hdu 3038 How Many Answers Are Wrong

 

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2589    Accepted Submission(s): 1003

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

分析：

这道题目和hdu（3047）内容完全一样。。。。。只是数据的范围改变了，另外要注意有多组测试数据。

代码复制过来 ，修改一下maxn和 b++;//转为左闭右开 [a..b) ，提交通过。

 

并查集维护的是节点之间相对的关系，子序列和不能表示头尾节点的关系，所以要做转换。

类似 hdu3047， 转换关系为两点间的距离差。dist存储相对于父节点的距离，所以距离差可以表示成 dist[b+1]-dist[a] ，也可表示成 dist[b]-dist[a-1] ，看你怎么理解了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=200000+5;
int p[maxn], dist[maxn];//dist存储的是相对于父节点的距离

void make_set()
{
    memset(p, -1, sizeof(p)); 
    memset(dist, 0, sizeof(dist)); 
}

int find_set(int x)
{
    if(p[x]==-1) return x;
    int fx=p[x];
    p[x]=find_set(p[x]);
    dist[x]+=dist[fx];
    return p[x];
}

void union_set(int x, int y, int d)
{
    int fx=find_set(x), fy=find_set(y);
    if(fx==fy) return;
    p[fy]=fx;
    dist[fy]=-dist[y]+d+dist[x];
}


int main()
{
    int n, m;
    int a, b, x;
    int ans;
    while(scanf("%d%d", &n, &m)!=EOF) {
        ans=0;
        make_set();
        while(m--)
        {
            scanf("%d%d%d", &a, &b, &x);
            b++;//转为左闭右开 [a..b)
            //a--;
            if(find_set(a)==find_set(b))
            {
                if(dist[b]-dist[a]!=x) 
                {
                    //printf("a=%d b=%d x=%d, dist[b]-dist[a]=%d
", a, b, x, dist[b]-dist[a]);
                    ans++;
                }
            }
            else union_set(a, b,x);
        }
        printf("%d
", ans);
    }

    return 0;
}