Codeforces Round #344 (Div. 2) C 题题解 (贪心+单调栈)
2016-03-05 10:03
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C. Report
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced
by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers.
Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers
in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has
number i = m).
Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of
length n and the description of each manager, that is value ri and
his favourite order. You are asked to speed up the process and determine how the final report will look like.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) —
the number of commodities in the report and the number of managers, respectively.
The second line contains n integers ai (|ai| ≤ 109) —
the initial report before it gets to the first manager.
Then follow m lines with the descriptions of the operations managers are going to perform. The i-th
of these lines contains two integers tiand ri (
, 1 ≤ ri ≤ n),
meaning that the i-th manager sorts the first ri numbers
either in the non-descending (if ti = 1)
or non-ascending (if ti = 2)
order.
Output
Print n integers — the final report, which will be passed to Blake by manager number m.
Examples
input
output
input
output
Note
In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2
1 3. The report got to Blake in this form.
In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4
2 1 3. After the second manager the report changed to: 2 4 1 3. This report was
handed over to Blake.
题目大意:给出一个长度为n的数组a,和m个操作。
操作有两种: 1 将前 r 个数字进行非降序排列
2 将前 r 个数字进行非升序排列
要求输出最后的数组状态。
题解:首先,我们知道,当我们对前 3 个数字进行排序后,如果在该操作完成后,又出现了对前 3 + 1 个数字进行排序的操作的话,那么曾经做过的,对前3个数字进行排序的操作就将变得无效。因此,我们首先可以对操作进行压缩。
假如我们要对序列的前 1 2 3 4 2 3 2 个数字进行操作,那么,我们可以将操作化简成对前 4 3 2 的操作。(这是第一次优化)
这样可以减少大部分的操作,但是,如果测试数据里面有一个对前 n , n -1 , n - 2......... 1 这样的操作的数据的话,优化便没有用了。
现在我们可以肯定我们优化后得到的操作序列一定是如上面的 4 , 3 , 2一样的降序。以此,我们可以发现,4 是对前4个数字进行排序(不管是升序降序,无所谓),3是对前3个数进行排序。 那么我们可以发现【4,3) 这段整数区间是可以确定下来的。如果4操作是要进行非降序排序,那么我们只要将最大的数字(原序列中前4个位置中最大的数字)依次从第4个位置填起即可,如果4操作是要进行非升序排序,则我们只要将最小的数字依次从第4个位置开始填起即可。(第二次优化)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <set>
#include <vector>
using namespace std;
stack<int> p;
const int maxx = 200000 + 10;
struct node{
int t;
int r;
}op[maxx];
int n, m;
int num[maxx];
int ans[maxx];
int main( ){
cin>>n>>m;
for(int i = 1;i <= n; ++i){
cin>>num[i];
}
for(int i = 1;i <= m; ++i){
cin>>op[i].t>>op[i].r;
}
p.push(0);
for(int i = m; i >= 1; --i){
if(op[i].r > op[p.top()].r){
p.push(i);
}
}
int l = 1;
int r = op[p.top()].r;
for(int i = r+1;i <= n; ++i){
ans[i] = num[i];
}
int nowr = r;
int nowt = op[p.top()].t;
sort(num+1,num+r+1);
p.pop();
while(!p.empty()) {
int nextr = op[p.top()].r;
int nextt = op[p.top()].t;
for(int i = nowr; i > nextr; --i) {
if(nowt == 1)ans[i] = num[r--];
else ans[i] = num[l++];
}
nowr = nextr;
nowt = nextt;
p.pop();
}
for(int i = 1; i <= n; ++i) {
printf("%d ",ans[i]);
}
puts("");
return 0;
}
如有BUG,欢迎指出~
联系方式:hh_0828@outlook.com
不胜感激~
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Each month Blake gets the report containing main economic indicators of the company "Blake Technologies". There are n commodities produced
by the company. For each of them there is exactly one integer in the final report, that denotes corresponding revenue. Before the report gets to Blake, it passes through the hands of m managers.
Each of them may reorder the elements in some order. Namely, the i-th manager either sorts first ri numbers
in non-descending or non-ascending order and then passes the report to the manager i + 1, or directly to Blake (if this manager has
number i = m).
Employees of the "Blake Technologies" are preparing the report right now. You know the initial sequence ai of
length n and the description of each manager, that is value ri and
his favourite order. You are asked to speed up the process and determine how the final report will look like.
Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 200 000) —
the number of commodities in the report and the number of managers, respectively.
The second line contains n integers ai (|ai| ≤ 109) —
the initial report before it gets to the first manager.
Then follow m lines with the descriptions of the operations managers are going to perform. The i-th
of these lines contains two integers tiand ri (
, 1 ≤ ri ≤ n),
meaning that the i-th manager sorts the first ri numbers
either in the non-descending (if ti = 1)
or non-ascending (if ti = 2)
order.
Output
Print n integers — the final report, which will be passed to Blake by manager number m.
Examples
input
3 1 1 2 3 2 2
output
2 1 3
input
4 2 1 2 4 3 2 3 1 2
output
2 4 1 3
Note
In the first sample, the initial report looked like: 1 2 3. After the first manager the first two numbers were transposed: 2
1 3. The report got to Blake in this form.
In the second sample the original report was like this: 1 2 4 3. After the first manager the report changed to: 4
2 1 3. After the second manager the report changed to: 2 4 1 3. This report was
handed over to Blake.
题目大意:给出一个长度为n的数组a,和m个操作。
操作有两种: 1 将前 r 个数字进行非降序排列
2 将前 r 个数字进行非升序排列
要求输出最后的数组状态。
题解:首先,我们知道,当我们对前 3 个数字进行排序后,如果在该操作完成后,又出现了对前 3 + 1 个数字进行排序的操作的话,那么曾经做过的,对前3个数字进行排序的操作就将变得无效。因此,我们首先可以对操作进行压缩。
假如我们要对序列的前 1 2 3 4 2 3 2 个数字进行操作,那么,我们可以将操作化简成对前 4 3 2 的操作。(这是第一次优化)
这样可以减少大部分的操作,但是,如果测试数据里面有一个对前 n , n -1 , n - 2......... 1 这样的操作的数据的话,优化便没有用了。
现在我们可以肯定我们优化后得到的操作序列一定是如上面的 4 , 3 , 2一样的降序。以此,我们可以发现,4 是对前4个数字进行排序(不管是升序降序,无所谓),3是对前3个数进行排序。 那么我们可以发现【4,3) 这段整数区间是可以确定下来的。如果4操作是要进行非降序排序,那么我们只要将最大的数字(原序列中前4个位置中最大的数字)依次从第4个位置填起即可,如果4操作是要进行非升序排序,则我们只要将最小的数字依次从第4个位置开始填起即可。(第二次优化)
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <set>
#include <vector>
using namespace std;
stack<int> p;
const int maxx = 200000 + 10;
struct node{
int t;
int r;
}op[maxx];
int n, m;
int num[maxx];
int ans[maxx];
int main( ){
cin>>n>>m;
for(int i = 1;i <= n; ++i){
cin>>num[i];
}
for(int i = 1;i <= m; ++i){
cin>>op[i].t>>op[i].r;
}
p.push(0);
for(int i = m; i >= 1; --i){
if(op[i].r > op[p.top()].r){
p.push(i);
}
}
int l = 1;
int r = op[p.top()].r;
for(int i = r+1;i <= n; ++i){
ans[i] = num[i];
}
int nowr = r;
int nowt = op[p.top()].t;
sort(num+1,num+r+1);
p.pop();
while(!p.empty()) {
int nextr = op[p.top()].r;
int nextt = op[p.top()].t;
for(int i = nowr; i > nextr; --i) {
if(nowt == 1)ans[i] = num[r--];
else ans[i] = num[l++];
}
nowr = nextr;
nowt = nextt;
p.pop();
}
for(int i = 1; i <= n; ++i) {
printf("%d ",ans[i]);
}
puts("");
return 0;
}
如有BUG,欢迎指出~
联系方式:hh_0828@outlook.com
不胜感激~
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