ZOJ3197 Google Book 【贪心】
来源:互联网 发布:黑客wifi破解软件 编辑:程序博客网 时间:2024/05/18 01:36
You, the best hacker in the world, want to download the books published on Google Book. After some investigation, you found that the address of each page consists of two parts. The first part is the page number, the second part is the signature which is unique for each page. To get the signature, you can send the query to the server. The query has one parameter, which indicates the page number. The server will return the signature of the required page, and it may also return the signature of some adjacent pages.
To minimize the bytes downloaded from the internet, and also make the server adminstrator hard to notice your "hack", you'd like to minimize the number of queries
Input
The input has multiple cases.
The first line of the input is a single integer T which is the number of test cases. Then T consecutive test cases follow. In each test case, the first line is a number N (1<=N<=5000), indicating the number of pages of the book. Then n lines follows. On the i-th line, there will be two integers ai and bi (ai<=i<=bi). They indicate that the query for the i-th page will return the signatures from page ai to page bi (inclusive)
Output
Results should be directed to standard output. The output of each test case should be a single integer, which is the minimum number of queries to get all the signatures.
Sample Input
231 12 23 331 11 33 3
Sample Output
31
#include <stdio.h>#include <string.h>#include <algorithm>#define maxn 5002using std::sort;struct Node{ int l, r;} E[maxn];bool cmp(Node a, Node b){ if(a.l == b.l) return a.r > b.r; return a.l < b.l;}int main(){ int t, n, i, j, ans, flag, tmp; scanf("%d", &t); while(t--){ scanf("%d", &n); for(i = 0; i < n; ++i) scanf("%d%d", &E[i].l, &E[i].r); sort(E, E + n, cmp); flag = E[0].r; ans = 1; for(i = 1; i < n; ++i){ tmp = flag; for(j = i; j < n; ++j) if(E[j].l <= flag + 1){ if( E[j].r > tmp) tmp = E[j].r; }else{ flag = tmp; i = j - 1; ++ans; break; } } if(flag != n) ++ans; printf("%d\n", ans); } return 0;}
- ZOJ3197 Google Book 【贪心】
- ZOJ3197
- 【bzoj3119】【book】【贪心】
- ZOJ 3197 Google Book
- google book downloader
- zoj-3197-Google Book
- ZOJ 3197 Google Book
- Firefox浏览器内google book
- New Year Book Reading(贪心)
- BZOJ 3119 Book (贪心+数学推导)
- Google Book Search APIs-----Data API
- zoj 3197 Google Book//最小区间覆盖
- ZOJ Problem Set - 3197 Google Book
- zoj 3197 Google Book 【区间覆盖】
- zoj 3197 Google Book 【区间覆盖】
- book
- book
- book
- GCC添加外部LIB的问题
- IOS 系统打电话事件的监听与状态获取
- 输入一行字符,分别统计出其中英文字母、空格、数字和其它字符的个数。
- centos6.5安装oracle 11g R2
- 的双方的首发你电话疯狂的临时工
- ZOJ3197 Google Book 【贪心】
- maven常见问题问答
- 可视化 笔记
- hdu 1863 kruskal
- Delphi dbgrideh使用方法
- 图像算法
- 排队买票找钱(50元, 100元)
- 2G到4G网络的演进过程
- html5的优势在哪里?