随笔分类 - 图论
摘要:Ice_cream’s world II Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5486 Accepted Submission(s):
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摘要:关于为什么不能用Prim求解此类问题,如下 Prim可以看成是维护两个顶点集或者看成维护一颗不断生成的树(感觉前一种说法好一点) 倘若是有向图有三个顶点1.2.3 边的情况如下 1->2: 5 1->3: 6 2->3: 1000861 3->2: 2 显然若是按照Prim算法来说,先将顶点一压入集
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摘要:http://www.cnblogs.com/kuangbin/p/3147329.html #include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N=110;const int INF=0
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摘要:Picnic Planning Time Limit: 5000MS Memory Limit: 10000K Total Submissions: 10642 Accepted: 3862 Description The Contortion Brothers are a famous set o
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摘要:Minimal Ratio TreeTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4462 Accepted Submission(s): 14
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摘要:After going through the receipts from your car trip through Europe this summer, you realised that the gas prices varied between the cities you visited
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摘要:Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their
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摘要:For their physical fitness program, N (2 ≤ N ≤ 1,000,000) cows have decided to run a relay race using the T (2 ≤ T ≤ 100) cow trails throughout the pa
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摘要:Description Tour operator Your Personal Holiday organises guided bus trips across the Benelux. Every day the bus moves from one city S to another city
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摘要:Description Christmas is coming to KCM city. Suby the loyal civilian in KCM city is preparing a big neat Christmas tree. The simple structure of the t
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摘要:Remmarguts' Date Time Limit: 4000MS Memory Limit: 65536K Total Submissions: 30017 Accepted: 8159 Description "Good man never makes girls wait or break
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摘要:1040: [ZJOI2008]骑士 Description Z国的骑士团是一个很有势力的组织,帮会中汇聚了来自各地的精英。他们劫富济贫,惩恶扬善,受到社会各 界的赞扬。最近发生了一件可怕的事情,邪恶的Y国发动了一场针对Z国的侵略战争。战火绵延五百里,在和平环境 中安逸了数百年的Z国又怎能抵挡的住Y
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摘要:https://wenku.baidu.com/view/ce296043192e45361066f575.html //仙人掌图基础知识3个判定条件 http://blog.csdn.net/yihuikang/article/details/7904347 //参考代码 题目:HDU 3594
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摘要:1016: [JSOI2008]最小生成树计数 Description 现在给出了一个简单无向加权图。你不满足于求出这个图的最小生成树,而希望知道这个图中有多少个不同的 最小生成树。(如果两颗最小生成树中至少有一条边不同,则这两个最小生成树就是不同的)。由于不同的最小生 成树可能很多,所以你只需要输
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