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2013年6月4日

Uva 1456 - Cellular Network 概率dp

摘要: A cellular network is a radio network made up of a number of cells each served by a base station located in the cell. The base station receives call signals from mobile users (mobiles) in the cell it serves, which then connects the calls to the wired land-line telephone network. When a call is reque 阅读全文

posted @ 2013-06-04 11:48 电子幼体 阅读(150) 评论(0) 推荐(0) 编辑

Uva 11552 - Fewest Flops 字符串dp

摘要: Problem FFEWEST FLOPSA common way to uniquely encode a string is by replacing its consecutive repeating characters (or “chunks”) by the number of times the character occurs followed by the character itself. For example, the string “aabbbaabaaaa” may be encoded as “2a3b2a1b4a”. (Note for this problem 阅读全文

posted @ 2013-06-04 10:41 电子幼体 阅读(188) 评论(0) 推荐(0) 编辑

2013年6月3日

Uva 10859 - Placing Lampposts 树形dp

摘要: Input: Standard InOutput: Standard OutNext Generation Contest 1Time Limit: 2 secondsProblem DPlacing LamppostsAs a part of the mission �Beautification of Dhaka City�, the government has decided to replace all the old lampposts with new expensive ones. Since the new ones are quite expensive and the b 阅读全文

posted @ 2013-06-03 15:56 电子幼体 阅读(170) 评论(0) 推荐(0) 编辑

Uva 11825 - Hackers' Crackdown 状态压缩

摘要: Problem HHackers’ CrackdownInput:Standard InputOutput:Standard OutputMiracle Corporations has a number of system services running in a distributed computer system which is a prime target for hackers. The system is basically a set ofNcomputer nodes with each of them running a set ofNservices. Note th 阅读全文

posted @ 2013-06-03 10:42 电子幼体 阅读(172) 评论(0) 推荐(0) 编辑

Uva 10891 - Game of Sum dp博弈

摘要: Problem EGame of SumInput File:e.inOutput:Standard OutputThis is a two player game. Initially there areninteger numbers in an array and playersAandBget chance to take them alternatively. Each player can take one or more numbers from the left or right end of the array but cannot take from both ends a 阅读全文

posted @ 2013-06-03 01:58 电子幼体 阅读(140) 评论(0) 推荐(0) 编辑

高斯消元模板

摘要: int gauss(int r, int c) { bool flag=false; int coe=1; int i=0,t=0; for(int j=0; j0) { index=k; break; } if(a[index][j]) { if(index != i) { for(int k=j; k=j; --l) ... 阅读全文

posted @ 2013-06-03 00:48 电子幼体 阅读(117) 评论(0) 推荐(0) 编辑

2013年6月2日

MUTC 2 E - Save the dwarfs DP?

摘要: Save the dwarfsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 934Accepted Submission(s): 285Problem DescriptionSeveral dwarfs are trapped in a deep well. They are not tall enough to climb out of the well, so they want to make a human-pyramid, that 阅读全文

posted @ 2013-06-02 12:58 电子幼体 阅读(162) 评论(0) 推荐(0) 编辑

MUTC 2 C - Meeting point-2 切比雪夫距离orz

摘要: Meeting point-2Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 889Accepted Submission(s): 485Problem DescriptionIt has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth 阅读全文

posted @ 2013-06-02 12:28 电子幼体 阅读(290) 评论(0) 推荐(0) 编辑

2013年6月1日

MUTC 2 D - Matrix 并查集

摘要: MatrixTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1613Accepted Submission(s): 606Problem DescriptionMachines have once again attacked the kingdom of Xions. The kingdom of Xions has N cities and N-1 bidirectional roads. The road network is such 阅读全文

posted @ 2013-06-01 17:00 电子幼体 阅读(165) 评论(0) 推荐(0) 编辑

MUTC 2 B - Meeting point-1 二分

摘要: Meeting point-1Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2324Accepted Submission(s): 735Problem DescriptionIt has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tent 阅读全文

posted @ 2013-06-01 16:49 电子幼体 阅读(171) 评论(0) 推荐(0) 编辑

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