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snowy_smile
终于在ICPC World Final里拿到了气球！
再努力一点吧，坚持下去吧，梦想是会实现的！ 人生会有低谷，但我更相信，未来也会有更多梦想能够实现~
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【HDU4474 2012成都现场赛K】【搜索-BFS】Yet Another Multiple Problem 最小的要求字符集的数字倍数 余数哈希BFS
#include#include#include#include#include#include#include#include#include#include#include#include#includeusing namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c:原创 2015-11-04 19:41:58 · 563 阅读 · 0 评论 -
【Codeforces Round 262 (Div 2)E】【爆搜 估价函数剪枝 搜索步长剪枝】Roland and Rose 整格放点 点对欧几里得距离平方和尽可能大
Roland and Rosetime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputRoland loves growing flowers. He has recent原创 2016-03-19 12:39:13 · 714 阅读 · 0 评论 -
【HDU5637 BestCoder Round 74 (div1)B】【bfs预处理】Transform 15个值+17个2的幂数最小异或步数使得s变成t
TransformTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 491 Accepted Submission(s): 171Problem DescriptionA list of n integers原创 2016-03-08 19:14:05 · 750 阅读 · 0 评论 -
【Educational Codeforces Round 10E】【双连通分量缩环 BFS】Pursuit For Artifacts 从ST到ED每条边最多经过一次能否经过任一特殊边
E. Pursuit For Artifactstime limit per test3 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputJohnny is playing a well-known com原创 2016-03-27 10:20:13 · 753 阅读 · 0 评论 -
【Codeforces Round 354 (Div 2)D】【迷宫搜索BFS】Theseus and labyrinth 门门互达 可做旋转操作 最少步数起点到终点
D. Theseus and labyrinthtime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputTheseus has just arrived to Crete原创 2016-05-27 13:23:05 · 866 阅读 · 0 评论 -
【HDU5925 2016CCPC东北地区大学生程序设计竞赛 - 重现赛 D】【离散化 BFS】Coconuts 大地图少数坏点输出每个联通块的大小
CoconutsTime Limit: 9000/4500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 277 Accepted Submission(s): 90Problem DescriptionTanBig, a friend of Mr.原创 2016-10-08 16:36:50 · 1385 阅读 · 1 评论