Time Limit: 2000/1000 MS (Java/Others)
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：
每次给出一个区间，和这个区间的所有数的和，检查有一次给出区间的值和给出的值矛盾
思路：
可以用一个数组num记录每个位置到其根节点区间内所有的元素和。每次判断是否给出的两个位置是否在一个根节点上，也就是说是否可比。如果可比，检查这两个点之间的元素和和给出的是否一致。如果不一致，就将根节点跟新为最的那个，然后更新距离。
然后令$B_i$向后移动增加一个，解决距离在计算的时候的问题。
#include "cmath"#include "cstring"#include "algorithm"#include "cstdio"using namespace std;const int maxn = 200000 + 100;int N, M;int par[maxn];int num[maxn];int ans;int find_root(int x){ if(x!=par[x]){ int t = par[x]; par[x] = find_root(par[x]); num[x] += num[t]; } return par[x];}void unite(int x, int y, int z) { int fx = find_root(x); int fy = find_root(y); if (fx != fy) { num[fy] = z - num[y] + num[x]; par[fy] = fx; } else { if (num[y] - num[x] != z) ans++; }}void init() { ans = 0; memset(num, 0, sizeof(num)); for (int i = 0;i < maxn; i++) par[i]=i;}int main(int argc, char const *argv[]) { while (scanf("%d%d", &N, &M) != EOF) { init(); for (int i = 0; i < M; i++) { int a, b, c; scanf("%d%d%d", &a, &b, &c); if (a < b) unite(a, b + 1, c); else unite(b, a + 1, c); } printf("%d\n", ans); } return 0;}