并查集------D - How Many Answers Are Wrong

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

还以为并查集简单呢，看来变形还是挺多的，以后涉及区间和的问题多转换为区间和考虑。。。。
区间和可以理解为前缀和相减。每个节点记录前缀和，对每个询问先判断两个节点是否连通，这便是带权并查集干的事了，若联通则权值相减看是否为给定值，若不为则矛盾；若不联通则两棵树合并为一棵（注意有顺序）同时计算出新的子节点的权值。
像A 到 B 的区间和为s，则相当于(A - 1) -> B的边权为 S,用r数组表示每个节点到根节点的和。。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1);
#define pb push_back
#define mp make_pair
#define fi first
#define se second
const int N = 2 * 1e5 + 5;

int b[N];
//记录与根节点的关系
int r[N];

int join1(int x)
{
    int k = b[x];
    if(b[x] != x){
        b[x] = join1(b[x]);
        r[x] = r[x] + r[k];
    }
    return b[x];
}

int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m)){
        for(int i = 0;i <= n;++i){
            b[i] = i;
            r[i] = 0;
        }
        int sum = 0;
        for(int i = 0;i < m;++i){
            int x,y,z;
            scanf("%d %d %d",&x,&y,&z);
            int p = join1(x - 1),q = join1(y);
            if(p != q){
                b[q] = p;
                r[q] = r[x - 1] - r[y] + z;
            }else if(p == q && (r[y] - r[x - 1]) != z){
                sum++;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}

并查集------D - How Many Answers Are Wrong

猜你喜欢