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摘要: hdu1232#define _CRT_SECURE_NO_DEPRECATE#include#define MAX 1100int par[MAX], n;int find(int x){ if (par[x] == x){ return x; } return par[x] = find(par[x]);}void united(int x, int y){ int a = find(x), b = find(y); if (a != b){ par[a] = b; }}int main(){ int m, a, b, i, res; ... 阅读全文
posted @ 2014-02-21 20:06 偶尔会寂寞 阅读(149) 评论(0) 推荐(0) 编辑
摘要: 悼念512汶川大地震遇难同胞——珍惜现在,感恩生活Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12417 Accepted Submission(s): 5250Problem Description急!灾区的食物依然短缺!为了挽救灾区同胞的生命,心系灾区同胞的你准备自己采购一些粮食支援灾区,现在假设你一共有资金n元,而市场有m种大米,每种大米都是袋装产品,其价格不等,并且只能整袋购买。请问:你用有限的资金最多能采购多少公斤粮... 阅读全文
posted @ 2014-02-21 18:18 偶尔会寂寞 阅读(694) 评论(0) 推荐(0) 编辑
摘要: Red and BlackTime Limit:1000MSMemory Limit:30000KTotal Submissions:20512Accepted:10986DescriptionThere is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can' 阅读全文
posted @ 2014-02-19 21:34 偶尔会寂寞 阅读(221) 评论(0) 推荐(0) 编辑
摘要: 算法提高 最小方差生成树 时间限制:1.0s 内存限制:256.0MB问题描述给定带权无向图,求出一颗方差最小的生成树。输入格式输入多组测试数据。第一行为N,M,依次是点数和边数。接下来M行,每行三个整数U,V,W,代表连接U,V的边,和权值W。保证图连通。n=m=0标志着测试文件的结束。输出格式对于每组数据,输出最小方差,四舍五入到0.01。输出格式按照样例。样例输入4 51 2 12 3 23 4 24 1 12 4 34 61 2 12 3 23 4 34 1 12 4 31 3 30 0样例输出Case 1: 0.22Case 2: 0.00数据规模与约定1#include#incl. 阅读全文
posted @ 2014-02-19 16:55 偶尔会寂寞 阅读(365) 评论(0) 推荐(0) 编辑
摘要: Prime Ring ProblemTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22702 Accepted Submission(s): 10108Problem DescriptionA ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the ... 阅读全文
posted @ 2014-02-19 08:10 偶尔会寂寞 阅读(204) 评论(0) 推荐(0) 编辑
摘要: SafecrackerTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7305 Accepted Submission(s): 3680Problem Description=== Op tech briefing, 2002/11/02 06:42 CST === "The item is locked in a Klein safe behind a painting in the second-floor library. . 阅读全文
posted @ 2014-02-18 21:50 偶尔会寂寞 阅读(206) 评论(0) 推荐(0) 编辑
摘要: A Walk Through the ForestTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5071 Accepted Submission(s): 1846Problem DescriptionJimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax... 阅读全文
posted @ 2014-02-18 15:49 偶尔会寂寞 阅读(359) 评论(0) 推荐(0) 编辑
摘要: 一个人的旅行Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 16100 Accepted Submission(s): 5521Problem Description虽然草儿是个路痴(就是在杭电待了一年多,居然还会在校园里迷路的人,汗~),但是草儿仍然很喜欢旅行,因为在旅途中 会遇见很多人(白马王子,^0^),很多事,还能丰富自己的阅历,还可以看美丽的风景……草儿想去很多地方,她想要去东京铁塔看夜景,去威尼斯看电影,去阳明山上看海... 阅读全文
posted @ 2014-02-18 14:31 偶尔会寂寞 阅读(1105) 评论(0) 推荐(0) 编辑
摘要: 最短路径问题Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9819 Accepted Submission(s): 2987Problem Description给你n个点,m条无向边,每条边都有长度d和花费p,给你起点s终点t,要求输出起点到终点的最短距离及其花费,如果最短距离有多条路线,则输出花费最少的。 Input输入n,m,点的编号是1~n,然后是m行,每行4个数 a,b,d,p,表示a和b之间有一条边,且其长度为d,花... 阅读全文
posted @ 2014-02-18 13:18 偶尔会寂寞 阅读(427) 评论(0) 推荐(0) 编辑
摘要: find the safest roadTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5628 Accepted Submission(s): 2005Problem DescriptionXX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1 间的实数(包括0,1),一条从u 到 v 的通道P 的安全度为Safe(P) = s(e1)*s(e2)…*... 阅读全文
posted @ 2014-02-18 08:56 偶尔会寂寞 阅读(231) 评论(0) 推荐(0) 编辑
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