欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

codeforces 1399D

程序员文章站 2022-03-27 16:03:09
题意描述You are given a binary string s consisting of n zeros and ones.Your task is to divide the given string into the minimum number of subsequences in such a way that each character of the string belongs to exactly one subsequence and each subsequence loo...

题意描述

You are given a binary string s consisting of n zeros and ones.

Your task is to divide the given string into the minimum number of subsequences in such a way that each character of the string belongs to exactly one subsequence and each subsequence looks like “010101 …” or “101010 …” (i.e. the subsequence should not contain two adjacent zeros or ones).

Recall that a subsequence is a sequence that can be derived from the given sequence by deleting zero or more elements without changing the order of the remaining elements. For example, subsequences of “1011101” are “0”, “1”, “11111”, “0111”, “101”, “1001”, but not “000”, “101010” and “11100”.

You have to answer t independent test cases.

要求以010101或者101010的形式分组,求每个0或1属于哪个分组

思路

我们使用两个队列来分别表示0和1的空间,如果遇到0时,判断1的空间内是否存在元素,如果存在的话,则说明0和该元素一组,否则需要重新开辟一个空间;遇到1的话同理。

AC代码

#include<bits/stdc++.h>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=2*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
int ans[N];
void solve(){
	mst(ans,0);
	int n;cin>>n;
	string s;cin>>s;
	queue<int> q0,q1;
	int cnt=1;
	rep(i,0,n){
		if(s[i]=='0'){
			if(q1.empty()) ans[i]=cnt++;
			else{
				ans[i]=ans[q1.front()];q1.pop();
			}
			q0.push(i);
		}else{
			if(q0.empty()) ans[i]=cnt++;
			else{
				ans[i]=ans[q0.front()];q0.pop();
			}
			q1.push(i);
		}
	}
	cout<<cnt-1<<endl;
	rep(i,0,n) cout<<ans[i]<<' ';
	cout<<endl;
}
int main(){
    IOS;
    int t;cin>>t;
    while(t--){
    	solve();
    }
    return 0;
}

本文地址:https://blog.csdn.net/weixin_45729946/article/details/107854181