HDU 3038 How Many Answers Are Wrong (带权并查集)

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4120 Accepted Submission(s): 1577

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



题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3038

题目大意：n个数字，m个命题，a b s，表示从a到b的闭区间的数值和为s，求从第一个命题开始有几个与之前的相矛盾，不矛盾的则认为是正确的命题

题目分析：s = Σ(1,b) - Σ(1,a-1)，因此利用带权并查集，把数字小的当作根，计算偏移量，如果根相同，偏移量不为s，则表示矛盾，根不相同时合并，合并是要改变偏移量

如果fa < fb，则fa做根，w[x]表示x对根的偏移量，都是从大数字到小数字的

w[a] = a -> fa，w[b] = b -> fb，s = b -> a，fb -> fa = -fb -> b + b -> a + a -> fa = w[a] + s - w[b]，fa > fb时同理

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 2e5 + 5;
int fa[MAX], w[MAX];
int n, m;

void UF_set()
{
    for(int i = 0; i <= n; i++)
    {
        fa[i] = i;
        w[i] = 0;
    }
}

int Find(int x)
{
    if(x == fa[x])
        return x;
    int tmp = fa[x];
    fa[x] = Find(fa[x]);
    w[x] += w[tmp];
    return fa[x];
}

bool Union(int a, int b, int x)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(r1 < r2)
    {
        fa[r2] = r1;
        w[r2] = w[a] - w[b] + x;
    }
    else if(r1 > r2)
    {
        fa[r1] = r2;
        w[r1] = w[b] - w[a] - x;
    }
    else if(r1 == r2 && w[b] - w[a] != x)
        return true;
    return false;
}

int main()
{
    while(scanf("%d %d", &n, &m) != EOF)
    {
        UF_set();
        int cnt = 0;
        while(m --)
        {
            int a, b, x;
            scanf("%d %d %d", &a, &b, &x);
            if(Union(a - 1, b, x))
                cnt ++;
        }
        printf("%d\n", cnt);
    }
}