How Many Answers Are Wrong

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24151    Accepted Submission(s): 8191

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  

 

Source 2009 Multi-University Training Contest 13 - Host by HIT  

 

Recommend gaojie   |   We have carefully selected several similar problems for you:  3033 3035 3036 3037 3031   

Statistic | Submit | Discuss | Note

 

题意

思路

CODE

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <iostream>
  5 #include <cmath>
  6 #include <assert.h>
  7 #include <vector>
  8 
  9 #define dbg(x) cout << #x << "=" << x << endl
 10 
 11 using namespace std;
 12 typedef long long LL;
 13 const int maxn = 4e5 + 7;
 14 
 15 int fa[maxn];
 16 int n,m,ans,cnt;
 17 int sum[maxn];
 18 //vector <int> v;
 19 
 20 void init()
 21 {
 22     for(int i = 0; i <= maxn; i++) {
 23         fa[i] = i;
 24     }
 25 }
 26 
 27 int fid(int x)
 28 {
 29     if(x != fa[x]) {
 30         int r = fa[x];
 31         fa[x] = fid(fa[x]);
 32         sum[x] += sum[r];
 33     }
 34     /*int i,j;///路径压缩
 35     i = x;
 36     while(fa[i] != r) {
 37         j = fa[i];
 38         fa[i] =  r;
 39         i = j;
 40     }
 41     return r;*/
 42     return fa[x];
 43 }
 44 
 45 void join(int r1, int r2, int x)///合并
 46 {
 47     int fidroot1 = fid(r1), fidroot2 = fid(r2);
 48     if(fidroot1 == fidroot2) {
 49         if(sum[r1] - sum[r2] != x) {
 50             ++ans;
 51         }
 52     }
 53     else {
 54         fa[fidroot1] = fidroot2;
 55         sum[fidroot1] = sum[r2] - sum[r1] + x;
 56     }
 57 }
 58 
 59 template<class T>inline void read(T &res)
 60 {
 61     char c;T flag=1;
 62     while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;res=c-'0';
 63     while((c=getchar())>='0'&&c<='9')res=res*10+c-'0';res*=flag;
 64 }
 65 
 66 namespace _buff {
 67     const size_t BUFF = 1 << 19;
 68     char ibuf[BUFF], *ib = ibuf, *ie = ibuf;
 69     char getc() {
 70         if (ib == ie) {
 71             ib = ibuf;
 72             ie = ibuf + fread(ibuf, 1, BUFF, stdin);
 73         }
 74         return ib == ie ? -1 : *ib++;
 75     }
 76 }
 77 
 78 int qread() {
 79     using namespace _buff;
 80     int ret = 0;
 81     bool pos = true;
 82     char c = getc();
 83     for (; (c < '0' || c > '9') && c != '-'; c = getc()) {
 84         assert(~c);
 85     }
 86     if (c == '-') {
 87         pos = false;
 88         c = getc();
 89     }
 90     for (; c >= '0' && c <= '9'; c = getc()) {
 91         ret = (ret << 3) + (ret << 1) + (c ^ 48);
 92     }
 93     return pos ? ret : -ret;
 94 }
 95 
 96 int main()
 97 {
 98     while(scanf("%d %d",&n,&m) != EOF) {
 99         init();
100         memset(sum, 0, sizeof(sum));
101         ans = 0;
102         for(int i = 1; i <= m; ++i) {
103             int a,b,x;
104             scanf("%d %d %d",&a, &b, &x);
105             a -= 1;
106             join(a,b,x);
107         }
108         cout << ans << endl;
109     }
110     return 0;
111 }

View Code