HDU 3038 How Many Answers Are Wrong 【YY && 带权并查集】

任意门：http://acm.hdu.edu.cn/showproblem.php?pid=3038

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15582    Accepted Submission(s): 5462

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

 

题意概括：

有 M 次信息，每次给出一个区间 u~v 的计数值，求错误的信息有多少条。

例如：

1 10 10

1 4 2

5 10 5

第一条信息说明 1~10 的和为 10

第二条信息的区间和与第三条信息的区间和 相加不等于 10

说明第三条有误。

解题思路：

这题巧妙之处在于把区间问题转化为并查集问题。

fa[ i ] ：表示 i 结点的最左端点。

val [ i ] ：表示结点 i 到 最左端点 fa[ i ] 的区间和

假设输入 u v w ；fa_u = getfa[ u ], fa_v = getfa[ v ];

合并区间（fa_u ！= fa_v）：

情况一：fa_u < fa_v

fa[ fa_v ] = fa_u (更新最左端点）;

val [ fa_v ] = val [ u ] + w - val [ v ];

情况二： fa_u > fa_v

fa[ fa_u ] = fa_v (更新最左端点）;

val [ fa_u ] = val [ v ] - w - val [ u ];

判断信息（fa_u == fa_v)：

判断 val [ u ] + w ?= val [ v ];

tip:

输入信息后左端点 u--，这样才能不重复合并区间。

最后要注意多测试样例

AC code:

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int MAXN = 2e5+;

 int fa[MAXN];
 int val[MAXN];
 int N, M;

 int getfa(int x)
 {
     if(fa[x] == x) return fa[x];
     int t = fa[x];
     fa[x] = getfa(fa[x]);
     val[x] += val[t];
     return fa[x];
 }

 void init()
 {
     memset(val, , sizeof(val));
     for(int i = ; i <= N; i++) fa[i] = i;
 }

 int main()
 {
     while(~scanf("%d%d", &N, &M)){
         int ans = ;
         init();
         for(int i = , u, v, w, ru, rv; i <= M; i++){
             scanf("%d%d%d", &u, &v, &w);
             u = u-;
             ru = getfa(u);
             rv = getfa(v);
             if(ru == rv && val[u]+w != val[v]) ans++;
             else if(ru < rv){
                 fa[rv] = ru;
                 val[rv] = val[u] - val[v] + w;
             }
             else if(ru > rv){
                 fa[ru] = rv;
                 val[ru] = val[v] - val[u] - w;
             }
         }
         printf("%d\n", ans);
     }
     return ;
 }