set, map, string, find(), string name[100],等的混合
There are n single boys and m single girls. Each of them may love none, one or several of other people unrequitedly and one-sidedly. For the comingq days, each night some of them will come together to hold a single party. In the party, if someone loves all the others, but is not loved by anyone, then he/she is called king/queen of unrequited love.
Input
There are multiple test cases. The first line of the input is an integer T ≈ 50 indicating the number of test cases.
Each test case starts with three positive integers no more than 30000
--nmq
. Then each of the nextn lines describes a boy, and each of the nextm lines describes a girl. Each line consists of the name, the number of unrequitedly loved people, and the list of these people's names. Each of the lastq lines describes a single party. It consists of the number of people who attend this party and their names. All people have different names whose lengths are no more than20
. But there are no restrictions that all of them are heterosexuals.
Output
For each query, print the number of kings/queens of unrequited love, followed by their names in lexicographical order, separated by a space. Print an empty line after each test case. See sample for more details.
Sample Input
2 2 1 4 BoyA 1 GirlC BoyB 1 GirlC GirlC 1 BoyA 2 BoyA BoyB 2 BoyA GirlC 2 BoyB GirlC 3 BoyA BoyB GirlC 2 2 2 H 2 O S He 0 O 1 H S 1 H 3 H O S 4 H He O S
Sample Output
0 0 1 BoyB 0 0 0
题目大意:给定一些关系,对于每个人(男孩或女孩),列出他所喜欢的人(允许同性恋),对于每次询问(聚会),求这样一种人:他喜欢所有人,但所有人都不喜欢他
分析:简单分析可知,这种人假如存在,最多只有一个。因为假设有2个这样的人,他们彼此就与题意矛盾。故可以枚举这个人,如何快速枚举?
对于一次聚会,先把第一个人假设为这种人,遍历其后的人,与当前这个人判断关系,若发现这个人不可能是这种人,则把当前遍历的更新为这种人。
扫一遍后,再判断这个人是否真的是,只要和他前面所有的人判断一下即可
#include<cstdio> #include<string> #include<set> #include<map> using namespace std; const int N=30005; map<string,int> M; map<string,int>::iterator it; set< pair<int,int> > S; string name[N]; int tol,party[N]; char na[22]; int hash(char *s){ //关键的一段函数,许多注意点 it=M.find(s); if(it!=M.end())return it->second; else { name[++tol]=s; return M[s]=tol; } } void Cin(int x){ int i,k,u,v; for(i=0;i<x;i++){ scanf("%s%d",na,&k); u=hash(na); while(k--){ scanf("%s",na); v=hash(na); S.insert(make_pair(u,v)); } } } int main(){ int T,n,m,q,i,k,ans; scanf("%d",&T); while(T--){ scanf("%d%d%d",&n,&m,&q); M.clear(),S.clear(),tol=0; Cin(n),Cin(m); while(q--){ scanf("%d%s",&k,na); ans=party[0]=M[na]; int p=0; for(i=1;i<k;i++){ scanf("%s",na),party[i]=M[na]; if(S.find(make_pair(ans,party[i]))==S.end()||S.find(make_pair(party[i],ans))!=S.end()){ //查找核心 ans=party[i],p=i; } } for(i=0;i<p;i++){ if(S.find(make_pair(ans,party[i]))==S.end()||S.find(make_pair(party[i],ans))!=S.end())break; } if(i!=p)puts("0"); else printf("1 %s\n",name[ans].c_str()); //输出注意点 } puts(""); } return 0; }
name[]数组的输出:不能直接写成 printf("1 %s\n",name[ans]),会出错,需要在后面加上“.c_str()”;
map<string,int>类型的容器,前面的string下标也可以用char类型的字符数组表示;
find(map和set都是可以用的)找不到就是==*.end(),找到就是!=*.end();
make_pair的用法,以及 map<string,int>::iterator it; 中it->second的用法
map清空也用clear;
传入函数中的字符数组用了“*”,并且后面没有“[]”;
摘自:http://blog.csdn.net/libing923/article/details/8902193
#include <iostream> #include<cstdio> #include<set> #include<map> using namespace std; map<string,int> mp; set< pair<int,int> > s; string name[30005]; int n,m,p,l; int party[30005]; int num(char *ch) { map<string,int>::iterator it; it=mp.find(ch); if (it!=mp.end()) return mp[ch]; else { name[++l]=ch; mp[ch]=l; return l; } } void read(int n) { char ch[26],str[26]; for(int i=1;i<=n;i++) { int k; scanf("%s%d",&ch,&k); int u=num(ch); for(;k>0;k--) { scanf("%s",&str); int v=num(str); s.insert(make_pair(u,v)); } } return; } int main() { int t; while(~scanf("%d",&t)) { for(;t>0;t--) { scanf("%d%d%d",&n,&m,&p); l=0; mp.clear(); s.clear(); read(n); read(m); for(int i=1;i<=p;i++) { int k; char ch[26]; scanf("%d",&k); for(int j=1;j<=k;j++) { scanf("%s",&ch); party[j]=mp[ch]; } bool flag; for(int j=1;j<=k;j++) { flag=1; for(int z=1;z<=k;z++) { if (j==z) continue; if (s.find(make_pair(party[j],party[z]))==s.end()) {flag=0;break;} if (s.find(make_pair(party[z],party[j]))!=s.end()) {flag=0;break;} } if (flag) {printf("1 %s\n",name[party[j]].c_str());break;} } if (!flag) printf("0\n"); } printf("\n"); } } return 0; }
上面是自己写的,在上一个程序的借鉴上。