查环操作,裸题.一次AC. #include <iostream> #include <cstring> #include <cstdlib> #include <cctype> #include <cmath> #include <string> #include <cstdio> #include <algorithm> #include <numeric> using namespace st…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2122 Description ice_cream’s world becomes stronger and stronger; every road is built as undirected. The queen enjoys traveling around her world;…

题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=2120 Ice_cream's world I Description ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are s…

题目链接: 题目 Ice_cream's world II Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 问题描述 After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it a…

Ice_cream's world I Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 741    Accepted Submission(s): 429 Problem Description ice_cream's world is a rich country, it has many fertile lands. Today,…

并查集+最小生成树 Ice_cream’s world III Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1121    Accepted Submission(s): 365 Problem Description ice_cream’s world becomes stronger and stronger; every roa…

Ice_cream’s world III Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 576    Accepted Submission(s): 185 Problem Description ice_cream’s world becomes stronger and stronger; every road is built…

Ice_cream's world I Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description ice_cream's world is a rich country, it has many fertile lands. Today, the q…

Ice_cream's world I Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 839    Accepted Submission(s): 488 Problem Description ice_cream's world is a rich country, it has many fertile lands. Today,…

Ice_cream’s world III Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 6   Accepted Submission(s) : 3 Problem Description ice_cream’s world becomes stronger and stronger; every road is built as un…

称号:pid=2121" target="_blank">hdoj 2121 Ice_cream's world II 题意:题目是一道躶题,给n个点,m条边的有向图.然后找一个点.到全部点的距离和最小.找出这个点并输入距离. 分析:非常明显是求一个最小树形图,可是没有说根节点.要找跟节点,我们能够虚拟一个节 点 x .x 到全部节点连边距离为前面全部距离和+1为 dis . 然后从x 节点求一次最小树形图为ans,则ans - dis 就是最小树形图的距离. 假设图不…

开始学习最小树形图,模板题. Ice_cream’s world II Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in…

hdu2121 Ice_cream's world II 给一个有向图,求最小树形图,并输出根节点 \(n\leq10^3,\ m\leq10^4\) 最小树形图 对于求无根最小树形图,可以建一个虚拟节点,连向其他所有节点,权值为 \(\inf\) ,最后的答案即为 \(ans-\inf\) .无解当且仅当 \(ans>\inf\times2\) 至于求根节点,可以考虑记下前驱为虚拟节点的节点,但由于节点编号不断改变,因此只需记下连接两个节点的边即可. 时间复杂度 \(O(nm)\) 代码 #i…

Problem Description ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partit…

Ice_cream’s world II http://acm.hdu.edu.cn/showproblem.php?pid=2121 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6736    Accepted Submission(s): 1779 Problem Description After awarded lands t…

点击打开链接 ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_c…

Ice_cream’s world II http://acm.hdu.edu.cn/showproblem.php?pid=2121 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description After awarded lands to ACMers, the queen want to choose a city be her capital.…

Ice_cream’s world II Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6849    Accepted Submission(s): 1818 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2121 Description: After awarded lands to…

Ice_cream’s world II Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2121 Description After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in…

Ice_cream's world I ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partit…

Description ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2121 Ice_cream’s world II Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5832    Accepted Submission(s): 1493 Problem Description After awarded la…

Ice_cream's world III Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 10   Accepted Submission(s) : 3 Problem Description ice_cream's world becomes stronger and stronger; every road is built as u…

解题思路:给出n对点的关系,求构成多少个环,如果对于点x和点y,它们本身就有一堵墙,即为它们本身就相连,如果find(x)=find(y),说明它们的根节点相同,它们之间肯定有直接或间接的相连,即形成环 样例的示意图 共3个环 Ice_cream's world I Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 642    Acce…

解题思路:基础的最小生成树反思:不明白为什么i从1开始取,就一直WA,难道是因为村庄的编号是从0开始的吗 Ice_cream’s world III Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1032    Accepted Submission(s): 335 Problem Description ice_cream’s wo…

Ice_cream's world III Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1237    Accepted Submission(s): 408 Problem Description ice_cream's world becomes stronger and stronger; every road is buil…

[解题思路]这题先看了NotOnlySuccess的解题思路,即设置虚根再处理的做法:弄了一个上午,再次有种赶脚的感觉~~如果需要找出为什么需要去比所有权值之和更大的数为新增的虚边的话,一开始我理解仅是找比最大边权值要大的数就够了,因为这样就已经满足了不会提前找到虚边的情况,但如果是这样的话那就很难判断这样一种情况:生成的最小树形图(含虚根)有两条边是从虚根出来的. #include<cstdio> #include<cstring> #define SIZE 1002 #defi…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2121 题目意思大概是要你在一些城市中选一个做首都 , 要求首都都能到其他城市 , 道路花费要最少 , 且道路都是单向的 , 这个时候就要用到最小树形图算法了 , 而且是不固定根. 不定根就是加一个虚根(原本不存在的点) , 可以让这个虚根到每个点的距离大于原本所有点连接的道路花费之和sum , 然后计算出的结果减去sum,如果比sum还大就可以认为通过这个虚拟节点我们连过原图中两个点,即原图是不连通…

这个题就是需要求整个有向带权图的最小树形图,没有指定根,那就需要加一个虚根 这个虚根到每个点的权值是总权值+1,然后就可以求了,如果求出来的权值大于等于二倍的总权值,就无解 有解的情况,还需要输出最根,多解的情况,肯定是某个环上所有的点为根都可以(比如所有的点构成一个环), 这样加边的时候虚边的时候按照点的标号从小到大编,这样第一个以虚根为前驱的点也是最小的点就可以标记(标记一下) #include <iostream> #include <algorithm> #include…

2坑,3次WA. 1.判断重边取小.2.自边舍去. (个人因为vis数组忘记初始化,WA了3次,晕死!!) #include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <cmath> #include <cctype> #include <algorithm> #include <numeric> #d…