How Many Answers Are Wrong |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 283 Accepted Submission(s): 134 |
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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) |
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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. |
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Output
A single line with a integer denotes how many answers are wrong. |
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Sample Input
10 5 1 10 100 7 10 28 1 3 32 4 6 41 6 6 1 |
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Sample Output
1 |
思路：好题好题好题好题，网上写的题解是用sum[i]表示前i个数的和，但是我没有这么做（估计是我没有看懂怎么做），那好我说说我的思路：
sum[i]表示i到i祖先的距离差，在一步一步更新时，由于前面的和可以相互抵消，所以距离之差就可以成为判断的依据（好好品味代码吧，我思考了一个晚上），但是要保证的一点是在更新祖先时，一定要让小的祖先指向大的祖先，才能保证不错误。
1 #include <cstdio> 2 #include <iostream> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <cstring> 6 #include <string> 7 using namespace std; 8 9 const int maxn=200100; 10 int f[maxn],sum[maxn],a,b,x,y,fx,fy,n,m,v,ans; 11 12 void close() 13 { 14 exit(0); 15 } 16 17 int find(int k) 18 { 19 if (f[k]==k) 20 return k; 21 int t=f[k]; 22 f[k]=find(f[k]); 23 sum[k]+=sum[t]; 24 return f[k]; 25 } 26 27 void work() 28 { 29 } 30 31 void init () 32 { 33 while(scanf("%d %d",&n,&m)!=EOF) 34 { 35 ans=0; 36 if (n==0 && m==0) break; 37 for (int i=0;i<=n+2;i++) 38 { 39 f[i]=i; 40 sum[i]=0; 41 } 42 for (int i=1;i<=m;i++) 43 { 44 scanf("%d %d %d",&x,&y,&v); 45 y++;//这句很关键 46 fx=find(x); 47 fy=find(y); 48 if (fx==fy && v+sum[y]!=sum[x]) 49 ans++; 50 else 51 if(fx!=fy) 52 { 53 if (fy>fx) 54 { 55 f[fx]=fy; 56 sum[fx]=sum[y]+v-sum[x]; 57 } 58 else 59 { 60 f[fy]=fx; 61 sum[fy]=sum[x]-v-sum[y]; 62 } 63 } 64 } 65 printf("%d\n",ans); 66 } 67 } 68 69 int main () 70 { 71 init(); 72 work(); 73 close(); 74 return 0; 75 }