hdu3038
How Many Answers Are Wrong
http://acm.hdu.edu.cn/showproblem.php?pid=3038TT 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.
BoringBoringa 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.
Input
10 5
1 10 100
7 10 28
1 3 32
4 6 41
6 6 1
Output
1
这一题的思路我也是看别人博客才知道的，学长说的果然没错，博客真是个好东西。
首先看题目大意。（看翻译吧！看翻译吧！）
然后本题每一次输入都需要判断Ai到Bi的Si是否与前面的有所冲突，如何去判断就是本体的思路。
首先看样例
查找0和10的pr是否相同（因为这里是从1到10，所以我们为了可以将数据联系起来，这里要Ai-1），1到10的和明显不是100，但看了翻译我们都知道，1到10我们需要将他的权赋值成为100，即sum[10]=100，把10的父亲也就是根变成1，即pr[10]=1，然后输入7 10 28，查找6和10的pr是否相同，明显不同，pr[6]=6，pr[10]=1,这里将pr[6]赋值为1，然后sum[6]=sum[10]-Si-sum[6]，为什么sum[Ai]也要被减去呢？因为这个公式其实是这样写的sum[ pr[Ai] ]=sum[Bi]-Si-sum[Ai]（此处6的pr大于10的pr才这样写），不懂可以看图。
然后输入1 3 32，（差不多的过程我就不写了，试着自己推一下吧！），然后是4 6 41，pr[3]=1,pr[6]=1,此时就需要判断了，因为3和6的pr相同且等于1，说明1到3是有权值的，判断sum[Ai]+s是否等于sum[Bi]（自己画一下吧，我懒得画了。），明显不等于所以ans++，然后输入6 6 1，pr[5]不等于pr[6]，更新sum[6]=1,输出答案ans。
#include
#include
using namespace std;
const int maxn=200005;
int pr[maxn],sum[maxn],ans,n,m;
void init()//初始化
{
memset(sum,0,sizeof(sum));
for(int i=0; i<=n; i++)
pr[i]=i;
ans=0;
}
int finds(int x)
{
if(x==pr[x])
return x;
int k=pr[x];
pr[x]=finds(pr[x]);
sum[x]+=sum[k];//增加量sum[t]是从根节点到当前节点路径上权值之和
return pr[x];
}
void unions(int a,int b,int s)
{
int x=finds(a);
int y=finds(b);
if(x==y) {
if(sum[a]+s!=sum[b])
ans++;
return;
}
else if(x