B. Minimum Ternary String——Educational Codeforces Round 47 (Rated for Div. 2)

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本文介绍了一种通过特定字符交换操作来最小化三元字符串的方法。对于由字符'0'、'1'和'2'组成的字符串，允许进行相邻字符'0'与'1'或者'1'与'2'的互换，目标是通过这些操作获得字典序最小的字符串。

You are given a ternary string (it is a string which consists only of characters '0', '1' and '2').

You can swap any two adjacent (consecutive) characters '0' and '1' (i.e. replace "01" with "10" or vice versa) or any two adjacent (consecutive) characters '1' and '2' (i.e. replace "12" with "21" or vice versa).

For example, for string "010210" we can perform the following moves:

"010210" $\to$ "100210";
"010210" $\to$ "001210";
"010210" $\to$ "010120";
"010210" $\to$ "010201".

Note than you cannot swap "02" $\to$ "20" and vice versa. You cannot perform any other operations with the given string excluding described above.

You task is to obtain the minimum possible (lexicographically) string by using these swaps arbitrary number of times (possibly, zero).

String $a$ is lexicographically less than string $b$ (if strings $a$ and $b$ have the same length) if there exists some position $i$ ( $1 \leq i \leq | a |$ , where $| s |$ is the length of the string $s$ ) such that for every $j < i$ holds $a_{j} = b_{j}$ , and $a_{i} < b_{i}$ .

Input

The first line of the input contains the string $s$ consisting only of characters '0', '1' and '2', its length is between $1$ and $10^{5}$ (inclusive).

Output

Print a single string — the minimum possible (lexicographically) string you can obtain by using the swaps described above arbitrary number of times (possibly, zero).

Examples
inputCopy
100210
outputCopy
001120
inputCopy
11222121
outputCopy
11112222
inputCopy
20
outputCopy
20

1可以和0和2交换位置，所以1的位置可以任意放置，0和2不能交换所以，0和2的相对位置不变。

每两个2之间0的个数不变，1全部在第一个2之前。

贪心思路：第一个2之前把所有1输出（1在0后），记录0的个数，一碰到2就输出全部0。

#include<iostream>
#include<set>
#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
char s[100010];
int main()
{
    while(scanf("%s",s)!=EOF)
    {
        int cnt1=0;
        for(int i=0;s[i]!='\0';++i)
            cnt1+=s[i]=='1';
        int cnt0=0;
        for(int i=0;s[i]!='\0';++i)
        {
            if(s[i]=='1') continue;
            if(s[i]=='0') cnt0++;
            else if(s[i]=='2')
            {
                for(int i=0;i<cnt0;++i)
                    printf("0");
                for(int i=0;i<cnt1;i++)
                    printf("1");
                printf("2");
                cnt1=cnt0=0;
            }
        }
        for(int i=0;i<cnt0;++i) printf("0");

        for(int i=0;i<cnt1;i++) printf("1");
        cout<<endl;
    }
    return 0;
}