NYOJ-115 城市平乱
摘要:城市平乱时间限制:1000 ms | 内存限制:65535 KB难度:4描述南将军统领着N个部队,这N个部队分别驻扎在N个不同的城市。他在用这N个部队维护着M个城市的治安,这M个城市分别编号从1到M。现在,小工军师告诉南将军,第K号城市发生了暴乱,南将军从各个部队都派遣了一个分队沿最近路去往暴乱城市平乱。现在已知在任意两个城市之间的路行军所需的时间,你作为南将军麾下最厉害的程序员,请你编写一个程序来告诉南将军第一个分队到达叛乱城市所需的时间。注意,两个城市之间可能不只一条路。输入第一行输入一个整数T,表示测试数据的组数。(T 2 #include 3 #include 4 5 using...
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posted @
2013-06-29 19:40
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NYOJ-203 三国志
摘要:三国志时间限制:3000ms | 内存限制:65535KB难度:5描述《三国志》是一款很经典的经营策略类游戏。我们的小白同学是这款游戏的忠实玩家。现在他把游戏简化一下,地图上只有他一方势力,现在他只有一个城池,而他周边有一些无人占的空城,但是这些空城中有很多不同数量的同种财宝。我们的小白同学虎视眈眈的看着这些城池中的财宝。按照游戏的规则,他只要指派一名武将攻占这座城池,里面的财宝就归他所有了。不过一量攻占这座城池,我们的武将就要留守,不能撤回。因为我们的小白手下有无数的武将,所以他不在乎这些。从小白的城池派出的武将,每走一公理的距离就要消耗一石的粮食,而他手上的粮食是有限的。现在小白统计出了地
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posted @
2013-03-16 23:13
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HDOJ-3790 最短路径问题
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): 5982Accepted Submission(s): 1803Problem 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,花费为p。最后一行是
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2012-12-01 10:27
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HDU 1869 六度分离 -----Floyd
摘要:六度分离Time Limit: 5000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1791Accepted Submission(s): 696Problem Description1967年,美国著名的社会学家斯坦利·米尔格兰姆提出了一个名为“小世界现象(small world phenomenon)”的著名假说,大意是说,任何2个素不相识的人中间最多只隔着6个人,即只用6个人就可以将他们联系在一起,因此他的理论也被称为“六度分离”理论(six degree
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2012-08-29 15:28
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POJ-1251 Jungle Roads
摘要:Jungle RoadsTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 15610Accepted: 6988DescriptionThe Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so
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2012-08-24 14:08
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HDOJ-1385 Minimum Transport Cost -----路径记录floyd
摘要:Minimum Transport CostTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4503Accepted Submission(s): 1121Problem DescriptionThese are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is s
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2012-08-22 13:05
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HDOJ-2066 一个人的旅行 ------floyd
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): 10134Accepted Submission(s): 3432Problem 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,去阳明山上看海芋,去纽约纯粹看
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posted @
2012-08-21 23:12
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HDOJ-1548 A strange lift -------BFS/dijkstra
摘要:A strange liftTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6572Accepted Submission(s): 2443Problem DescriptionThere is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.T
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2012-08-21 18:30
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POJ-3767 I Wanna Go Home -----有限制的dijkstra
摘要:I Wanna Go HomeTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 2481Accepted: 1035DescriptionThe country is facing a terrible civil war----cities in the country are divided into two parts supporting different leaders. As a merchant, Mr. M does not pay attention to politics but he actually kn
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2012-08-21 17:39
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POJ 1125 Stockbroker Grapevine 弗洛伊德算法
摘要:Stockbroker GrapevineTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 21012Accepted: 11402DescriptionStockbrokers are known to overreact to rumours. You have been contracted to develop a method of spreading disinformation amongst the stockbrokers to give your employer the tactical edge in th
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2012-08-08 15:54
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POJ 1135 Domino Effect 最短路径+枚举
摘要:Domino EffectTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 5539Accepted: 1396DescriptionDid you know that you can use domino bones for other things besides playing Dominoes? Take a number of dominoes and build a row by standing them on end with only a small distance in between. If you do
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posted @
2012-08-07 20:07
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POJ 2387 Til the Cows Come Home --单源最短路径
摘要:Til the Cows Come HomeTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 21302Accepted: 7090DescriptionBessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants
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2012-08-07 20:03
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HDOJ-1142 A Walk Through the Forest
摘要:A Walk Through the ForestTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3140Accepted Submission(s): 1147Problem DescriptionJimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a
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2012-07-22 22:22
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HDOJ-1233 还是畅通工程
摘要:还是畅通工程Time Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13192Accepted Submission(s): 6030Problem Description某省调查乡村交通状况,得到的统计表中列出了任意两村庄间的距离。省政府“畅通工程”的目标是使全省任何两个村庄间都可以实现公路交通(但不一定有直接的公路相连,只要能间接通过公路可达即可),并要求铺设的公路总长度为最小。请计算最小的公路总长度。Input测试输入包含若干测试用例。每个测试用
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posted @
2012-06-30 23:19
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HDOJ-2544 最短路
摘要:#include<stdio.h>int map[101][101];int dijsk(int n){ int visit[101]={0}; int dis[101]; int i,j,k,min; for(i=1;i<=n;++i) dis[i]=map[1][i]; visit[1]=1; for(i=1;i<n;++i) { min=0x7ffffff; k=0; for(j=1;j<=n;++j) { if(!visit[j]&&min>dis[j]) { min=dis[j]; k=j; } } visit[k]=1; for(
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posted @
2012-06-26 23:51
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HDOJ-1874 畅通工程续
摘要:畅通工程续Time Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11370Accepted Submission(s): 3814Problem Description某省自从实行了很多年的畅通工程计划后,终于修建了很多路。不过路多了也不好,每次要从一个城镇到另一个城镇时,都有许多种道路方案可以选择,而某些方案要比另一些方案行走的距离要短很多。这让行人很困扰。现在,已知起点和终点,请你计算出要从起点到终点,最短需要行走多少距离。Input本题目包含多
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posted @
2012-06-14 16:22
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POJ-2502 Subway
摘要:http://poj.org/problem?id=2502这题本来应该用迪杰斯特拉做的 ,因为想试试弗洛伊德所以就试试 谁知道代码一一直wrong ,路过的麻烦指点一下代码一:#include<stdio.h>#include<math.h>struct node{ double x; double y;}node[205];double dis[205][205];double cal(struct node a,struct node b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int mai
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posted @
2012-06-12 21:15
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POJ-1502 MPI Maelstrom
摘要:http://poj.org/problem?id=1502第一篇C++代码学到几个函数:括号里边都是数组名atof() 将字符串转换成浮点数值atoi() 将字符串转换成整数值atol() 将字符串转换成长整数值strtod() 将字符串转换成双精度型数值strtol() 将字符串转换成长型数值用迪杰斯特拉求最短路中的最大值#include<iostream>#include<cstring>using namespace std;int G[102][102],dis[102];void dijs(int n){int k,i,j,t,min;int visit[1
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posted @
2012-06-07 18:19
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