http://acm.hdu.edu.cn/showproblem.php?pid=3038
How Many Answers Are Wrong
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3648 Accepted Submission(s):
1401
-________-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.
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)
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.
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<ctype.h>
#define N 200010 int f[N], r[N]; int Find(int x)
{
int k = f[x];
if(x != f[x])
f[x] = Find(f[x]);
r[x] = r[x] + r[k];
return f[x];
} int main()
{
int i, n, m, x, y, w, rx, ry, ans;
while(scanf("%d%d", &n, &m) != EOF)
{
ans = ;
for(i = ; i <= n ; i++)
{
f[i] = i;
r[i] = ;
}
while(m)
{
m--;
scanf("%d%d%d", &x, &y, &w);
x--;
rx = Find(x), ry = Find(y);
if(rx != ry)
{
f[ry] = rx;
r[ry] = w + r[x] - r[y];
}
else
{
if(w != r[y] - r[x])
ans++;
}
}
printf("%d\n", ans);
}
return ;
}