1263.水果
#include<iostream>
#include<string>
#include<map>
using namespace std;
//map<string,map<string,int>>这种比较新奇，解决了三个变量的键值对应
struct MyStruct
{
map <string, int>MyStructma; //存放水果名以及该种水果的数量
};
int main()
{
map <string, MyStruct>ma; //地名
map <string, MyStruct>::iterator it;
map <string, int>::iterator MyStructmait;
string fruit,place;
int count;
int n,t;
cin>>t;
while(t--)
{
ma.clear();
cin>>n;
while(n--)
{
cin>>fruit>>place>>count;
ma[place].MyStructma[fruit] += count;
}
for (it = ma.begin(); it != ma.end(); it++)
{
cout<<it->first<<endl;
for (MyStructmait = it->second.MyStructma.begin(); MyStructmait != it->second.MyStructma.end(); MyStructmait++)
{
cout<<" |----"<<MyStructmait->first<<"("<<MyStructmait->second<<")"<<endl;
}
}
if(t != 0)cout<<endl;
}
return 0;
}
双重map：参考
错误：
#include <iostream>
#include <string>
#include <map>
using namespace std;
int main()
{
int n,m,i;
cin>>n;
while(n--){
cin>>m;
string fruit,city;
int num;
map<string, int> str1;//fruit,num
map<string,string> str2;//fruit,city
set<string> cstr;
for(i=0;i<m;i++){
cin>>fruit>>city>>num;
str1[fruit]+=num;
cstr.insert(city);
str2[fruit]=city;
}
for (auto iter = cstr.begin(); iter != cstr.end(); ++iter) {
cout <<*iter<<endl;
for (auto it = str2.begin(); it != str2.end(); ++it){
if(it->second == *iter){
cout <<" |----" <<it->first<<"("<<str1[it->first]<<")"<<endl;
}
}
}
if(n!=0) cout<<endl;
}
return 0;
}
1075.What Are You Talking About
#include<string>
#include<iterator>
#include<map>
#include<iostream>
using namespace std;
int main()
{
string a,b;
map<string,string> val;
getline(cin, a);
while(cin>>a){
if(a=="END") break;
else {
cin>>b;
val[b]=a;
}
}
getline(cin, a);
getline(cin, a);
string ss="";
while(1)
{
getline(cin, a);//a=mar
if(a=="END") break;
else
ss=ss+a+"\n";
}
string temp="";
for(int i=0;i<ss.length();i++){
//mar的长度
if('a'<=ss[i]&&ss[i]<='z'){
temp+=ss[i];//可以翻译的mar存在temp中
}else{
map<string,string>::iterator it=val.find(temp);
if(it!=val.end())
cout<<it->second;
else
cout<<temp;
temp="";
cout<<ss[i];
}
}
}
1006.Let the Balloon Rise
#include<bits/stdc++.h>
using namespace std;
map<string, int> count1;
int main()
{
int n;
int max = 0;
string pop;
while (cin>>n&&n) {
while (n--) {
string color;
cin >> color;
count1[color]++;
if (count1[color] > max) {
max = count1[color];
pop = color;
}
}
cout << pop << endl;
max = 0;
}
}
set
set容器中只能存储键，是单纯的键的集合，其中键是不能重复的。
set支持大部分的map的操作，但是set不支持下标的操作，而且没有定义mapped_type类型。
按照个数升序排列。
#include <cstdio>
#include <set>
using namespace std;
int main()
{
int n,i;
while(scanf("%d", &n)==1 && n!=0){
string s;
set<string> str;
for(i=0;i<n;i++){
cin >> s;
str.insert(s);
}
cout << *str.rbegin()<<endl;
}
return 0;
}
1029.Ignatius and the Princess IV
#include<stdio.h>
#include<vector>
#include<algorithm>
#define N 1010000
using namespace std;
int p[N],f[N];
int main(void)
{
int n,i;
vector<int>a;//定义数组
while(scanf("%d",&n)==1){
for(i=1;i<=n;i++){
scanf("%d",&p[i]);
a.push_back(p[i]);//尾部插入数组
}
for(i=1;i<=n;i++){
//count函数用来统计字符串中某个字符的个数
f[i]=count(a.begin(),a.end(),p[i]);//起始地址，结束地址，需要查找的字符
if(f[i]>=(n+1)/2){
printf("%d\n",p[i]);
break;
}
}
a.clear();
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int i,n,cnt,anss,x;
int main()
{
while(scanf("%d",&n)!=EOF)
{
cnt=0;
for(i=1;i<=n;i++)
{
cin>>x;
if(cnt==0)
{
cnt++;
anss=x;
}
else
{
if(x==anss)
cnt++;
else
cnt--;
}
}
cout<<anss<<endl;
}
}
1412.{A} + {B}
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
using namespace std;
int main()
{
int n,m;
while(cin>>n>>m){
set<int> q;
int t,i;
for(i=0;i<n;i++){
scanf("%d",&t);
q.insert(t);
}
for(i=0;i<m;i++){
scanf("%d",&t);
q.insert(t);
}
/*set<int>::iterator it;
for (it = q.begin(); it != q.end(); it++)
cout << *it <<" ";
*/
//输出set，格外注意输出格式
int f=0;
for(auto i:q){
if(f==0)
printf("%d",i);
else
printf(" %d",i);
f=1;
}
cout<<"\n";
}
return 0;
}
4286.Data Handler
方法1
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 500000+10;
#define REP(_,a,b) for(int _ = (a); _ <= (b); _++)
int n,m,dir;
deque<int> dqi,lft,rgt;
int Lp,Rp;
int num[maxn];
void init() {
dqi.clear();
lft.clear();
rgt.clear();
dir = 1;
}
void input() {
scanf("%d",&n);
REP(i,1,n) scanf("%d",&num[i]);
scanf("%d%d",&Lp,&Rp);
REP(i,1,Lp-1 ) lft.push_back(num[i]);
REP(i,Lp,Rp) dqi.push_back(num[i]);
REP(i,Rp+1,n ) rgt.push_back(num[i]);
scanf("%d",&m);
}
void Reverse() {
dir = 1-dir;
}
void Insert() {
char op[20];
int x;
scanf("%s%d",op,&x);
if( (op[0]=='R' && dir) || (op[0] =='L' && !dir) ) {
dqi.push_back(x);
} else {
dqi.push_front(x);
}
}
void Delete() {
char op[20];
scanf("%s",op);
if( (op[0]=='R' && dir) || (op[0] =='L' && !dir) ) {
dqi.pop_back();
} else {
dqi.pop_front();
}
}
void MoveLeft() {
char op[20];
scanf("%s",op);
if( op[0]=='L' && dir ){
int x = lft.back();
lft.pop_back();
dqi.push_front(x);
}else if( op[0]=='R' && !dir ){
int x = dqi.front();
dqi.pop_front();
rgt.push_front(x);
}
else if(op[0] == 'R' && dir){
int x = dqi.back();
dqi.pop_back();
rgt.push_front(x);
} else {
int x = lft.back();
lft.pop_back();
dqi.push_back(x);
}
}
void MoveRight() {
char op[20];
scanf("%s",op);
if( (op[0]=='L' && dir)) {
int x = dqi.front();
dqi.pop_front();
lft.push_back(x);
}
else if(op[0]=='R' && dir) {
int x = rgt.front();
rgt.pop_front();
dqi.push_back(x);
}
else if(op[0]=='L' && !dir){
int x = dqi.back();
dqi.pop_back();
lft.push_back(x);
}else{
int x = rgt.front();
rgt.pop_front();
dqi.push_front(x);
}
}
void output() {
bool flag = false;
for(deque<int>::iterator it = lft.begin(); it != lft.end(); it++) {
if(!flag) {
printf("%d",*it);
flag = true;
} else {
printf(" %d",*it);
}
}
if(!dir) reverse(dqi.begin(),dqi.end());
for(deque<int>::iterator it = dqi.begin(); it != dqi.end(); it++) {
if(!flag) {
printf("%d",*it);
flag = true;
} else {
printf(" %d",*it);
}
}
for(deque<int>::iterator it = rgt.begin(); it != rgt.end(); it++) {
if(!flag) {
printf("%d",*it);
flag = true;
} else {
printf(" %d",*it);
}
}
puts("");
}
void solve() {
char cmd[20];
while(m--) {
scanf("%s",cmd);
if(cmd[0]=='R') Reverse();
else if(cmd[0]=='I') Insert();
else if(cmd[0] == 'D') Delete();
else if(cmd[0]=='M' && cmd[4]=='R') MoveRight();
else MoveLeft();
}
output();
}
int main(){
int ncase;
cin >> ncase;
while(ncase--) {
init();
input();
solve();
}
return 0;
}
方法2
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#define PI acos(-1.0)
#define M 1000005 //10^6
#define eps 1e-8
#define LL long long
#define moo 1000000007
#define INF -999999999
#define LL long long
using namespace std;
struct aaa
{
int val;
int le;
int ri;
}aa[M]; //双向链表
int stle[M];//左栈
int stri[M];//右栈
int b[M]; //初始读入数组
char a[10];
char d[10];
int judge()
{
if(strcmp(a,"MoveLeft")==0)
return 1;
if(strcmp(a,"MoveRight")==0)
return 2;
if(strcmp(a,"Insert")==0)
return 3;
if(strcmp(a,"Delete")==0)
return 4;
if(strcmp(a,"Reverse")==0)
return 5;
}
int main()
{
int m,n,T;
while(scanf("%d",&T)!=EOF)
{
while(T--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&b[i]);
}
int lee,rii;
scanf("%d%d",&lee,&rii);
int hele=0,heri=0;
for(int i=1;i<lee;i++)
{
hele++;
stle[hele]=b[i];
}
for(int i=n;i>rii;i--)
{
heri++;
stri[heri]=b[i];
}
int now=rii-lee;
for(int i=lee;i<=rii;i++)
{
aa[i-lee].le=i-lee-1;
aa[i-lee].ri=(i==rii?-1:i-lee+1);
aa[i-lee].val=b[i];
}
rii=rii-lee;
lee=0;
//以上是把三部分分别储存好的操作。
scanf("%d",&m);
int flag=0;
while(m--)
{
scanf("%s",a);
int tt=judge();
if(tt==5)
{
flag++;
continue;
}
if(tt==1)
{
scanf("%s",d);
if(strcmp(d,"L")==0)
{
aa[++now].val=stle[hele--];
if(flag%2==0)
{
aa[now].le=-1;
aa[lee].le=now;
aa[now].ri=lee;
lee=now;
}
else
{
aa[now].ri=-1;
aa[rii].ri=now;
aa[now].le=rii;
rii=now;
}
}
else
{
if(flag%2==0)
{
stri[++heri]=aa[rii].val;
aa[aa[rii].le].ri=-1;
rii=aa[rii].le;
}
else
{
stri[++heri]=aa[lee].val;
aa[aa[lee].ri].le=-1;
lee=aa[lee].ri;
}
}
continue;
}
if(tt==2)
{
scanf("%s",d);
if(strcmp(d,"R")==0)
{
aa[++now].val=stri[heri--];
if(flag%2==0)
{
aa[now].ri=-1;
aa[rii].ri=now;
aa[now].le=rii;
rii=now;
}
else
{
aa[now].le=-1;
aa[lee].le=now;
aa[now].ri=lee;
lee=now;
}
}
else
{
if(flag%2==0)
{
stle[++hele]=aa[lee].val;
aa[aa[lee].ri].le=-1;
lee=aa[lee].ri;
}
else
{
stle[++hele]=aa[rii].val;
aa[aa[rii].le].ri=-1;
rii=aa[rii].le;
}
}
continue;
}
if(tt==3)
{
scanf("%s",d);
int kkk;
scanf("%d",&kkk);
now++;
aa[now].val=kkk;
if(strcmp(d,"L")==0)
{
if(flag%2==0)
{
aa[now].ri=lee;
aa[now].le=-1;
aa[lee].le=now;
lee=now;
}
else
{
aa[now].le=rii;
aa[now].ri=-1;
aa[rii].ri=now;
rii=now;
}
}
else
{
if(flag%2==0)
{
aa[now].le=rii;
aa[now].ri=-1;
aa[rii].ri=now;
rii=now;
}
else
{
aa[now].ri=lee;
aa[now].le=-1;
aa[lee].le=now;
lee=now;
}
}
continue;
}
if(tt==4)
{
scanf("%s",d);
if(strcmp(d,"L")==0)
{
if(flag%2==0)
{
lee=aa[lee].ri;
aa[lee].le=-1;
}
else
{
rii=aa[rii].le;
aa[rii].ri=-1;
}
}
else
{
if(flag%2==0)
{
rii=aa[rii].le;
aa[rii].ri=-1;
}
else
{
lee=aa[lee].ri;
aa[lee].le=-1;
}
}
}
}
//以上是分别处理7种不同操作
//下面是输出部分，因为flag的原因，在输出链表部分的时候也得判断正反
int fuck=0;
for(int i=1;i<=hele;i++)
{
if(fuck==0)
{
cout<<stle[i];
fuck=1;
}
else
cout<<" "<<stle[i];
}
if(flag%2==0)
{
while(lee!=-1)
{
if(fuck==0)
{
cout<<aa[lee].val;
fuck=1;
}
else
cout<<" "<<aa[lee].val;
lee=aa[lee].ri;
}
}
else
{
while(rii!=-1)
{
if(fuck==0)
{
cout<<aa[rii].val;
fuck=1;
}
else
cout<<" "<<aa[rii].val;
rii=aa[rii].le;
}
}
for(int i=heri;i>0;i--)
{
if(fuck==0)
{
cout<<stri[i];
fuck=1;
}
else
cout<<" "<<stri[i];
}
cout<<endl;
}
}
}
1051.Wooden Sticks
#include<algorithm>
#include<stdio.h>
#include<string.h>
using namespace std;
struct sticks{
int l;
int w;
int flag;
}s[5002];/*运用结构体使条理更清晰*/
//返回正数就是说 cmp 传入参数第一个要放在第二个后面, 负数就是传入参数第一个要放第二个前面, 如果是 0, 那就无所谓谁前谁后..
int cmp(const void *a,const void *b){
/*struct的二级排序，先按l排若相等，则按w排，注意大小关系*/
struct sticks *c=(sticks*)a;
struct sticks *d=(sticks*)b;
if(c->l!=d->l)
return c->l - d->l;
else
return c->w - d->w;
}
int main(){
int T,n;
int i,j,count,cl,cw;/*count是计数的，ccurrent是当前的，所以cl为当前
的长，同理cw*/
scanf("%d",&T);
while(T--){
scanf("%d",&n);
memset(s,0,sizeof(s));/*清零防干扰*/
for(i=0;i<n;i++)
scanf("%d%d",&s[i].l,&s[i].w);
qsort(s,n,sizeof(s[0]),cmp);/*这里排序的数目是n，从小到大排*/
count=1;/*第一根木棍上来肯定要花1*/
s[0].flag=1;/*标记下是否被容纳过*/
cl=s[0].l;/*记录当前的l*/
cw=s[0].w;/*记录当前的w*/
/*用循环去跟当前比较，如果都小的话，就是能被容纳，不需要多加时间*/
for(i=1;i<n;i++){
for(j=i;j<n;j++){
if(s[j].flag!=1&&s[j].l>=cl&&s[j].w>=cw){
s[j].flag=1;
cl=s[j].l;
cw=s[j].w;
}
}
j=1;
while(s[j].flag==1)/*判断是否全被容纳过，若是，则完成*/
j++;
i=j;/*找到最前边的没有被标记过的一组数，即剩下中最小的一组数，再次找*/
if(i==n)
break;/*跳出*/
count++;/*每找过一次，就说明一次可能，所以count计数是所求结果*/
s[i].flag=1;
cl=s[i].l;/*重新刷新下当前数*/
cw=s[i].w;
}
printf("%d\n",count);
}
return 0;
}
1050.Moving Tables
方法1
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
int t,n,count[410],i,start,end,k;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
memset(count,0,sizeof(count));
while(n--){
scanf("%d%d",&start,&end);
//可能出发位置比目的地房间大。
if(start>end){
//无论大小，我们都可以看做从小的房间移动到大的房间
k=start;
start=end;
end=k;
}
if(start%2==0)//考虑实际情况，出发房间为偶数是减一，可参照题中给出的图一
start-=1;
if(end%2==1)//目的地房间为奇数时加一
end+=1;
for(i=start;i<=end;++i)
count[i]+=10;
}
printf("%d\n",*max_element(count,count+400));//STL中寻找数列最大值函数
}
return 0;
}
方法2
#include <iostream>
using namespace std;
struct table{
int front,back;
bool use;
}tb[10001];
int main()
{
int T,N,total; //T为有几组数据，N为每组数据有几行数据,total为总用时
int i,j,temp;
table tem;
cin>>T;
while(T--)
{
//************************** 输入
cin>>N;
for(i=0;i<N;++i)
{
cin>>tb[i].front>>tb[i].back;
if(tb[i].front%2==1)
tb[i].front+=1;
if(tb[i].back%2==1)
tb[i].back+=1;
if(tb[i].front>tb[i].back)
{
temp=tb[i].front;
tb[i].front=tb[i].back;
tb[i].back=temp;
}
tb[i].use=true;
}
//************************* 排序
for(i=0;i<N-1;++i)
for(j=i+1;j<N;++j)
if(tb[i].front>tb[j].front)
{
tem=tb[i];
tb[i]=tb[j];
tb[j]=tem;
}
//************************* 计算值
total=0;
for(i=0;i<N;++i)
{
if(tb[i].use==false)
continue;
temp=tb[i].back;
for(j=i+1;j<N;++j)
{
if(tb[j].use==false)
continue;
if(tb[j].front>temp)
{
tb[j].use=false;
temp=tb[j].back;
}
}
total+=10;
}
//************************* 输出
cout<<total<<endl;
}
return 0;
}
5127.Dogs’ Candies
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f3f3f3f3f
#define mod 10000007
#define debug() puts("what the fuck!!!")
#define N 500010
#define M 1000000
#define ll long long
using namespace std;
ll n;
struct node
{
ll p,q;
ll num;
} zz[N];
int main()
{
ll t,a,b;
while(scanf("%lld",&n)&&n)
{
ll len=0;
for(ll i=0; i<n; i++)
{
scanf("%lld%lld%lld",&t,&a,&b);
if(t==1)
{
zz[len].p=a;
zz[len].q=b;
zz[len++].num=1;
}
else if(t==0)
{
ll maxx=-inf;
for(ll j=0; j<len; j++)
{
if(zz[j].num)
{
maxx=max(maxx,a*zz[j].p+b*zz[j].q);
}
}
printf("%lld\n",maxx);
}
else if(t==-1)
{
for(ll j=0; j<len; j++)
if(zz[j].p==a&&zz[j].q==b&&zz[j].num)
{
zz[j].num=0;
break;
}
}
}
}
return 0;
}