*HDU3038 并查集

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6349    Accepted Submission(s): 2395

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

 

题意：

共n个数每次给出a~b的区间的和 c，问在前面给出的区间和的前提下新给出的区间和是否正确，统计错误的个数。

代码：

 //这题处理时将输入的前一个数减一，就可以把他们连在一起了，如：1,2 3,4 ——>0,2 2,4 ——>0,4 ;并查集之后计算每个点到根节点的距离
 //如果输入的两个点有相同的根节点就看他们的距离是否与并查集里的相矛盾。合并时将大的并到小的里面就可以看成每个点到1点的距离
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 using namespace std;
 int n,m;
 int fat[],num[];
 int find(int x)
 {
     if(fat[x]!=x)
     {
         int f=fat[x];
         fat[x]=find(fat[x]);
         num[x]+=num[f];       // !!!!!!!!
     }
     return fat[x];
 }
 void connect(int x,int y,int c)
 {
     int a=find(x),b=find(y);
     if(a<b)                // !!!!!!!! 将大的并到小的里面。
     {
         fat[b]=a;
         num[b]=num[x]-num[y]+c;
     }
     else if(b<a)
     {
         fat[a]=b;
         num[a]=num[y]-num[x]-c;
     }
 }
 int main()
 {
     int a,b,c;
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         for(int i=;i<=n;i++)
         {
             fat[i]=i;
             num[i]=;
         }
         int ans=;
         for(int i=;i<=m;i++)
         {
             scanf("%d%d%d",&a,&b,&c);
             a--;
             if(find(a)==find(b))
             {
                 if(num[b]-num[a]!=c)
                 ans++;
             }
             else connect(a,b,c);
         }
         printf("%d\n",ans);
     }
     return ;
 }