Codeforces-1256-D. Binary String Minimizing（贪心）

Input

3
8 5
11011010
7 9
1111100
7 11
1111100

Output

01011110
0101111
0011111

题意：t组样例，给一个长度为n的字符串，要求在k步以内进行字符交换（1步只能交换i和i+1）使得字符串字典序最小。
思路：贪心，存下0的位置，让前面的0尽可能往前移，然后就是模拟计算操作。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#pragma GCC optimize(2)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
#include<cmath>
#include<string>
#include<map>
#include<vector>
#include<ctime>
using namespace std;
#define mm(a,b) memset(a,b,sizeof(a))
typedef long long ll;
const int maxn = 1e6+10;;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;
int gcd(int a,int b)
{if(b==0) return a;return gcd(b,a%b);}
char s[maxn];
int main(){
    int q;
    scanf("%d",&q);
    while(q--)
    {
        vector<int>v;
        int n;
ll k;
        scanf("%d %lld",&n,&k);
        scanf("%s",s);
        for(int i=0;i<n;i++)
        {
            if(s[i]=='0')
                v.push_back(i);
        }
        int len=v.size();
        if(v.size()==n||v.size()==0)
        {
            printf("%s\n",s);
            continue;
        }
        int pos1=0,pos2=0;
        while(k>0&&pos2<len&&pos1<=n-1)
        {
            if(s[pos1]=='0')
            {
                pos1++;
            }
            else if(v[pos2]<=pos1)
            {
                pos2++;
            }
            else
            {
                int cnt=v[pos2]-pos1;
                if(k<cnt)
                {
                    pos1=v[pos2]-k;
                    k=0;
                }
                else k=k-(cnt);
                s[pos1]='0';
                s[v[pos2]]='1';
                pos1++;
                pos2++;
                
            }
        }
        printf("%s\n",s);
    }
    return 0;
}
/*
 9
 8 1
 11011010
 8 2
 11011010
 8 3
 11011010
 8 4
 11011010
 8 5
 11011010
 8 6
 11011010
 8 7
 11011010
 8 10
 11011010
 8 50
 11011010
 */