二分贪心专题E
2017-01-30 00:49
399 查看
FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year" competition. In this contest every farmer arranges his cows in a line and herds them past the judges.
The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase
ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows' names.
FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.
FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he's
finished, FJ takes his cows for registration in this new order.
Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single initial ('A'..'Z') of the cow in theith position in the original line
Output
The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows ('A'..'Z') in the new line.
Sample Input
Sample Output
每次从两头选字母 不一样时选小的 一样时往里找第一次出现不一样选小的 注意一下特殊情况如AABB这类的
源代码:
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int n,tot=0;
char a[2222];
cin>>n;
for (int i=1;i<=n;i++) cin>>a[i];
int head=1,tail=n;char s[2222];
while (head<=tail)
{
if (a[head]<a[tail]) s[++tot]=a[head++];
if (a[head]>a[tail]) s[++tot]=a[tail--];
if (a[head]==a[tail])
{
int l=head,r=tail;
while (a[l]==a[r]&&l!=r)
{
l++;r--;
}
if (a[l]<a[r]) s[++tot]=a[head++];
if (a[r]<a[l]) s[++tot]=a[tail--];
if (r==l) s[++tot]=a[tail--];
if (l>r) s[++tot]=a[tail--];
}
}
if (tot!=n)
{
s[++tot]=a[head];
}
for (int i=1;i<=tot;i++)
{
cout<<s[i];if (i%80==0) cout<<endl;
}
return 0;
}
The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase
ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows' names.
FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.
FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he's
finished, FJ takes his cows for registration in this new order.
Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.
Input
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single initial ('A'..'Z') of the cow in theith position in the original line
Output
The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows ('A'..'Z') in the new line.
Sample Input
6 A C D B C B
Sample Output
ABCBCD
每次从两头选字母 不一样时选小的 一样时往里找第一次出现不一样选小的 注意一下特殊情况如AABB这类的
源代码:
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
int n,tot=0;
char a[2222];
cin>>n;
for (int i=1;i<=n;i++) cin>>a[i];
int head=1,tail=n;char s[2222];
while (head<=tail)
{
if (a[head]<a[tail]) s[++tot]=a[head++];
if (a[head]>a[tail]) s[++tot]=a[tail--];
if (a[head]==a[tail])
{
int l=head,r=tail;
while (a[l]==a[r]&&l!=r)
{
l++;r--;
}
if (a[l]<a[r]) s[++tot]=a[head++];
if (a[r]<a[l]) s[++tot]=a[tail--];
if (r==l) s[++tot]=a[tail--];
if (l>r) s[++tot]=a[tail--];
}
}
if (tot!=n)
{
s[++tot]=a[head];
}
for (int i=1;i<=tot;i++)
{
cout<<s[i];if (i%80==0) cout<<endl;
}
return 0;
}
相关文章推荐
- 【贪心专题】POJ 2456 Aggressive cows && NYOJ 586 疯牛(最大化最小值 贪心+二分搜索)
- 【贪心专题】POJ 3258 River Hopscotch (最大化最小值 贪心+二分搜索)
- 二分贪心专题D
- 二分贪心专题F
- 二分贪心练习题专题总结
- 二分贪心专题A
- 二分贪心专题B
- 二分贪心专题总结
- 二分贪心专题C
- UVa 714 Copying Books 二分 + 贪心 (最大值最小化问题)
- CodeForces - 732D Exams(二分+贪心)
- 贪心专题 HDU 2111
- POJ3497 Assemble 二分+贪心
- SDAU 贪心专题 16 中位数
- 【BZOJ3048】Cow lineup,贪心+队列维护(或二分答案)
- 贪心,二分,半平面交(丛林警戒队,LA 4992)
- POJ 2456 Aggressive cows(二分+贪心 计算个数)
- 【二分答案+贪心】UVa 1335 - Beijing Guards
- River Hopscotch poj3258 (二分+贪心思想+最小值最大化)
- [NWPU][2014][TRN][5]二分和贪心 HDU 4296