首页 > 代码库 > poj2021 字符串处理 BFS
poj2021 字符串处理 BFS
Relative Relatives
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 3430 | Accepted: 1504 |
Description
Today is Ted‘s 100th birthday. A few weeks ago, you were selected by the family to contact all of Ted‘s descendants and organize a surprise party. To make this task easier, you created an age-prioritized list of everyone descended from Ted. Descendants of the same age are listed in dictionary order.
The only materials you had to aid you were birth certificates. Oddly enough, these birth certificates were not dated. They simply listed the father‘s name, the child‘s name, and the father‘s exact age when the baby was born.
The only materials you had to aid you were birth certificates. Oddly enough, these birth certificates were not dated. They simply listed the father‘s name, the child‘s name, and the father‘s exact age when the baby was born.
Input
Input to this problem will begin with line containing a single integer n indicating the number of data sets. Each data set will be formatted according to the following description.
A single data set has 2 components:
Note:
A single data set has 2 components:
- Descendant Count - A line containing a single integer X (where 0 < X < 100) indicating the number of Ted‘s descendants.
- Birth Certificate List - Data for X birth certificates, with one certificate‘s data per line. Each certificate‘s data will be of the format "FNAME CNAME FAGE" where:
- FNAME is the father‘s name.
- CNAME is the child‘s name.
- FAGE is the integer age of the father on the date of CNAMEs birth.
- FNAME is the father‘s name.
Note:
- Names are unique identifiers of individuals and contain no embedded white space.
- All of Ted‘s descendants share Ted‘s birthday. Therefore, the age difference between any two is an integer number of years. (For those of you that are really picky, assume they were all born at the exact same hour, minute, second, etc... of their birth year.)
- You have a birth certificate for all of Ted‘s descendants (a complete collection).
Output
For each data set, there will be X+1 lines of output. The first will read, "DATASET Y", where Y is 1 for the first data set, 2 for the second, etc. The subsequent X lines constitute your age-prioritized list of Ted‘s descendants along with their ages using the format "NAME AGE". Descendants of the same age will be listed in dictionary order.
Sample Input
2 1 Ted Bill 25 4 Ray James 40 James Beelzebub 17 Ray Mark 75 Ted Ray 20
Sample Output
DATASET 1 Bill 75 DATASET 2 Ray 80 James 40 Beelzebub 23 Mark 5
这题自己感觉也不难,编译成功提交后老是不过,写了几个版本的。后来看了别人AC的代码,发现自己想多了。今晚重改后一次AC.当然这题有多种写法。dfs、用stl里的list写成bfs都可以
#include<iostream> #include<algorithm> #include<cstring> #include<map> using namespace std; typedef struct{ char name[20]; int age; }Name; typedef struct{ char father[20]; char son[20]; int fage; }Origin; typedef map<string,int> record; int n,k,i,j,len,x; string a,b; bool cmp(const Name& a,const Name& b){ if(a.age>b.age)return true; if(a.age==b.age&&strcmp(a.name,b.name)<0)return true; return false; } int main(){ cin>>n; for(i=1;i<=n;i++){ cin>>x; record re; string s="Ted"; re[s]=100; len=0; Origin ori[100]; Name name[100]; for(j=0;j<x;j++){ cin>>ori[j].father>>ori[j].son>>ori[j].fage; if(re[ori[j].father]&&!re[ori[j].son]){ re[ori[j].son]=re[ori[j].father]-ori[j].fage; len++; } } while(len<x){ for(j=0;j<x&&len<x;j++){ if(re[ori[j].father]&&!re[ori[j].son]){ re[ori[j].son]=re[ori[j].father]-ori[j].fage; len++; } } } for(j=0;j<x;j++){ strcpy(name[j].name,ori[j].son); name[j].age=re[ori[j].son]; } sort(name,name+x,cmp); cout<<"DATASET "<<i<<endl; for(j=0;j<x;j++)cout<<name[j].name<<" "<<name[j].age<<endl; } return 0; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。