codeforces 358C
2016-10-01 20:10
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题意:有三种数据结构,栈 队列 双端队列, 每次遇到一个非0的数就可以把这个数放到 栈顶or队列尾or 双端队列的前面或者尾部, 当遇到0的时候最多可以提取三种数据结构中的一个数字, 栈顶的数 + 队列的头部 + 双端的头或尾, 使得每次提取的结果最大,且提取完后三种数据结构全部清空。请模拟整个操作过程。
思路:因为每次遇到0之后数据会全部清空,那也就是说我们只要找到0之前的最大的三个数,分别放进栈,队列,双端的头, 其余的数全部放进双端的尾部即可、
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
const int qq = 1e5+10;
int command[qq];
string pop[4];
string push[4];
int main(){
pop[0]="popStack",pop[1]="popQueue",pop[2]="popFront",pop[3]="popBack";
push[0]="pushStack",push[1]="pushQueue",push[2]="pushFront",push[3]="pushBack";
int n;scanf("%d",&n);
for(int i=0; i<n; ++i) scanf("%d",command+i);
int a, b, c;
for(int i=0; i<n; ++i){
a = b = c = 0;
int j = i;
while(command[j]!=0&&j<n){
if(command[j]>a){
c = b;
b = a;
a = command[j];
}else if(command[j]>b){
c = b;
b = command[j];
}else if(command[j]>c)
c = command[j];
j++;
}
if(a==0&&b==0&&c==0) cout << 0 << endl;
else{
int count=3;
if(b==0) count--;
if(c==0) count--;
for(int k=i; k<j; ++k)
if(command[k]==a){
cout << push[0] << endl;
a = 0;
}else if(command[k]==b){
cout << push[1] << endl;
b = 0;
}else if(command[k]==c){
cout << push[2] << endl;
c = 0;
}else cout << push[3] << endl;
if(j>=n) break;
cout << count;
for(int k=0; k<count; ++k)
cout << " " << pop[k];
cout << endl;
i = j;
}
}
return 0;
}
思路:因为每次遇到0之后数据会全部清空,那也就是说我们只要找到0之前的最大的三个数,分别放进栈,队列,双端的头, 其余的数全部放进双端的尾部即可、
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<iostream>
using namespace std;
const int qq = 1e5+10;
int command[qq];
string pop[4];
string push[4];
int main(){
pop[0]="popStack",pop[1]="popQueue",pop[2]="popFront",pop[3]="popBack";
push[0]="pushStack",push[1]="pushQueue",push[2]="pushFront",push[3]="pushBack";
int n;scanf("%d",&n);
for(int i=0; i<n; ++i) scanf("%d",command+i);
int a, b, c;
for(int i=0; i<n; ++i){
a = b = c = 0;
int j = i;
while(command[j]!=0&&j<n){
if(command[j]>a){
c = b;
b = a;
a = command[j];
}else if(command[j]>b){
c = b;
b = command[j];
}else if(command[j]>c)
c = command[j];
j++;
}
if(a==0&&b==0&&c==0) cout << 0 << endl;
else{
int count=3;
if(b==0) count--;
if(c==0) count--;
for(int k=i; k<j; ++k)
if(command[k]==a){
cout << push[0] << endl;
a = 0;
}else if(command[k]==b){
cout << push[1] << endl;
b = 0;
}else if(command[k]==c){
cout << push[2] << endl;
c = 0;
}else cout << push[3] << endl;
if(j>=n) break;
cout << count;
for(int k=0; k<count; ++k)
cout << " " << pop[k];
cout << endl;
i = j;
}
}
return 0;
}
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