(并查集) hdu 3038

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2856    Accepted Submission(s): 1100

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

2009 Multi-University Training Contest 13 - Host by HIT

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
#define maxn 200010
int n,m,fa[maxn],dist[maxn],ans;
int find(int x)
{
      if(x==fa[x])
            return fa[x];
      int k=fa[x];
      fa[x]=find(fa[x]);
      dist[x]+=dist[k];
      return fa[x];
}
void Union(int x,int y,int z)
{
      int fx,fy;
      fx=find(x),fy=find(y);
      if(fx!=fy)
      {
            fa[fx]=fy;
            dist[fx]=dist[y]-dist[x]+z;
      }
      else
      {
            if(dist[x]-dist[y]!=z)
                  ans++;
      }
}
int main()
{
      int x,y,z;
      while(scanf("%d%d",&n,&m)!=EOF)
      {
            ans=0;
            for(int i=0;i<=n+1;i++)
                  fa[i]=i,dist[i]=0;
            for(int i=0;i<m;i++)
            {
                  scanf("%d%d%d",&x,&y,&z);
                  Union(x,y+1,z);
            }
            printf("%d\n",ans);
      }
      return 0;
}

  

时间: 2024-10-24 22:29:32

(并查集) hdu 3038的相关文章

并查集 -- HDU 1232 UVALA 3644

并查集: 1 int pa[maxn],Rank[maxn]; 2 ///初始化 x 集合 3 void make_set(int x) 4 { 5 pa[x]=x; 6 Rank[x]=0; 7 } 8 ///递归查找 x 所在的集合 9 int find_set(int x) 10 { 11 if(pa[x]!=x) pa[x]=find_set(pa[x]); 12 return pa[x]; 13 } 14 ///更新根节点,如果不更新可能会暴栈 15 void mix(int x,in

集合问题 离线+并查集 HDU 3938

题目大意:给你n个点,m条边,q个询问,每条边有一个val,每次询问也询问一个val,定义:这样条件的两个点(u,v),使得u->v的的价值就是所有的通路中的的最长的边最短.问满足这样的点对有几个. 思路:我们先将询问和边全部都按照val排序,然后我们知道,并查集是可以用来划分集合的,所以我们就用并查集来维护每一个集合就行了. //看看会不会爆int!数组会不会少了一维! //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespac

并查集 hdu 1856

More is better Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others) Total Submission(s): 16863    Accepted Submission(s): 6205 Problem Description Mr Wang wants some boys to help him with a project. Because the projec

(并查集) hdu 2473

Junk-Mail Filter Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6389    Accepted Submission(s): 2025 Problem Description Recognizing junk mails is a tough task. The method used here consists o

分组并查集 hdu 1829

A Bug's Life Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 10063    Accepted Submission(s): 3288 Problem Description Background Professor Hopper is researching the sexual behavior of a rare

(枚举+并查集) hdu 1598

find the most comfortable road Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4096    Accepted Submission(s): 1768 Problem Description XX星有许多城市,城市之间通过一种奇怪的高速公路SARS(Super Air Roam Structure---超级

(并查集) hdu 4750

Count The Pairs Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 952    Accepted Submission(s): 422 Problem Description   With the 60th anniversary celebration of Nanjing University of Science

(并查集) hdu 3635

Dragon Balls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3443    Accepted Submission(s): 1342 Problem Description Five hundred years later, the number of dragon balls will increase unexpecte

(并查集) hdu 3461

Code Lock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 1308    Accepted Submission(s): 471 Problem Description A lock you use has a code system to be opened instead of a key. The lock contai