标签：HDU 3038 very Many TT answers wrong FF include
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
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
标签：HDU,3038,very,Many,TT,answers,wrong,FF,include 来源： https://www.cnblogs.com/orangeko/p/12268284.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。