HDU1027,CF501B,HDU1716,HDU4277)
1.泛型程序设计简介与迭代器的介绍
1.1 泛型程序设计简介
泛型程序设计,简单地说就是使用模板的程序设计法。将一些常用的数据结构(比如链表,数组,二叉树)和算法(比如排序,查找)写成模板,以后则不论数据结构里放的是什么对象,算法针对什么样的对象,则都不必重新实现数据结构,重新编写算法。
总而言之,不多赘述,有了STL,不必再从头写大多的标准数据结构和算法,并且可获得非常高的性能。
1.2迭代器
操作 | 效果 |
begin() | 返回一个迭代器,指向第一个元素 |
end() | 返回一个迭代器,指向最后一个元素之后 |
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 10474 - Where is the Marble? 总结:sort使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 10000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; int arr[MAXN]; int hash[MAXN]; int main() { // freopen("in.txt","r",stdin); int n,q,ca=1; while(scanf("%d %d",&n,&q)&&(n+q)) { mem(hash,-1); bool flag=false; printf("CASE# %d:\n",ca++); Rep(i,n) scanf("%d",&arr[i]); sort(arr,arr+n); Rep(i,n) { if(arr[i]!=arr[i-1])//不加的话会wa hash[arr[i]]=i+1; } int ques; Rep(i,q) { flag=false; scanf("%d",&ques); if(hash[ques]!=-1) printf("%d found at %d\n",ques,hash[ques]); else printf("%d not found\n",ques); } } return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 10815 - Andy's First Dictionary 总结:set使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; set<string>jihe; int main() { // freopen("in.txt","r",stdin); string sen; while(getline(cin,sen)) { for(int i=0;i<sen.size();i++) { if(!isalpha(sen[i])) continue; string tmp; while(isalpha(sen[i])) { tmp+=tolower(sen[i]); i++; } jihe.insert(tmp); } } for(set<string>::iterator it=jihe.begin();it!=jihe.end();it++) { cout<<*it<<endl; } return 0; }<strong> </strong>
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 156 - Ananagrams 总结:map+set+vector使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; map<string,int> hash;//标准句子出现次数 string record[MAXN];//所有原来的句子 set<string> ans;//所有答案 void standard(string& s1) { string ans; Rep(i,s1.size()) s1[i]=tolower(s1[i]); sort(s1.begin(),s1.end()); } int main() { // freopen("in.txt","r",stdin); string sen; int index=0; while(getline(cin,sen)) { if(sen[0]=='#') break; Rep(i,sen.size()) { string tmp; if(!isalpha(sen[i]))continue; while(isalpha(sen[i])) { tmp+=sen[i]; i++; } record[index++]=tmp; standard(tmp); hash[tmp]++; } } Rep(i,index) { string tmp=record[i]; standard(tmp); if(hash[tmp]==1) ans.insert(record[i]); } for(set<string>::iterator it=ans.begin(); it!=ans.end(); it++) cout<<*it<<endl; return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 156 - Ananagrams 总结:multimap使用练习 Tricks:string利用构造函数拷贝 **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; multimap<string,string> hash;//从标准到原句子 set<string> ans;//所有标准句子 void standard(string& s1) { string ans; Rep(i,s1.size()) { s1[i]=tolower(s1[i]); } sort(s1.begin(),s1.end()); } int main() { freopen("in.txt","r",stdin); string sen; while(getline(cin,sen)) { if(sen[0]=='#') break; Rep(i,sen.size()) { string tmp; if(!isalpha(sen[i]))continue; while(isalpha(sen[i])) { tmp+=sen[i]; i++; } string ttmp=tmp; standard(tmp); hash.insert(MP(tmp,ttmp)); } } for(map<string,string>::iterator it=hash.begin();it!=hash.end(); it++) { // cout<<it->second<<" "<<hash.count(it->second)<<endl; if(hash.count(it->first)==1) ans.insert(it->second); } for(set<string>::iterator it=ans.begin(); it!=ans.end(); it++) { cout<<*it<<endl; } return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 540 - Team Queue 总结:queue使用练习 Tricks:可能重复入队出队 **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; int main() { // freopen("in.txt","r",stdin); int T; int ca=1; while(scanf("%d",&T)&&T) { printf("Scenario #%d\n",ca++); queue<int>zhudui; queue<int>meige[MAXN]; vector<int>record[MAXN]; bool zhuvis[MAXN]; mem(zhuvis,0); Rep(j,T) { int n; scanf("%d",&n); Rep(i,n) { int tmp; scanf("%d",&tmp); record[j].PB(tmp); } } string s1; int num; while(cin>>s1) { if(s1=="STOP") break; if(s1=="ENQUEUE") { cin>>num; int index=-1; Rep(i,T) { vector<int>::iterator result=find(record[i].begin(),record[i].end(),num); if(result==record[i].end()) //没找到 continue; else { index=i; break; } } if(index!=-1) { meige[index].push(num);//某人入某队 if(!zhuvis[index]) { zhuvis[index]=true; zhudui.push(index);//该队入主队 } } } else if(s1=="DEQUEUE") { int duii=zhudui.front(); int ansi=meige[duii].front(); meige[duii].pop(); if(!meige[duii].size()) { zhudui.pop(); zhuvis[duii]=false;//没有的话会RE } cout<<ansi<<endl; } } cout<<endl; } return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 136 - Ugly Numbers 总结:优先队列使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; int ref[3]={2,3,5}; using namespace std; int main() { // freopen("in.txt","r",stdin); priority_queue<LL,vector<LL> ,greater<LL> >pq; set<LL>record; pq.push(1); record.insert(1); int ca=1; while(!pq.empty()) { LL tmp=pq.top(); pq.pop(); if(ca==1500) { printf("The 1500'th ugly number is %lld.\n",tmp); break; } Rep(i,3) { LL ttmp=tmp*ref[i]; if(!record.count(ttmp)) { record.insert(ttmp); pq.push(ttmp); } } ca++; } return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: UVA 136 - Ugly Numbers 总结:set使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define viti (vector<int>::iterator) #define MP(a, b) make_pair(a, b) #define PB push_back #define MID(a, b) (a + ((b - a) >> 1)) 1 #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int INF = 0x3f3f3f3f; const double eps = 1e-8; const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; set<LL>Ch; int ref[3]={2,3,5}; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int ca=1; Ch.insert(1); set<LL>::iterator it=Ch.begin(); while(ca<=2000) { it=Ch.begin(); for(it;it!=Ch.end()&&ca<=2000;it++) { Rep(j,3) { LL tmp=(*it)*ref[j]; if(Ch.find(tmp)==Ch.end()) { Ch.insert(tmp); ca++; } } } } ca=1; it=Ch.begin(); for(it;it!=Ch.end()&&ca++<1500;it++); printf("The 1500'th ugly number is %lld.\n",*it); return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: HDU 1027 - Ignatius and the Princess II 总结:next_permutation使用练习 Tricks: **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; int arr[MAXN]; using namespace std; int main() { // freopen("in.txt","r",stdin); int n,m; while(~scanf("%d%d",&n,&m)) { for(int i=1;i<=n;i++) arr[i]=i; int ca=1; do { if(ca==m) { int i; for(i=1;i<n;i++) { cout<<arr[i]<<" "; } cout<<arr[i]<<endl; break; } ca++; }while(next_permutation(arr+1,arr+n+1)); } return 0; }
/********************************* 日期:2015-04-03 作者: 题号: HDU 1716 - 排列2 总结:next_permutation的使用 Tricks:0开始的忽略 重复的不计 写吐了,while里面乱搞的。。。出题不善,大家跳过这个题吧 **********************************/ #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <cstring> #include <set> #include <vector> #include <map> #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define N 10010 using namespace std; int vis[12]; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); int a[6]; scanf("%d %d %d %d",&a[1],&a[2],&a[3],&a[4]); if(!(a[1]+a[2]+a[3]+a[4])) return 0; do { bool mark=false; mem(vis,0); for(int i=1; i<=4; i++) { vis[a[i]]++; } sort(a+1,a+5); int tmp=a[1]; int cnt=1; do { if(a[1]==0) continue; if(a[1]!=tmp) { mark=false; if(tmp!=0) cout<<endl; tmp=a[1]; } else if(vis[a[1]]>1) { vis[a[1]]--; cnt++; } if(mark) { cout<<" "; } else mark=true; cout<<a[1]<<a[2]<<a[3]<<a[4]; }while(next_permutation(a+1,a+5)); cout<<endl; scanf("%d %d %d %d",&a[1],&a[2],&a[3],&a[4]); if(a[1]+a[2]+a[3]+a[4]) cout<<endl; else break; }while(1); return 0; }
/********************************* 日期:2015-04-05 作者:matrix68 题号: Codefores 501B - Misha and Changing Handles 总结:map使用练习 Tricks:建立新名字到旧名字的映射,把所有新名字放在set里排重。 **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long const double PI = acos(-0); const int MAXN = 1000 + 10; const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; map<string,string>x_j;//新名字到旧名字的映射 set<string>newn;//新名字 int main() { // freopen("in.txt","r",stdin); int T; scanf("%d",&T); string s1,s2; while(T--) { cin>>s1>>s2; if(!(newn.find(s1)==newn.end()))//新名字集合里找到s1 { x_j[s2]=x_j[s1]; x_j[s1]=x_j[s1]; x_j.erase(s1); newn.erase(s1); newn.insert(s2); } else { newn.insert(s2); x_j[s2]=s1; } } cout<<newn.size()<<endl; for(set<string>::iterator it=newn.begin();it!=newn.end();it++) { cout<<x_j[*it]<<" "<<*it<<endl; } return 0; }
/********************************* 日期:2015-04-03 作者:matrix68 题号: HDU 4277 - USACO ORZ 总结:DFS+set Tricks:注意用long long **********************************/ #include <cstdio> #include <set> #include <iostream> #include <string> #include <vector> #include <queue> #include <cstring> #include <iomanip> #include <algorithm> #include <cctype> #include <string> #include <map> #include <cmath> #define SZ(x) (int)x.size() #define MP(a, b) make_pair(a, b) #define PB push_back #define Lowbit(x) ((x) & (-x)) #define Rep(i,n) for(int i=0;i<n;i++) #define mem(arr,val) memset((arr),(val),(sizeof (arr))) #define LL long long #define N 20 #define M 1000000 const double PI = acos(-0); const int MOD = 1000007; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; using namespace std; set<LL>jihe; bool ok(int a,int b,int c) { //这里可以防止重复也可以存的时候sort一下 return a+b>c && b+c>a && a+c>b && a>=b && b>=c && a!=0 && b!=0 && c!=0; } int n; int side[N]; void DFS(int t,int a,int b,int c) { if(t==n) { if(ok(a,b,c)) { LL p1=M*a+b; jihe.insert(p1); } return; } a+=side[t]; DFS(t+1,a,b,c); a-=side[t]; b+=side[t]; DFS(t+1,a,b,c); b-=side[t]; c+=side[t]; DFS(t+1,a,b,c); c-=side[t]; } int main() { // freopen("in.txt","r",stdin); int T; scanf("%d",&T); while(T--) { jihe.clear(); scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&side[i]); } DFS(0,0,0,0); cout<<jihe.size()<<endl; } return 0; }