HDU3038 How Many Answers Are Wrong

题目链接

题目

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

题解

知识点：并查集。

因为给出的是区间和，我们拆分成两端前缀和方便用带权并查集维护。

依然用向量的思想会很好理解合并和压缩过程中的计算。

时间复杂度 $O(n+m\log n)$

空间复杂度 $O(n)$

代码

#include <bits/stdc++.h>
#define ll long long
 
using namespace std;
 
int fa[200007], sum[200007];///sum[i] 指 (i,根节点]的和
///可以理解为一个点到点的向量长度，这样就可以用向量加减理解
 
int find(int x) {
    if (fa[x] == x) return x;
    int pre = fa[x];
    fa[x] = find(fa[x]);
    sum[x] += sum[pre];///递归求和，路径压缩
    return fa[x];
}
 
 
int main() {
    std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n, m;
    while (cin >> n >> m) {///不是我的锅，是他没写
        for (int i = 0;i <= n;i++) fa[i] = i, sum[i] = 0;
        int cnt = 0;
        for (int i = 1;i <= m;i++) {
            int a, b, S;
            cin >> a >> b >> S;
            a--;
            int ra = find(a);
            int rb = find(b);
            ll A = sum[a];
            ll B = sum[b];
            if (ra == rb) {
                if (A - B != S) cnt++;
            }
            else {
                sum[ra] = B + S - A;///(ra,rb]的大小,可以是逆序的，这样和就是相反数
                fa[ra] = rb;
            }
        }
        cout << cnt << '\n';
    }
    return 0;
}