Codeforces 950C Zebras (greedy && emulation)

Question meaning: Given a 01 string, you are required to separate it into several subsequences (each number can only belong to a certain subsequence). If this is not possible, output -1.

 

analyze: 

It's easy to think of the need to greedily pair 0 and 1

That is, if you can construct alternately 0 and 1, try to choose this scheme, so that you can use fewer 0s to pair as many 1s as possible.

Use the two-dimensional array vector<int> idx[] to hold the information of each subsequence

Use two sets to hold the subsequences of the now 0-terminated and 1-terminated subsequences

Then O(n) sweep from left to right, for 0 elements to check if there is currently a subsequence that ends with 1

If there are any, fill in it, otherwise create a new subsequence, starting with this 0. if there is not,

Indicates that it cannot be done, output -1

 

#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 2e5 + 10;
int num0, num1, len, ans, arr[maxn];
vector<int> idx[maxn];
set<int> zero, one;

int main(void)
{
    while(true){
        char ch;
        scanf("%c", &ch);
        if(ch == '\n') break;
        if(ch == '0') num0++, arr[++len] = 0;
        else num1++, arr[++len] = 1;
    }

    if(num0 <= num1){
        puts("-1");
        return 0;
    }

    for(int i=1; i<=len; i++){
        if(arr[i] == 0){
            num0--;
            if(one.size() == 0){
                idx[ans].push_back(i);
                zero.insert(ans++);
            }else{
                int ii = *one.begin();
                one.erase(one.begin());
                idx[ii].push_back(i);
                zero.insert(ii);
            }
        }else{
            num1--;
            if(num0 <= 0){
                puts("-1");
                return 0;
            }
            if(zero.size() == 0){
                puts("-1");
                return 0;
            }else{
                int ii = *zero.begin();
                zero.erase(zero.begin());
                idx[ii].push_back(i);
                one.insert(ii);
            }
        }
    }

    if(one.size() > 0){
        puts("-1");
        return 0;
    }

    printf("%d\n", ans);
    for(int i=0; i<ans; i++){
        printf("%d ", idx[i].size());
        for(int j=0; j<idx[i].size(); j++){
            printf("%d ", idx[i][j]);
        }puts("");
    }

    return 0;
}

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