HDU 3038 How Many Answers Are Wrong
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
就是給你 n,m, n個數，m給操作，a,b,c 表示a到b的和是c，如果通過以前的操作已知或者可以推出來a，b的值就判斷該操作描述的正誤，
輸出錯誤的次數
並査集，解釋在代碼中
先說一個重要的地方 假如知道 a->b->c d->e->f
輸入 a,d,len
那麼當函數cha（）找到c時，c->f f時num[c]=-num[a]+len+num[d]; ; 自己好好體會吧；
在一個需要理解的地方就是cha（）函數中num[x]+=num[t]; 一定要注意遞歸的順序，拿上面例子來說當我們把c指向f時，
下一步cha（b）時就會更改num[b]為f到f的距離，並且father[b]=f;(好好想想吧)
過程就是住先改變父親節點，後改變子節點（遞歸）
要說的已經說完了，還有一些細節在代碼中
#include
#include
#include
#include
using namespace std;
#define N 200005
int father[N],num[N];
int n,m;
int cha(int x)
{
if(x!=father[x])
{
int t=father[x]; //先保存父親節點
father[x]=cha(father[x]);
num[x]+=num[t]; //等到父親節點改變再改變此節點，重要的地方
}
return father[x];
}
int main()
{
int i;
int a,b,len;
while(~scanf(%d%d,&n,&m))
{
for(i=0;i<=n;i++)
{
father[i]=i;
num[i]=0;
}
int ans=0;
while(m--)
{
scanf(%d%d%d,&a,&b,&len);
a--; //a到b的和等於0到b的和減去0到（a-1）的和
int aa=cha(a);
int bb=cha(b);
if(aa!=bb)
{
father[bb]=aa;
num[bb]=-num[b]+num[a]+len;//此步如上面解釋，是一大重點
}
else
{
if(num[bb]!=(-num[b]+num[a]+len))
ans++;
}
}
printf(%d
,ans);
}
return 0;
}
