HDU 3038 How Many Answers Are Wrong

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题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3038

解题思路:首先,由于序列数据可正可负,因此实际不满足情况的只有已知区间样例中所给出的那种情况类似。所以,使用并查集判断区间是否连通,使用前缀和来计算是否满足条件即可。

使用带权并查集,val数组记录前缀和,对于[a, b],fa = find(a), fb = find(b),合并的时候val[fb]只表示[fa, fb]的距离,而在每次find的时候才完成了 val值转换为前缀和+路径压缩。

如果区间两个端点不连通,说明他们的值未知,合并,更新val

如果区间两个端点连通,那么通过前缀和(已经在find同时算出)判断是否与题目中描述相同。

注意多组数据!

代码:

 1 const int inf = 0x3f3f3f3f;
 2 const int maxn = 200010;
 3 int fa[maxn], val[maxn], n, m;
 4 
 5 int find(int x){
 6     if(x == fa[x]) return x;
 7     int tmp = find(fa[x]);
 8     val[x] += val[fa[x]];
 9     fa[x] = tmp;
10     return tmp;
11 }
12 int main(){
13     while(scanf("%d %d", &n, &m) != EOF){
14         for(int i = 0; i <= n; i++){
15             fa[i] = i;
16             val[i] = 0;
17         }
18         int ans = 0;
19         while(m--){
20             int a, b, x;
21             scanf("%d %d %d", &a, &b, &x);
22             a -= 1;
23             int ra = find(a);
24             int rb = find(b);
25             if(ra != rb){
26                 fa[rb] = ra;
27                 val[rb] = val[a] + x - val[b]; 
28             }
29             else{
30                 if(val[b] - val[a] != x) ans++;
31             }
32         }
33         printf("%d\n", ans);
34     }
35 }

题目:

How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9297    Accepted Submission(s): 3377


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).
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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

 























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