HDU 1025 Constructing Roads In JGShining's Kingdom LIS题解
来源:互联网 发布:宫崎骏隐喻知乎 编辑:程序博客网 时间:2024/06/10 02:33
本题是LIS题解。主要是理解他的题意。他的题意都好像比较隐晦,比如每个poor city和rich city一定是需要对应起来的,比如poor city和rich city并不是按顺序给出的。
其实是可以把数列按照poor city排序,然后求rich city城市号的最大递增子序列。
不过这里不用排序,利用hash的思想直接对应起来就可以了。
然后就是本题是卡DP的O(n*n)的解法的,这里需要O(nlgn)的LIS解法。
二分好已经找到的递增序列,插入新的数值就可以了。是利用了一个单调队列的思想,故此这样的方法成立。
<span style="white-space:pre"></span>const int MAX_N = 500001;int arr[MAX_N], N;int getIndex(int v, int low, int up){while (low <= up){int mid = low + ((up-low)>>1);if (arr[mid] > v) up = mid-1;else low = mid+1;}return low;}int LIS(){if (N < 1) return 0;int j = 1;for (int i = 2; i <= N; i++){if (arr[i] >= arr[j]) arr[++j] = arr[i];else arr[getIndex(arr[i], 1, j)] = arr[i];}return j;//注意下标由1开始,故此j不用+1}int main(){int p, r, t = 1;while (scanf("%d", &N) != EOF){for (int i = 0; i < N; i++){scanf("%d %d", &p, &r);arr[p] = r;}int lis = LIS();printf("Case %d:\nMy king, at most %d road", t++, lis);if (lis > 1) putchar('s');puts(" can be built.");}return 0;}
1 0
- HDU 1025 Constructing Roads In JGShining's Kingdom LIS题解
- hdu 1025Constructing Roads In JGShining's Kingdom(LIS)
- hdu 1025 Constructing Roads In JGShining's Kingdom(LIS)
- hdu 1025 Constructing Roads In JGShining's Kingdom (LIS)
- hdu 1025 Constructing Roads In JGShining's Kingdom (LIS)
- 【DP|LIS】HDU-1025 Constructing Roads In JGShining's Kingdom
- HDU 1025 Constructing Roads In JGShining's Kingdom(LIS)
- hdu 1025 Constructing Roads In JGShining's Kingdom(lis)
- HDU Problem 1025 Constructing Roads In JGShining's Kingdom 【LIS】
- HDU-1025-Constructing Roads In JGShining's Kingdom【LIS】【二分】
- HDU 1025 Constructing Roads In JGShining's Kingdom(LIS nlogn)
- HDU 1025Constructing Roads In JGShining's Kingdom(LIS)
- HDU 1025 Constructing Roads In JGShining's Kingdom(LIS)
- HDU 1025 Constructing Roads In JGShining's Kingdom ( LIS )
- HDU 1025 Constructing Roads In JGShining's Kingdom LIS -
- Hdu-1025 Constructing Roads In JGShining's Kingdom (LIS)
- hdoj 1025 Constructing Roads In JGShining's Kingdom 【LIS】
- 【HD 1025】Constructing Roads In JGShining's Kingdom(LIS)
- POJ 3280 Cheapest Palindrome (区间dp)
- qq空间尾巴怎么修改成别的机型
- 进程和线程的区别
- 2014 Multi-University Training Contest 8题解
- 韩顺平PHP学习视频笔记整理027apache服务器使用及配置② apache目录结构
- HDU 1025 Constructing Roads In JGShining's Kingdom LIS题解
- Android程序员必备精品资源
- 鼠标右键菜单怎么截图
- 【转】C#:String.Format数字格式化输出
- 技术走向管理一些思考(5)-安静的办公环境
- Codeforces 77C 树形dp + 贪心
- 模仿最新微信主页面UI
- POJ 1226后缀数组:求出现或反转后出现在每个字符串中的最长子串
- 黑马程序员——Java的File类的简单应用,文件的拷贝