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 -________-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

 

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gaojie

并查集看不出来系列

建立一个并查集, 并查集中每个节点的值是指这个节点到其根节点的距离(与根节点的差),

若要求区间 [a,b] 的和, 即为求sum[b]-sum[a-1];

代码:

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<iomanip>
 #define INF 0x7ffffff
 #define MAXN 200010
 using namespace std;
 const double eps=1e-;
 int b[MAXN];
 int val[MAXN];
     int n,m;
 void init()
 {
     memset(val,,sizeof(val));
     for(int i=;i<=n;i++){
         b[i]=i;
     }
 }
 int find(int x)
 {
 //    int tem;
 //    int t=x;
 //    while(b[t]!=t){
 //        t=b[t];
 //    }
 //    while(b[x]!=x){
 //        val[t]=val[t]+val[b[t]];
 //        x=b[x];
 //        b[x]=t;
 //    }
 //    return t;
     if(b[x]==x) return x;
     int t=b[x];
     //val[x]+=val[b[x]];
     b[x]=find(b[x]);
     val[x]+=val[t];//注意此操作与DFS的顺序!!!这个操作的顺序是将根节点的值传下来, 注意有此操作必须结合路径压缩, 不然会出错误,why?
     return b[x];
 }
 void uni(int x,int y,int c)
 {
     int xx=find(x);
     int yy=find(y);
     if(xx<yy){
         b[yy]=xx;
         val[yy]=val[x]+c-val[y];
     }
     else{
         b[xx]=yy;
         val[xx]=val[y]-val[x]-c;
     }
 }
 int main()
 {
     #ifndef ONLINE_JUDGE
     freopen("data.in", "r", stdin);
     #endif
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     int ans=;
     int a,b,c;
     int xx,yy;
     while(cin>>n>>m){
         ans=;
         init();
         for(int i=;i<m;i++){
             cin>>a>>b>>c;
             a--;//注意--操作, 因为要求的是a到b的距离, 因此需要 sum[b]-sum[a-1]
             if(find(a)!=find(b)){
                 uni(a,b,c);
             }
             else{
                 if(val[b]-val[a]!=c){
                     ans++;
                 }
             }
         }
         cout<<ans<<endl;
     }
 }