{
FATHER[fy] = fx;
SUM[fy] = -SUM[y] - w + SUM[x];
}
}

HDU 3038 How Many Answers Are Wrong (并查集)的更多相关文章

  1. HDU 3038 How Many Answers Are Wrong (并查集)---并查集看不出来系列-1

    Problem Description TT and FF are ... friends. Uh... very very good friends -________-bFF is a bad b ...

  2. HDU 3038 How Many Answers Are Wrong 并查集带权路径压缩

    思路跟 LA 6187 完全一样. 我是乍一看没反应过来这是个并查集,知道之后就好做了. d[i]代表节点 i 到根节点的距离,即每次的sum. #include <cstdio> #in ...

  3. hdu 3038 How Many Answers Are Wrong

    http://acm.hdu.edu.cn/showproblem.php?pid=3038 How Many Answers Are Wrong Time Limit: 2000/1000 MS ( ...

  4. HDU 3038 - How Many Answers Are Wrong - [经典带权并查集]

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3038 Time Limit: 2000/1000 MS (Java/Others) Memory Li ...

  5. HDU 3038 How Many Answers Are Wrong 【YY && 带权并查集】

    任意门:http://acm.hdu.edu.cn/showproblem.php?pid=3038 How Many Answers Are Wrong Time Limit: 2000/1000 ...

  6. HDU 5458 Stability(双连通分量+LCA+并查集+树状数组)(2015 ACM/ICPC Asia Regional Shenyang Online)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5458 Problem Description Given an undirected connecte ...

  7. ACM: hdu 1811 Rank of Tetris - 拓扑排序-并查集-离线

    hdu 1811 Rank of Tetris Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & % ...

  8. HDU(1856),裸的带权并查集

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1856 题意:朋友圈问题,A和B是朋友,B和C是朋友则A和C也是朋友,依次类推,题目的意思就是求最大的朋 ...

  9. HDU 1598 find the most comfortable road 并查集+贪心

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1598 find the most comfortable road Time Limit: 1000 ...

随机推荐

  1. Kettle 创建 Transformation

    1.第一步,先准备数据和工具 安装好mysql以及客户端工具 数据: USE `test`; CREATE TABLE `account` (   `id` int(11) NOT NULL AUTO ...

  2. Linux编译安装RTL8192CU芯片驱动,使用TP_LINK wn823n无线网卡

    前几天给自己的台式电脑安装了Window 7+CentOS 6.4 Linux双系统,发现在Windows 7下面可以正常使用TP_LINK wn823n无线网卡来连接无线网络,但是在Linux下面, ...

  3. 利用HTML5开发Android(2)---Android中构建HTML5应用

    使用WebView控件 与其他控件的使用方法相同 在layout中使用一个<WebView>标签 WebView不包括导航栏,地址栏等完整浏览器功能,只用于显示一个网页 在WebView中 ...

  4. Matrix对bitmap的一些操作

    本篇文章是对使用Matrix对bitmap的旋转与镜像水平垂直翻转进行了详细的分析介绍,需要的朋友参考下  Bitmap convert(Bitmap a, int width, int height ...

  5. 切取图片的一部分(利用CCRenderTexture)

    转自:http://blog.csdn.net/wcjwdq/article/details/37932769 显示图片时,在项目中经常会用只读取图片的一部分,而不是全部. 错误方式:很多人这时候会采 ...

  6. Cocos2d-x利用CCHttpRequest获取网络图片并显示

    利用CCHttpRequest获取网上http地址的图片并缓存到本地生成CCSprite用于显示 //图片结构class imgstruct : public CCObject { public: i ...

  7. CodeForces 164 B. Ancient Berland Hieroglyphs 单调队列

    B. Ancient Berland Hieroglyphs 题目连接: http://codeforces.com/problemset/problem/164/B Descriptionww.co ...

  8. Android端百度地图API使用详解

    百度地图API简介 百度地图移动版API(Android)是一套基于Android设备的应用程序接口,通过该接口,可以轻松的访问百度服务和数据,构建功能丰富.交互性强的地图应用程序. 百度地图移动版A ...

  9. 关于去除Dialog的黑色背景框

    Dialog有两种形式的,一个是Dialog及其子类,还有一种是Activity的Dialog显示方式. 不管怎样,在自定义Dialog的时候,如果不做一些处理,都会出现黑色背景边框,这个问题动不动就 ...

  10. OC和JS之间的交互

    OC和JS之间的交互 目录 对OC和JS之间交互的理解 JS调用OC OC调用JS 对OC和JS之间交互的理解 JS调用OC JS文件 function sendCommand(cmd,param){ ...